Blood, ph and Logarithms Unit

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Blood, pH and Logarithms

Unit

(Level IV Graduate Math)

Draft

(NSSAL)

C. David Pilmer

(Last Updated: June 2012)

Important Note:

This unit is only available to learners who are taking the ALP Graduate Level IV Math and ACC Health Math Dual Accreditation program.

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This resource is the intellectual property of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education.

The following are permitted to use and reproduce this resource for classroom purposes.  Nova Scotia instructors delivering the Nova Scotia Adult Learning Program  Canadian public school teachers delivering public school curriculum

 Canadian nonprofit tuition-free adult basic education programs

The following are not permitted to use or reproduce this resource without the written

authorization of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education.

 Upgrading programs at post-secondary institutions  Core programs at post-secondary institutions  Public or private schools outside of Canada  Basic adult education programs outside of Canada

Individuals, not including teachers or instructors, are permitted to use this resource for their own learning. They are not permitted to make multiple copies of the resource for distribution. Nor are they permitted to use this resource under the direction of a teacher or instructor at a learning institution.

Acknowledgments

The Adult Education Division would also like to thank the following NSCC instructors for piloting this resource and offering suggestions during its development.

Elliott Churchill (Waterfront Campus) Donald Shay (Akerley Campus) Eric Tetford (Burridge Campus) Jeff Vroom (Truro Campus)

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Introduction

Normally in Level IV Graduate Math, learners must complete five required units and choose two of four optional units (Linear Functions and Systems of Equations Unit, Trigonometry Unit, Statistics Unit). There are, however, learners who wish to pursue careers in the health sciences and will also need to take the NSCC's ACC Health Math courses. There is significant overlap between ALP Graduate Math and ACC Health Math courses, so the Department of Labour and Advanced Education has worked collaboratively with the college to create a pathway between the two courses so learners can move through the two programs as quickly as possible and as seamlessly as possible. For this reason, we have created this particular unit. Learners interested in the health sciences will now take the original five required units for Graduate Math, the ALP Statistics Unit, the ALP Blood, pH, and Logarithm Unit, and the ACC Dosages, Solutions, and IV Therapy Unit. By doing so, the learner will receive the ALP Graduate Math credit which can be applied to their diploma, and the ACC Health Math credits that will officially be recognized following the learner's graduation from ALP. We stress that the ACC credits will not be applied to the learner's high school graduation diploma.

Negotiated Completion Date

After working for a few days on this unit, sit down with your instructor and negotiate a completion date for this unit.

Start Date: _________________ Completion Date: _________________

Instructor Signature: __________________________ Student Signature: __________________________

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The Big Picture

This flow chart only applies to learners who are enrolled in the ALP Graduate Level IV Math and ACC Health Math dual credit program.

Math in the Real World Unit

Solving Equations Unit

Consumer Finance Unit

Graphs and Functions Unit

Measurement Unit

Statistics Unit

Blood, pH and Logarithms Unit

Completion of ALP Graduate Math

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How are these Terms Related?

You are probably looking at the three terms in the unit title, The Blood, pH and Logarithms Unit, and asking, "How are these terms related?"

How are Blood and pH related?

Health care workers monitor a host of patient vitals which include respiration, heart rate, and blood pressure. Many do not know they also monitor patients' blood pH levels.

The kidneys and lungs maintain the proper balance of chemicals, called acids and bases, in the body. If the body correctly maintains this balance then the blood pH range for an adult should be between 7.35 and 7.45. For infants and children (16 yrs.), the blood pH levels should be between 7.35 and 7.42.

What is pH?

When some substances are dissolved in water, they alter the hydrogen ion (H+) concentration.

Acidic solutions (e.g. lemon juice) increase the hydrogen ion concentration. The pH is, however,

low for these solutions. Alkaline or basic solutions (e.g. Soapy Water) decrease the hydrogen ion concentration. The pH is, however, high for these alkaline or basic solutions.

Acidic Solutions Increase the H+ Concentration Low pH Alkaline Solutions Decrease the H+ Concentration High pH

In conclusion, pH tells you whether are dealing with an acidic solutions, basic solutions, or something in between (i.e. a neutral solution).

How do our bodies regulate pH?

There are five mechanisms that the body uses to regulate pH levels. 1. The lungs vent carbon dioxide (CO2) from our systems. This carbon

dioxide is a waste product of the metabolism of oxygen. This waste product gets excreted into our blood, then moved to our lungs, and finally exhaled. When still in the blood, this carbon dioxide can react with the water in our blood to create carbonic acid. It is important that the lungs function properly to ensure that CO2 can be removed in a timely fashion so

that carbonic acid does not rise to dangerous levels.

2. The kidneys filter blood and remove excess acids and bases. They do have the ability to alter the amount of acid or base that is excreted, but these adjustments occur slowly generally taking several days.

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4. The skin excretes a mild acid through perspiration. This acid also maintains the strength and integrity of the skin and prevents infection by inhibiting the growth of bacteria

5. A buffer is a solution that can maintain a nearly constant pH if it is diluted, or if strong acids or bases are added. The body uses chemical buffers in fluids and bones to balance pH levels. The buffers are released to neutralize excess acid or base.

When blood pH levels are off

If the blood pH level for an infant, child, or adult falls below 7.35 then the patient is said to have acid-base disorder acidosis. The condition is considered extreme if the pH drops to 7.2 or lower. When blood pH levels drop this low, coma and even death can occur.

If the blood pH level for an adult rises above 7.45 or the blood pH level for an enfant or child rises above 7.42, then the patient is said to have the acid-base disorder alkalosis. This condition is considered extreme if it rises to 7.5 or higher in adults and 7.6 or higher in infants or children. When these elevated pH levels occur, the patients will likely convulse and these convulsions may ultimately lead to death if the condition is not addressed.

Extreme Acidosis Normal Range Extreme Alkalosis

Adult  7.2 7.35 to 7.45  7.5

Infant or Child  7.2 7.35 to 7.42  7.6

It is important to note that these two acid-base disorders (acidosis and alkalosis) are not diseases themselves; rather they are signs of an underlying disease or condition.

