CHAPTER 16: ACIDS AND BASES
Active Learning: 4, 6, 14; End-of-Chapter Problems: 2-25, 27-58, 66-68, 70, 75-77, 83, 90-91, 93-104 Chapter 15 End-of-Chapter Problems: 69-74, 125, 129, 133
16.1 ACIDS AND BASES
ACID: A substance that produces H+ ions in water
– Some acids are monoprotic (release only H+ per molecule) – e.g. HCl, HBr, HI, HNO3, HClO4
– Some acids are polyprotic (release more than one H+ per molecule) – e.g. H2SO4 and H2CO3 are both diprotic; H3PO4 is triprotic.
– properties of acids: taste sour, neutralize bases, react with CO32− or HCO3− to produce CO2(g), react with metals to produce H2, turn blue litmus paper red We can show the ionization or dissociation of acids as follows:
HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq) Note: H3O+ is the hydronium ion, a hydrated proton: H+ + H2O = H3O+
Since H3O+ ⇔ H+ + H2O, we will abbreviate it as H+, but recognize that it
actually exists as H3O+.
HCl(aq) → H+(aq) + Cl–(aq)
HC2H3O2(aq) H+(aq) + C2H3O2–(aq)
BASE: A substance that produces OH– ions in water—e.g. NaOH, Ba(OH)2
– properties of bases: taste bitter, feel soapy or slippery, neutralize acids, turn red litmus paper blue
We can show the ionization of a base as follows:
NaOH(aq) → Na+(aq) + OH–(aq) Ba(OH)2(aq) → Ba+2(aq) + 2 OH–(aq) ARRHENIUS DEFINITIONS OF ACIDS AND BASES
Arrhenius acid: A substance that produces H+ ions in water—e.g. HCl, HNO3, H2SO4
Arrhenius base: A substance that produces OH– ions in water—e.g. NaOH, Ba(OH)2
The general equation for an Arrhenius acid-base neutralization reaction is shown below:
HX(aq) + MOH(aq) → H2O(l) + MX(aq)
acid base water salt
BRØNSTED-LOWRY DEFINITIONS OF ACIDS AND BASES Why is H+ called a proton?
Brønsted-Lowry acid: A substance that donates a proton (H+)—i.e., a proton donor Brønsted-Lowry base: A substance that accepts a proton (H+)—i.e., a proton acceptor
– Unlike an Arrhenius base, a B-L base need not contain OH–. A Brønsted-Lowry acid-base reaction simply involves a proton (H+) transfer, as shown in the example below:
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
Note in this reaction simply involves H2O donating a H+ ion to NH3 to produce NH4+ and OH–.
→ In this reaction, H2O is the Brønsted-Lowry acid, and NH3 is the Brønsted-Lowry base.
– The conjugate acid-base pairs differ only by a H+.
→ In this reaction, the conjugate acid-base pairs are NH3 and NH4+ and H2O and OH–.
conjugate acid-base pairs: a Brønsted-Lowry acid/base and its conjugate differ by a H+
– For the reaction above, when HA donates H+ to H2O, it leaves behind A–, which can act as a base for the reverse reaction.
– An acid and base that differ only by the presence of H+ are conjugate acid-base pairs.
– Thus, HA is the conjugate acid of A–, and A– is the conjugate base of HA.
The general reaction for the dissociation (or ionization) of an acid can be represented as, where the double-arrow indicates both the forward and reverse reactions occur:
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
Brønsted-Lowry acid = ____________ Brønsted-Lowry base= ____________
Note: The double arrow ( ) indicates the reaction is reversible (goes in both directions).
Example: Determine the Brønsted-Lowry acid and base in each of the following reactions, and fill in the blanks to below indicate the conjugate acid/base pairs in the reaction:
a. H2O(l) + H2SO4(aq) → HSO4–(aq) + H3O+(aq)
Brønsted-Lowry acid = ____________ Brønsted-Lowry base= ____________
H2O(l) is the conjugate acid base of _______________________.
circle one
H2SO4(aq) is the conjugate acid base of _______________________.
circle one
b. H2O(l) + H2PO4–(aq) H3PO4(aq) + OH–(aq)
Brønsted-Lowry acid = ____________ Brønsted-Lowry base= ____________
H3PO4(aq) is the conjugate ____________ of _______________________.
OH–(aq) is the conjugate ____________ of _______________________.
Note that H2O is amphoteric since it can behave as an acid or a base.
16.2 ACID STRENGTH
A strong acid ionizes completely in water to form hydronium ion.
→ Almost all of the acid is broken up into ions in solution.
→ A forward arrow (not a double arrow) is used for the dissociation of a strong acid like HNO3(aq) since virtually all the HNO3 ionizes to H+ and NO3–, leaving almost not HNO3.
HNO3(aq) + H2O(l) → H3O+(aq) + NO3–(aq)
Similarly, a strong base dissociates (breaks up) completely.
→ Almost all of the base is broken up into ions in solution.
→ A forward arrow is used for the dissociation of a strong base like Ca(OH)2 (aq) since virtually all the Ca(OH)2 ionizes to Ca2+ and OH–, leaving almost not Ca(OH)2.