Both acidosis and alkalosis are found in respiratory and metabolic forms. Respiratory acidosis or alkalosis is caused by various malfunctions of the lungs. Metabolic acidosis or alkalosis is caused by various metabolic disorders which result in an excessive build up or loss of acids or bases. Metabolic disorders are problems associated with the various processes within the body that convert food and other substances into energy and other metabolic byproducts used by the body.

Respiratory Acidosis

Respiratory acidosis results from a build-up of carbon dioxide (CO2) due to hypoventilation

(i.e. slow shallow breathing). This depressed breathing can be caused by drugs, central nervous system trauma, or lung diseases (e.g. pneumonia). Carbon dioxide reacts with the water in our blood and forms carbonic acid (H2CO3). Excess levels of CO2, result in higher

levels of the acid H2CO3, and correspondingly lower blood pH readings.

Metabolic Acidosis

Metabolic acidosis results from an increased production of acid or from the decreased

availability of bicarbonate (HCO3, a base) due to diabetes, kidney disease, poisoning (aspirin,

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Respiratory Alkalosis

Respiratory alkalosis results from decreased CO2 levels due to hyperventilation (i.e. rapid

deep breathing). This over breathing can be caused by anxiety, pain, or fever. Metabolic Alkalosis

Metabolic alkalosis results when the body loses too much acid or from the increased availability of bicarbonate (HCO3, a base). Prolonged vomiting is one way a body can lose

excessive amounts of acid. In some cases (e.g. poisoning), hospital staff have to suction out stomach contents, reducing acid levels which can lead to alkalosis. In addition, metabolic alkalosis can occur if sodium or potassium levels fall too low. When this happens, the kidneys are unable to maintain the proper acid base balance.

How are logarithms and pH related?

We have learned that pH is a way of distinguishing acidic solutions from basic solutions. The next obvious question is why does the pH rules seem to be backwards. For example, if a solution has a high concentration of H+, it has a low pH. Similarly, if the solution has a low concentration of H+, the pH is high. The answer to this apparent backwardness is how pH is described using logarithms.

You may have also noticed that the range of blood pH levels from dangerously low to normal to dangerously high seems quite small. For example the difference between an extreme low (7.2) to the low normal (7.35) is only 0.15. Similarly the difference between a high normal (7.45) in adults and an extreme high (7.5) in adults is only 0.05. These differences seem very small but the implications for the patients are huge. Why are these differences so small? Again, the answer is logarithms.

Now we cannot just dive into the topic of logarithms because there is some fundamental work with exponents and scientific notation that must first be completed. In the next few sections we will work on these fundamental concepts with the understanding that they will ultimately lead to logarithms and further discussions of blood pH.

Important Note:

You will have to be familiar with these four terms and the blood pH ranges in later sections of this unit.

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Introduction to Exponents

Before you start this unit, it is expected that you are familiar with powers of 2, 3, 4, and 5. You should know that:

4 22  and conversely 42 8 23  and conversely 3 8 2 16 24  and conversely 416 2 32 25  and conversely 5 322 9 32  and conversely 93 27 33  and conversely 3 27 3 81 34  and conversely 4 813 16 42  and conversely 16 4 64 43  and conversely 3 64 4 25 52  and conversely 25 5 125 53  and conversely 3 1255

Make sure you are familiar with these before proceeding.

The Laws of Exponents

When given the expression an, we say that an is the power, n is the exponent, and a is the base.

Therefore when given the power 2 , we state that the base is 2 and the exponent is 3. This 3 power can be written as 23 222.

The Law of Products of Powers

To multiply two powers that have the same base, keep the base and add the exponents.

n m n m a a a    Examples: 7 5 2 x x x   r3r8r r12 2423 27

The Law of Quotients of Powers

To divide two powers that have the same base, keep the base and subtract the exponents.

n m n m a a a    Examples: 3 2 5 x x x   r8 rr7 25 23 22

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Examples: (a) 3x27x3 21x5 (b) 2a2b9a4b3 18a6b4 (c) 24k6 8k2 3k4 (d) 45c8d9 5cd7 9c7d2 (e)

 

7 2 2 14 14 25 5 5x  x  x (f)



3 2



4 4 12 8 12 8 16 2 2m n  m n  m n (g) ₈ 4 8 4 2 2 4 2 9 3 3 v u v u v u _ _       (h) ₁₅ 3 9 15 3 3 9 3 3 5 3 27 8 3 2 3 2 z y x z y x z y x _ _       (i)





19 10 2 20 12 2 4 5 3 2 8 16 8 2 y x y x y x y x y x   (j) 9 9 9 10 9 4 8 6 9 2 18 2 3 6 a ab b a ab b a b a   

The Law of Powers of Powers

2r4 3 23r12 8r12

The Law of Power of Quotients

If a quotient in parentheses is raised to an exponent, the parentheses indicate that both the numerator and denominator must be raised to that power.

n n n b a b a _       where b0 Examples: 3 3 3 y x y x _       12 8 4 3 2 b a b a _       10 10 2 2 5 9 3 3 y y y _      

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Questions 1. Evaluate. (a) 72  (b) 23  (c) 34  (d) 43  (e) 25  (f) 36  (g) 100  (h) 3  27 (i) 3  125 (j) 4  16

2. Simplify each of the following expressions.

₂ 2  6 7 5 3 3 4 9 ab b a b a (z)





  x y y x y x 3 3 2 2 6 5 4 6

3. Determine x. No work needs to be shown.

(a) x2 36 x = _____ (b) 3x 27 x = _____ (c) x24 x = _____ (d) 64x3 x = _____ (e) 82x x = _____ (f) 5x 25 x = _____ (g) 53 x x = _____ (h) x4 81 x = _____

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More Exponents

Zero Exponents Consider _n n x x .

 We know that when we take one quantity and divide it by the same quantity, the quotient is always 1. For example: 1

3 3_ , 1 47 47 _ , 1 8 8_   , and 1 2 2 3 3  . Therefore we can conclude that _n 1 n x x

. The only exception to this rule is if the two quantities are equal to

zero       1 0 0 .

 Based on the Law of Quotients of Powers, we know that to divide two powers that have the same base, we keep the base and subtract the exponents. Therefore our expression

n n

x x

can now be expressed in the following manner.

0 x x x x n n n n   

 In the first bullet we stated that _n 1

n

x x

. In the second bullet we stated that x0 x x

n n

 . Therefore we can conclude x0 1 (as long as x does not equal zero).

Examples: 1 70 

 

40 1 1 5 3 0        1 7 9 0        1 40 

  (Note: In this question only 4, not -4, is raised to the exponent 0.)

1 8 . 0 0 

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Negative Exponents Consider ₅ 2 x x .  The expression ₅ 2 x x

can be simplified by dividing through (i.e. canceling out) like quantities. 3 5 2 1 x x x x x x x x x x x x x x x x x                 

 Using the Law of Quotients of Powers, our expression ₅

2 x x

can now be expressed in the following manner. 3 5 2 5 2   _ x x x x

 In the first bullet we stated that ₅ ₃

2

1

x x x

 . In the second bullet we stated that ₅ 3

2  x x x .

Therefore we can conclude

3 3 3 1 or 1         x x

x . If we research this further we could

generalize that n n x x         1

where x is not equal to zero. Therefore the negative exponent just means that we deal with the reciprocal of the base.

Examples: 25 1 5 1 5 2 2          8 1 2 1 2 3 3         

 

81 1 3 1 3 4 4 _          81 1 3 1 3 4 4            49 25 7 5 5 7 2 2 _               8 27 2 3 3 2 3 3 _               16 2 2 1 4 4          2 2 5 6 36 6 5 25  _  _{ }  _         2 2 5 6 36 6 5 25      _{ }  _{ }      

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Fractional Exponents

Part 1: The Fractional Exponent 2 1

 Suppose we were asked to find two identical numbers that multiplied to give us the product 9. Most people would quickly respond 3 because 339. This mathematical statement could also be written as the following.

9 9 9 

Now let’s work with the variable x. If we needed to find two identical terms that multiply to give us the product x, we would end up with the following statement.

x x

x 

 Again we are going to find two identical terms that multiply to give us the product x but we are not going to rely on radicals (i.e. square roots, cube roots, fourth roots,…). Instead we are going to rely on our understanding of the Law of Products of Powers. We need to find two powers with the base of x and the same exponent whose product is x.

x x

x? ? 

The only exponent that makes sense is one half. x x x x 2 1or 1 2 1  

 In the first bullet we stated that x x x. In the second bullet we stated thatx x2 x

1 2 1

. Therefore we can conclude that x2  x

1

.

Part 2: The Fractional Exponent 3 1

 Suppose we were asked to find three identical numbers that multiplied to give us the product 8. Most people would quickly respond 2 because 2228. This

mathematical statement could also be written as the following. 8

8 8 8 3 3

3   

Now let’s work with the variable x. If we needed to find three identical terms that multiply to give us the product x, we would end up with the following statement.

x x x

x3 3 

3

 Again we are going to find three identical terms that multiply to give us the product x but we are not going to rely on radicals (i.e. square roots, cube roots, fourth roots,…). Instead we are going to rely on our understanding of the Law of Products of Powers. We need to find three powers with the base of x and the same exponent whose product is x.

x x x

x? ? ? 

The only exponent that makes sense is one third. x x x x x 3 1or 1 3 1 3 1   

 In the first bullet we stated that 3 x3 x3 x  x

. In the second bullet we stated that

x x x x   3  1 3 1 3 1

. Therefore we can conclude that 3 3 1

x

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Conclusion: If x2  x 1 and 3 3 1 x

x  , its fair to conclude that _xn n _x

1 . Examples: 5 25 252 1   273 3 27 3 1   164 416 2 1   6 5 36 25 36 25 2 1         3 2 27 8 27 8 3 3 1           2 1 32 1 32 1 5 5 1         number.) negative a of root square the e cannot tak (We solution real no 4 1 4 1 2 1         

Operations with Zero, Negative, and Fractional Exponents

The laws of exponents that we learned in the previous section still apply when dealing with powers with zero, negative, and fractional exponents

Examples: 7 7 0 7 0 x x x x     c4c5 c4 5 c9 5p9q23p1q5 15p8q7 To Sum It Up:

Zero Exponents: x0 1 where x0 Example: 50 1 Negative Exponents: n n x x         1 where x0 Example: 9 1 3 1 3 2 2 _         Fractional Exponents: _xn n _x 1 Example: 83 3 8 2 1   Note:

We can only take the square roots and fouth roots of positive numbers. 7 49  100 10 416 2 5 625 4   49 no real solution 4   16 no real solution

We can take the cube roots and fifth roots of both positive and negative numbers. 2 8 3  2 8 3   1 1 5  5 11

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 2 6 8 2 8        x x x x 3 2 5 2 3      d d d d 7 3 9 4 6 3 2 7 14 q p q p q p _   

 

3 4 _ 34 _ 12 x x x 3 2 6 6 3 1 f f f _         2 8 2 8 0 4 2 1 2 0 z y z y x z y x           Questions:

1. Evaluate each of the following. Do not use a calculator.

(a) 31 (b) 60  (c) 2  1 81 (d) 2  4 (e) 3  1 27 (f)        0 7 5 (g) 52  (h)        1 3 2 (i)        0 3 2 (j) 34  (k)        2 1 25 4 (l) 24  (m)        2 4 3 (n) 90  (o) 3  1 8 (p)        2 1 4 1 (q)        2 1 16 9 (r)        3 2 3 (s)        3 1 8 27 (t)        2 5 6

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2. Express each as a power. (a) 3 _p_(b) 2 1 b (c) h  (d) 1₅  d

3. Determine x. No work needs to be shown. (a) 9 1 2 _  x x = _____ (b) 6x 1 x = _____ (c) 10x 10 1 _ x = _____ (d) 2  x 1 25 x = _____ (e) x50 x = _____ (f) 416x x = _____ (g) 3 2 1   x x = _____ (h) 125x 5 x = _____ (i) 2x 8 x = _____ (j) 8 1 2x  x = _____

(a) x5x3  (b)   2 5 x x (c)

 

x4 5  (d) h5h7  (e) b1b0 b3  (f) _   2 6 k k (g)



a3b2



6  (h) 2c411c9  (i) _        _ 3 3 1 6 2p q (j) _₁  5 3 6 x x (k) 5x3y22x7y3  (l)           8 4 1 3 2 c b a

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(m) _ _   7 1 2 3 2 18 d c d c (n) _        2 5 0 7 2 3 t s r (o) _   1 4 5 35 gh h g (p) 4p4q23pq8 

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Scientific Notation

You have probably done some calculations on your calculator and ended up with pretty weird looking answers. The screen shot shown on the right contains two such calculations. In one case we took 2000 and raised it to the exponent 4, and in the other case we took 0.07 and divided it by 50 000. You might not have realized that your calculator is expressing the answer in scientific notation.

   13 4 10 6 . 1

2000 16 000 000 000 000 (This is a huge number!)

0000014 . 0 10 4 . 1 50000 07 . 0   6 

(This is a tiny number!)

When scientists (and people in health care professions) work with very large or very small numbers, they often express these numbers using scientific notation.

A positive number is in scientific notation when it is in the form n

a10 , where: 1. a is a number greater than or equal to 1 and less than 10, and

2. n is an integer (…-3, -2, -1, 0, 1, 2, 3,…)

Before we can start, we have to look at powers with a base of 10 and exponents that are integers. In last section project we learned that negative exponents meant we were dealing with

reciprocals (i.e. n n x x         1

where x0). If we apply this to powers with a base of 10, we get the following. 1000 1 10 1 10 100 1 10 1 10 10 1 10 1 10 1 10 10 10 100 10 1000 10 3 3 2 2 1 1 0 1 2 3             

This knowledge can now be used to convert numbers from standard form to scientific notation and vise versa. For example, 6103 means 61000 or 6000.

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Example 1

Convert the following numbers in standard form to scientific notation.

(a) 6 400 (b) 134 000

(c) 0.0047 (d) 0.00008

Answers:

We will show you two very similar methods for tackling these questions. You choose the one you prefer.

Method 1 Method 2

(a) 6 400 = 6.4  1000 =6.4103

We change the given number to a number greater than or equal to 1 and less than 10. Therefore we change 6400 to 6.4. In doing this, we moved the decimal point three places to the left.

So now we have the following but we do not know the exponent for the power.

6 400 = 6.410?

Since the decimal point was moved three places to the left, we know that the exponent for the power must be 3.

6 400 = 6.4103 (b) 134 000 = 1.34  100 000

= 1.34 10 5

We change 134 000 to 1.34. In doing this, we moved the decimal point five places to the left. Now we have the following.

134 000 = 1.3410?

Since the decimal point was moved five places to the left, we know that the exponent for the power must be 5. 134 000 = 1.34105 (c) 0.0047 = 4.7  0.001 = 4.7  1000 1 = 4.7103

We change the given number to a number greater than or equal to 1 and less than 10. Therefore we change 0.0047 to 4.7. In doing this, we moved the decimal point three places to the right. Now we have the following.

? 10 7 . 4 0047 . 0  

Since the decimal point was moved three places to the right, we know that the exponent for the power must be -3. 3 10 7 . 4 0047 . 0   

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(d) 0.00008 = 8  0.00001 = 8 

100000 1

= 8105

We change 0.00008 to 8. In doing this, we moved the decimal point five places to the right. Now we have the following.

? 10 8 00008 . 0  

Since the decimal point was moved five places to the right, we know that the exponent for the power must be -5. 5 10 8 00008 . 0    Example 2

Convert the following numbers in scientific notation to standard form.

(a) 9.2104 (b) 3.57106

(c) 2.4102 (d) 7104

Answers:

Again we will show you two very similar methods; choose the one you prefer.

Method 1 Method 2

(a) 9.2104 9.210000 = 92 000

Since the exponent of our power is 4, we move the decimal point, for the number 9.2, four places to the right and add zeros as placeholders.

4 10 2 . 9  = 92 000 (b) ₃_.₅₇₁₀6 ₃_.₅₇_1000000 = 3 570 000

Since the exponent of our power is 6, we move the decimal point, for the number 3.57, six places to the right and add zeros as placeholders.

6 10 57 . 3  = 3 570 000 (c) 100 1 4 . 2 10 4 . 2  2   024 . 0 01 . 0 4 . 2   

Since the exponent of our power is -2, we move the decimal point, for the number 2.4, two places to the left and add zeros as placeholders.

0242.4102 0. (d) 10000 1 7 10 7 4   0007 . 0 0001 . 0 7   

Since the exponent of our power is -4, we move the decimal point, for the number 7, four places to the left and add zeros as placeholders.

0007 . 0 10 7 4 

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Questions:

1. Convert the following numbers in standard form to scientific notation.

(a) 80 000 (b) 7 000 000

(c) 900 000 000 (d) 0.05

(e) 0.0006 (f) 0.000002

(g) 820 000 (h) 0.0048

(i) 12 500 000 (j) 0.0000731

2. Convert the following numbers in scientific notation to standard form.

(a) 5104 (b) 3107 (c) 3 10 4  (d) 6 10 9  (e) 1.2103 (f) 4.2105 (g) 3.75101 (h) 4.07105 (i) 7.9108 (j) 1.93102

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3. A certain thyroid tablet contains 0.00000025 g of medication. Express this amount in scientific notation.

4. Circle the appropriate response.

When a number is expressed in scientific notation and the exponent of the 10 is a negative, then the number is (less than or greater than) one.

5. Order these numbers from smallest to largest. (a) 3 200, 0.00095, 912.5, 0.001 (b) 0.01002, 32.005, 32.01, 0.008 (c) 4.3103, 2.1106, 5104 (d) 9.1101, 8102, 1.32102 (e) 7102, 6.4108, 3.8106 (f) 2.49104, 8103, 5.1106, 4.16104 (g) 5.99103, 9.3103, 4102, 1.93102 (h) 6.74104, 3103, 9.3103, 8.3104

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More Scientific Notation

In this section we will learn how to complete operations (addition, subtraction, multiplication, and division) with numbers in scientific notation.

Adding and Subtracting Numbers in Scientific Notation

Numbers in scientific notation can only be added if they have the same exponent on the power of base 10. We add or subtract the decimal part and leave the exponent unchanged.

Example 1

Complete the operation.

(a) 4.2103 3.5103 (b) 7.98104 1.49104 (c) 5107 8107 (d) 8.91106 3.42106 (e) 7106 4105 (f) 3.2105 9.2106

Answers:

The first four questions are a little easier because the exponents of the powers are the same in each case. (a) 3 3 3 10 7 . 7 10 5 . 3 10 2 . 4     (b) 4 4 4 10 49 . 6 10 49 . 1 10 98 . 7        (c) 7 7 7 10 13 10 8 10 5    

Our answer is not in the proper scientific notation form because the first number (13) is not greater than or equal to 1 and less than 10. We have to do more work.





8 7 1 7 10 3 . 1 10 10 3 . 1 10 13     (d) 6 6 6 10 33 . 12 10 42 . 3 10 91 . 8       

As with the last question, our answer has to be changed to the proper scientific notation form.





5 6 1 6 10 233 . 1 10 10 233 . 1 10 33 . 12       

With the two remaining questions, the exponents are different. This means that we must change one of the numbers (the smaller number) such that the two numbers end up with the same exponent.

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(e) 6 5

10 4 10 7  

Change the smaller number, 4105.





6 5 1 5 10 4 . 0 10 10 4 . 0 10 4    

Now we can do the addition.

6 6 6 10 4 . 7 10 4 . 0 10 7     (f) 5 6 10 2 . 9 10 2 . 3     

Change the smaller number, 9.2106.





5 6 1 6 10 92 . 0 10 10 92 . 0 10 2 . 9       

Now we can do the subtraction.

5 5 5 10 28 . 2 10 92 . 0 10 2 . 3       

Multiplying Numbers in Scientific Notation

One can multiply any two numbers written in scientific notation. Unlike addition and

subtraction, we do not need the same exponent on the powers with a base of 10. To multiply numbers in scientific notation we merely multiply the numbers to the left of the 10 notation and then multiply the powers with a base of 10 (by adding the exponents). We also have to make sure that the final answer is in the proper scientific notation form.





  4 2 6 2 6 2 6 10 82 . 29 10 82 . 29 10 10 1 . 7 2 . 4 10 1 . 7 10 2 . 4          

Not in the proper form.





5 4 1 4 10 982 . 2 10 10 982 . 2 10 82 . 29     (d)















3 5 8 5 8 5 8 10 86 . 38 10 86 . 38 10 10 8 . 5 7 . 6 10 8 . 5 10 7 . 6           





2 3 1 3 10 886 . 3 10 10 886 . 3 10 86 . 38       

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Dividing Numbers in Scientific Notation

Similar to multiplication, one can divide any two numbers written in scientific notation. We merely divide the numbers to the left of the 10 notation and then divide the powers with a base of 10 (by subtracting the exponents). Again one has to make sure the final answer is in the proper scientific notation form.

10 2 8 2 8 2 8 10 79 . 0 10 79 . 0 10 10 7 . 6 3 . 5 10 7 . 6 10 3 . 5            





11 10 1 10 10 9 . 7 10 10 9 . 7 10 79 . 0         (d)



 











  13 5 8 5 8 5 8 10 455 . 0 10 455 . 0 10 10 75 . 2 25 . 1 10 75 . 2 10 25 . 1           

Not in the proper form





12 13 1 13 10 55 . 4 10 10 55 . 4 10 455 . 0      Questions

1. Complete the operations. Write your answer in scientific notation.

(a) 4 4 10 2 . 6 10 7 . 1    (b) 7 7 10 1 . 3 10 6 . 8    (c) 8103 4.6103 (d) 7.4106 5.3106

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When we first learned how to solve simple linear equations like 3x57, we also learned about inverse operations. Consider the examples below.

3 7 10 7 7 10 7        x x

x The x and 7 are being added together on the left hand side of the equation. In order to solve for x, we had to “undo” that addition. This was accomplished by subtracting 7 from both sides of the equation. Subtraction is the inverse operation to addition. The subtraction reverses the work already done by the addition.

We can conclude that:

10 7  x is equivalent to x107 8 3 24 3 3 24 3    x x

x The 3 and x are being multiplied together. In order to reverse this process of multiplication we used the inverse operation division.

We can conclude that:

24 3x is equivalent to 3 24  x

In the two examples above, we were able to use some of the basic algebraic operations (i.e. addition, subtraction, multiplication, division, and roots) to solve for x. Unfortunately these same basic algebraic operations can not be used to solve exponential equations. Mathematicians had to create a new inverse operation to handle exponential equations. Complete the following using a calculator (i.e. Use the 10x button.).

(a) Evaluate 102.5, then find the log of that solution. Your Answer: _____ (b) Evaluate 100.7, then find the log of that solution. Your Answer: _____ (c) Evaluate 101.85, then find the log of that solution. Your Answer: _____ (d) Did the log button “undo” the work done by the exponent button? Yes or No

Here are the equivalent forms of equations we previously learned.

Example: b a x  is equivalent to xba x710 is equivalent to x107 b a x  is equivalent to xba x58 is equivalent to x85 b ax is equivalent to a b x 3x24 is equivalent to 3 24  x b a x _ is equivalent to xab 4 5  x is equivalent to x45 a xn  is equivalent to n a x x3 8 is equivalent to 3 8  x

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Now that we know that logs “undo” the work done by exponents, we can now add a new equivalent form of equations to our list.

Example: b ax  is equivalent to xlogab 10 8 x is equivalent to xlog108 Example 1

Change the following from exponential to logarithmic form. (a) 12553  (b) 8 1 23  (c) 2 1 9 3 Answers: (a) log₅1253 (b) 3 8 1 log₂       (c) 2 1 3 log₉  We would say, “Log

base 5 of 125 is 3.”

Example 2

Change the following from logarithmic to exponential form. (a) log₄162 (b) log 2 3 1 8  (c) 1 9 1 log₉       Answers: (a) 42 16 (b) 83 2 1  (c) 9 1 91  Example 3

Evaluate each of the following logarithms. (a) 49log₇ (b)       16 1 log₂ (c) log₄2 Answers:

Do not use your log button on your calculator to solve any of these problems. The log button works only with logarithms with a base of 10. Our questions have the bases 7, 2 and 4. The easiest approach is to make each of these expressions equal to x, change from logarithmic form to exponential form, and then evaluate.

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(a) 2 49 log 2 49 7 49 log 7 7     x x x (b) 4 16 1 log 4 16 1 2 16 1 log 2 2                   x x x (c) 2 1 2 log 2 1 2 4 2 log 4 4     x x x Example 4

Evaluate each of the following logarithms. (a) log 270

(b) log 0.045 (c) log (-100)

Answers:

When the base is not supplied on a logarithm, one is to assume that the base is 10. Therefore: logxlog₁₀ x

Since the base is 10, we can use the log button on our calculator to evaluate each of these.

(a) log 270 = 2.431

This could be restated as 102.431 270. The answer 2.431 seems reasonable because 100

102  and 103 1000. Since 270 is between 100 and 1000, then we would expect our answer to be between 2 and 3.

(b) log 0.045 = -1.347

This could be restated as 101.347 0.045. The answer -1.347 seems reasonable because 10

1

101  or 0.1, and

100 1

102  or 0.01. Since 0.045 is between 0.1 and 0.01, then we would expect our answer to be between -1 and -2.

(c) log (-100) no solution

The calculator is showing an error because this question can be restated as 10x 100. It is impossible to take 10 and raise it to an exponent such that a negative number is generated.

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Example 5 Solve for x. (a) log2 x3 (b) 2 1 log9 x (c) log4 x1 (d) logx2

(e) 7logx3. (f) logx1.4

(g) 2log_x25 (h) 2 9 1 log       x Answers:

With all these questions, we will start by changing the equation from its logarithmic form to its exponential form. Once this is done, the first four questions (a to d) can be quickly solved in your head. The last two questions (e and f) will require one to use a calculator's 10 x button. (a) 8 2 3 log 3 2    x x x (b) 3 9 9 2 1 log 2 1 9     x x x x (c) 4 1 4 1 log 1 4      x x x (d) 100 10 2 log 2 log 2 10     x x x x (e) x x x    7 . 3 10 10 7 . 3 log 7 . 3 log (calculator) 9 . 5011  x (f) x x x      1.4 10 10 4 . 1 log 4 . 1 log (calculator) 0.0398 x (g) 5 25 2 25 log 2    x x x (h) 3 9 1 2 9 1 log 2            x x x

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Questions:

1. Change the following from exponential to logarithmic form.

(a) 34 81 (b) 6443 (c) 3 2 9 1 _  (d) 7 1 71  (e) 362 6 1  (f) 270 1

2. Change the following from logarithmic to exponential form.

(a) log121442 (b) 4log216

(c) 4 81 1 log₃       (d)         36 1 log 2 ₆ (e) log

 

3 2 1 9  (f) 3log0.001 (g) 1 25 1 log25       (h) log100004

3. Evaluate each of the following logarithms.

(a) 9log3 (b) log232

(c) 8log8 (d)       27 1 log3

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(e) log₉ x 2 1 _ (f) 3 1 log₈ x (g) xlog5125 (h) 3logx8

(i) logx5 (j) logx0.72

(k) 3log₃ x (l) 3.15logx (m) x      64 1 log₈ (n) 1 3 1 log       x

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Logarithms and pH

When an acid is added to water, it releases hydrogen ions (H+) which logically drives up the concentration of hydrogen ions (H+). This resulting acidic solution is said to have a pH lower than 7. For example, black coffee (a weak acidic solution) has a pH of 5 while stomach acid (a strong acidic solution) has a pH of 1.

When a base is added to water, it releases hydroxide ions (OH-) which in turns drives down the concentration of hydrogen ions. This resulting basic (or alkaline) solution is said to have a pH greater than 7. For example, milk of magnesia (a weak basic solution) has a pH of 10 while bleach (a strong basic solution) has a pH of 13.

This pH system was developed in 1909 by the Swedish chemist S.P.L. Sorensen. He defined pH as the logarithm (base 10) of the reciprocal of the concentration of the hydrogen ions, measured in moles per litre, in a solution.

       _ ] H [ 1 log

pH Note: [H+] represents the concentration of hydrogen ions From chemistry you have learned that one mole of a substance represents 6.0221023 representative particles (atoms, molecules or ions) of that substance. The concentration measured in moles per litre (mol/L) is referred to as molarity.

Example 1

If the concentration of hydrogen ions in a solution is 0.0078 moles per litre, then what is the corresponding pH of that solution?

Answer:        _ ] H [ 1 log pH        0.0078 1 log pH



128.2



log pH 1 . 2 pH Note:

We will talk more about hydroxide ions later in the unit and explain why it affects hydrogen ion concentrations.

Please Note:

Many chemistry textbooks define pH as the following.

 

[H ] log pH 

This is equivalent to the formula that we are using. To prove this, we can answer example 1 using this alternate formula and show that we get the same answer.

 

[H ] log pH 



0.0078



log pH



2.1



pH  1 . 2 pH

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Example 2

If the pH of a solution is 8.3, then what is the concentration of hydrogen ions in moles per litre?

Answer:        _ ] H [ 1 log pH        _ ] H [ 1 log 8.3 _       _ ] H [ 1 log 8.3 10 ] H [ 1 108.3  _ 1 10 ] H [   8.3  3 . 8 10 1 ] H [   9 10 01 . 5 ] H

[     moles per litre (It can also be written as 0.00000000501 mol/L.)

Example 3

The blood for one adult patient is showing hydrogen ion concentrations of 7.94108 moles per litre.

(a) Determine the pH for this patient's blood.

(b) Is this patient's blood in the normal pH range or is he/she experiencing acidosis or alkalosis?

Answer: (a) _       _ ] H [ 1 log pH         _-₈ 10 7.94 1 log pH



12594458



log pH 1 . 7 pH

(b) The normal blood pH range for adults is between 7.35 and 7.45 (See pages 2 and 3.). In this case the blood pH is below this range, therefore the patient is experiencing acidosis.

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Example 4

The blood pH for a 9 year old child is 7.68. The child has also been vomiting for the last two days and as a result can not keep down food or fluids.

(a) Based on the blood pH levels and the patient's symptom, what can one conclude? (b) Determine the concentration of hydrogen ions in moles per litre in this patient's blood.

Answer:

(a) The normal blood pH range for children (< 16 years) is between 7.35 and 7.42 (See pages 2 and 3.). In this case, the pH reading is higher than this meaning the patient is experiencing alkalosis. The prolonged vomiting indicates that we are dealing with metabolic alkalosis, rather than respiratory alkalosis (Again see pages 2 and 3.).

(b) _       _ ] H [ 1 log pH        _ ] H [ 1 log 7.68        _ ] H [ 1 log 7.68 ₁₀ ] H [ 1 107.68   1 10 ] H [   7.68  68 . 7 10 1 ] H [   8 10 09 . 2 ] H

[     moles per litre (It can also be written as 0.0000000209 mol/L.)

Questions:

1. The hydrogen ion concentrations for different solutions have been provided below. Determine the corresponding pH for each.

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(c) cider: 5.01104 mol/L (d) urine: 7.94107 mol/L

2. The pH readings of various solutions have been provided below. Determine the corresponding hydrogen ion concentrations in moles per litre.

(a) saliva: pH 6.7 (b) sea water: pH 8.2

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3. The blood pH for an adult male is 7.05. The patient's breathing is slow and shallow.. (a) Based on the blood pH levels and the patient's symptom, what can one conclude? (b) Determine the concentration of hydrogen ions in moles per litre in this patient's blood.

4. The blood for one child patient (<16 years) is showing hydrogen ion concentrations of

8

10 17 .

4   moles per litre.

5. The blood pH for a 6 year old female is 7.13. The child has also been having severe diarrhea over the last three days.

(a) Based on the blood pH levels and the patient's symptom, what can one conclude? (b) Determine the concentration of hydrogen ions in moles per litre in this patient's blood.

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6. The blood for an adult female is showing hydrogen ion concentrations of 1.99108 moles per litre.

(b) Based on the blood pH readings and that the patient suffers from anxiety attacks that result in prolonged bouts of rapid deep breathing, what can one conclude?

7. Complete the following pH Table (i.e. fill in the hydrogen ion concentrations). Initially you might think that you have to do lots of calculations to complete this table, but after you have done a few, you should be able to see a pattern that will allow you to quickly complete the table.

pH Hydrogen Ion

Concentration (in moles per litre)

Examples

14 Liquid Drain Cleaner

13 Bleach 12 Soapy Water 11 Ammonia 10 Milk of Magnesia 9 Baking Soda 8 Egg Whites 7 Pure Water 6 Milk 5 Black Coffee 4 Tomato Juice 3 Orange Juice 2 Lemon Juice 1 Stomach Acid 0 Battery Acid

8. Look at your completed pH table from the previous question and answer the following. (a) As the pH increases by one, the hydrogen ion concentration decreases by a factor of ___. (b) As the pH decreases by one, the hydrogen ion concentration increases by a factor of ___.

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Logarithms and pOH

Up to this point, we have talked about using the concentration of hydrogen ions as a means of describing the acidity or alkalinity of a solution. This, however, is not the complete story. There is another ion, the hydroxide ion, which we must consider. The obvious question is what is the relationship between the two ions (hydrogen ions and hydroxide ions)?

If you were to examine pure water, you would probably think that is totally comprised of molecules of H2O. However, approximately two out of every billion water molecules ionize to

form hydrogen ions (H+) and hydroxide ions (OH-). H2O(l)  H+(aq) + OH-(aq)

Many chemistry textbooks define pOH using the following formula.



[OH ]



log

pOH 

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Add an Acidic Solution to Pure Water

So what happens when an acidic solution is added to pure water? For example, when hydrochloric acid (HCL) is added to pure water, it increases the amount of H+ in the new solution while the amount of OH- remains the same. However, the concentration of H+ multiplied by the concentration of the OH- remains constant at 11014moles per litre. [H+][OH-] = 11014

Initially, one might think that these two statements do not seem to align. How can one increase the amount of one ion (H+), keep the amount of the other ion (OH-) the same, yet end up with the same constant



11014



when the concentrations are multiplied? The key to understanding this is recognizing the difference between the amount of an ion (measured in moles) and the

concentration of an ion (measured in moles per litre). Consider the following situation. Let's say that we have one litre of pure water. This water would have 1107 moles of hydrogen ions and 1107 moles of hydroxide ions.

Suppose we add one litre of a specific acidic solution to the pure water such that the amount of hydrogen ions will change from 1107 moles to 4107 moles. Adding an acidic solution to the water will not change the amount of hydroxide ions, therefore they will remain at 1107 moles.

For this new two litre solution, let's look at the ions as concentrations (moles per litre) rather than amounts (moles)

For the hydrogen ions (H+) we have: 4107 moles per 2 litres

7 10 4  moles of H+ 2 litres of solution 7 10 1  moles of OH -7 10 1  moles of H+ 1 litre of pure water

7

10

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-This can be simplified to: 2107 moles per 1 litres For the hydroxide ions (OH-) we have: 1107 moles per 2 litres This can be simplified to: 0.5107 moles per 1 litre

(or 5108 moles per litre)

Add a Basic Solution to Pure Water

So what happens when a basic solution is added to pure water? For example, when sodium hydroxide (NaOH) is added to pure water, it increases the amount of OH- in the new solution while the amount of H+ remains the same. However, the concentration of H+ multiplied by the concentration of the OH- remains constant at 11014moles per litre.

[H+][OH-] = 11014

The logic behind this is very similar to that discussed when we looked at adding an acidic solution to pure water.

Similarly there is a relationship between the pH and pOH pH + pOH = 14

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Questions

1. Complete the following table (i.e. complete the last two columns). pH Hydrogen Ion

Concentration (mol/L)

[H+]

Examples Hydroxide Ion

Concentration (mol/L)

[OH-]

pOH [H+][OH-]

14 14

10 Liquid Drain Cleaner 1 or 10 0

13 13 10 Bleach 1 10 12 12 10 Soapy Water 2 10 11 11 10 Ammonia 3 10 10 10 10 Milk of Magnesia 4 10 9 9 10 Baking Soda 5 10 8 8 10 Egg Whites 6 10 7 7 10 Pure Water 7 10 6 6 10 Milk 8 10 5 5 10 Black Coffee 9 10 4 4 10 Tomato Juice 10 10 3 3 10 Orange Juice 11 10 2 2 10 Lemon Juice 12 10 1 1 10 Stomach Acid 13 10 0 1 or 0 10 Battery Acid 14 10

2. Look at your completed table from the previous question and answer the following. (a) As the pOH increases by one, the hydroxide ion concentration decreases by a factor of

___.

(b) As the pOH decreases by one, the hydroxide ion concentration increases by a factor of ___.

(c) What can you say about the pH and pOH of the same solutions?

(d) What can you say about the hydrogen ion concentration and hydroxide ion concentration for the same solution?

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More pOH

In this section, we will apply the four formulas we have learned.

       _ ] H [ 1 log pH _       _ ] OH [ 1 log

pOH [H+][OH-] = 11014 pH + pOH = 14 Example 1

What is the pH of a solution that has a pOH of 2.7? Are we dealing with an acidic, basic or neutral solution? Answer: pH + pOH = 14 pH + 2.7 = 14 pH = 14 - 2.7 pH = 11.3

When the pH is greater than 7 (or the corresponding pOH is less than 7), then we are dealing with a basic solution.

Example 2

What is the hydrogen ion concentration for a solution that has a hydroxide ion concentration of

6

10 8 .

5   moles per litre?

Answer: [H+][OH-] = 14 10 1 

   

 

6 14 --14 10 8 . 5 10 1 H OH 10 1 H        

 

9 10 56 . 2

H    moles per litre Example 3

What is the pOH of a solution with a hydroxide ion concentration of 6.1104 moles per litre?

Answer:        _ ] OH [ 1 log pOH         _-₄ 10 6.1 1 log pOH



1639



log pOH 2 . 3 pOH

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Example 4

What is the hydroxide ion concentration in a solution having a pOH of 9.1?

Answer:        _ ] OH [ 1 log pOH        _ ] OH [ 1 log 9.1 _       _ ] OH [ 1 log 9.1 ₁₀ ] OH [ 1 109.1  _ 1 10 ] OH [   9.1  1 . 9 10 1 ] OH [   10 10 94 . 7 ] OH

[     moles per litre Example 5

What is the pOH of a solution having [H+] = 3.9106 moles per litre?

Answer:

Method A: First find the pH and then find the pOH.

       _ ] H [ 1 log pH         _-₆ 10 3.9 1 log pH



256410



log pH 4 . 5 pH pH + pOH = 14 5.4 + pOH = 14 pOH = 14 - 5.4 pOH = 8.6

Method B: First find the [OH-] and then find the pOH. [H+][OH-] = 11014

   

 

- 9 6 14 --14 -10 56 . 2 OH 10 9 . 3 10 1 OH H 10 1 OH                 _ ] OH [ 1 log pOH         _-₉ 10 2.56 1 log pOH



390625000



log pOH 6 . 8 pOH

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Example 6:

If the pH of a solution is 9.2, determine the concentration of hydroxide ions in moles per litre.

Answer:

Method A: First find the pOH and then find [OH-] pH + pOH = 14 9.2 + pOH = 14 pOH = 14 - 9.2 pOH = 4.8        _ ] OH [ 1 log pOH        _ ] OH [ 1 log 4.8 _       _ ] OH [ 1 log 4.8 ₁₀ ] OH [ 1 104.8  _ 1 10 ] OH [   4.8  8 . 4 10 1 ] OH [   5 10 58 . 1 ] OH

[     moles per litre Method B: First find [H+] and then find [OH-].

       _ ] H [ 1 log pH _       _ ] H [ 1 log 9.2 10 ] H [ 1 109.2  _ 1 10 ] H [   9.2  2 . 9 10 1 ] H [   10 10 31 . 6 ] H

[     moles per litre [H+][OH-] = 11014

   

   H 10 1 OH -14

-It was changed from logarithmic to exponential form.

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 

OH 1.58 10 molesper litre 10 31 . 6 10 1 OH 5 -10 -14 -       Questions

1. If the pH of saliva is 7.4, find its pOH.

2. What is the pOH of a solution which has a hydroxide ion concentration of 1.8106 moles per litre?

3. If the hydrogen ion concentration of milk is 1.8108 moles per litre, determine the concentration of hydroxide ions in moles per litre.

4. What is the pH of a solution which has a hydrogen ion concentration of 4.91011 moles per litre?

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5. A certain hair conditioner has a pOH of 7.9. What is the hydroxide ion concentration in that conditioner?

6. If the pOH of a glass of lemonade is 11.2, what is the corresponding pH?

7. If the hydroxide ion concentration of a solution is 3.8107 moles per litre, determine the concentration of hydrogen ions in moles per litre.

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9. If the pOH of a solution is 7.6, determine the concentration of hydrogen ions in moles per litre.

10. What is the pH of a solution having [OH-] = 4.9108 moles per litre?

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12. Determine the pOH of a solution that has a hydrogen ion concentration of 6.4109 moles per litre? Are we dealing with an acidic, basic, or neutral solution?

13. The hydrogen ion concentration in an adult blood sample in 4.2108 moles per litre. (a) What is the pH of this blood sample?

(c) What is the pOH of this blood sample?

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Putting It Together

Questions

(a) 5x37x6 (b) 4a6b3a5b2 (c) 50n9 5n6 (d) 28c10d6 7c4d (e)

 

8h9 2 (f)



2p5q3



4 (g) 2 5 3 7       d c (h) 3 3 4 2 3       z xy (i)





n m n m 5 3 5 2 4 2 (j) ₄ 2 10 7 12 6 8 ab b a b a 

2. Evaluate each of the following. Do not use a calculator.

(a) 72 (b) 4 0 (c) 41 (d) 2 1 100 (e) 0 2 5       (f) 3 1 125 (g) 34 (h) 1 11 8        (i) 0 7 4       (j) 92 (k) 2 1 9 25       (l) 2 8 7       

Blood, ph and Logarithms Unit