Ca(OH)2(aq) → Ca2+(aq) + 2 OH–(aq)
In contrast, a weak acid ionizes only to a very small extent.
→ Most of the acid remains intact as a molecule in solution.
→ Thus, a double arrow is used for the dissociation of a weak acid like HF(aq) since mostly HF exists at equilibrium, with very few H+ or F– ions in solution.
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
Know the following acids and bases. All other acids and bases are weak!
Strong Acids Strong Bases
HCl, HBr , HI, HNO3, HClO4, H2SO4 LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
Note: H2SO4(aq) is a strong acid and diprotic (able to release 2 H+ ions), but it generally ionizes to release only one H+, as follows:
H2SO4(aq) + H2O(l) → H3O+(aq) + HSO4–(aq)
The resulting HSO4–(aq) is a weak acid:
HSO4–(aq) + H2O(l) H3O+(aq) + SO4–2(aq)
The strength of an acid is inversely related to the strength of its conjugate base.
→ The weaker an acid, the stronger its conjugate base.
– The conjugate bases of weak acids (F–,NO2–, C2H3O2–, etc.) are much stronger bases compared to H2O.
→ These bases can react with a H2O molecule by removing a H+ ion from the water molecule to form its conjugate acid and OH–.
F–(aq) + H2O(l) HF(aq) + OH–(aq)
– The conjugate bases of strong acids (Cl–, Br–, I–, NO3–, ClO3–, ClO4–) are much weaker bases compared to H2O.
→ These bases are not strong enough to react with a H2O molecule.
Cl–(aq) + H2O(l) → No Reaction
Ex. 1: Circle all of the ions below that would react with water to produce hydroxide ion:
Br– C2H3O2– NO2– NO3– SO3–2 PO4–3
16.3 WATER AS AN ACID AND A BASE Consider the autoionization of water,
H2O + H2O H3O+ + OH−
which has the equilibrium expression: Kw = [H3O+] [ OH−] = [H+] [ OH−]
where Kw is called the ion-product constant or dissocation constant for water.
Experiment shows that at 25°C, [H+]=[ OH−]=1.0×10-7 M
so Kw = [H+] [ OH−] = (1.0×10-7) (1.0×10-7) = 1.0×10-14. (2 sig figs)
Acidic, Basic, and Neutral
A solution is acidic when A solution is basic when – it contains H+ ions. – it contains OH– ions.
– [H+] > [OH–] – [OH–] > [H+]
A substance is neutral when [H+]=[OH–], so water is neutral.
– If a substance does not produce H+or OH– ions in water, it will have [H+]=[OH–].
→ The substance is neutral—e.g., saline solution, dilute NaCl(aq), is neutral.
Ex. 1: Calculate the hydrogen or hydroxide ion concentration for each of the following, and indicate if the solution is acidic, basic, or neutral.
a. milk, [H+]=3.0×10-7M [OH−]=_____________ acidic basic neutral
b. carrots, [H+]=7.9×10-6M [OH−]=_____________ acidic basic neutral
c. eggs, [OH−]=6.3×10-7M [H+]=_____________ acidic basic neutral
d. urine, [H+]=1.8×10-5 M [OH−]=_____________ acidic basic neutral
Reviewing Logarithms
Logarithms provides a more convenient way to deal with very large and very small numbers.
– e.g., using log (or log10 = log base-10) the number 2.50×106 can be expressed as log 2.50×106 = 6.39794… or 106.39794…=2.50×106
The components of a logarithm are called the character and the mantissa.
– In the example above, the character is 6, corresponding to the exponent of the number when it’s expressed in scientific notation, and the mantissa is 39794…, which
– When a number is expressed in scientific notation,
– The exponent in scientific notation = the character of the logarithm, and the digit term in scientific notation = the mantissa of the logarithm.
Consider the following examples,
log 2.50×103 = 3.39794… log 2.50×10-1 = 0.39794… − 1 log 2.50×102 = 2.39794… log 2.50×10-2 = 0.39794… − 2 log 2.50×101 = 1.39794… log 2.50×10-3 = 0.39794… − 3 log 2.50×100 = 0.39794… log 2.50×10-4 = 0.39794… − 4
log 2.50×10-5 = 0.39794… − 5 log 2.50×10-6 = 0.39794… − 6
Note that the digit term of 2.50 in all of these examples always has the same mantissa, 0.39794… for non-negative exponents, and for negative exponents, the mantissa is the difference between the negative of the exponent and 0.39794….
Thus, when dealing with logarithms, the number of significant figures is determined by the number of digits in the mantissa.
→ For example, log 2.50×106 = 6.398 and 103.18 = 1.5×103
16.4 The pH Scale
pH Scale: a measure of the concentration of hydrogen ions, [H+]
The value of Kw at 25°C (1.0×10-14) is so small, indicating very few water molecules ionize to form H+ and OH−.
→The hydrogen ion concentration, [H+], is typically quite small, so the pH scale provides a convenient measure of a solution’s acidity.
The pH is a log10 scale and is defined as pH = −log [H+] so [H+] = 10-pH Similarly, pOH = −log [OH−] and [OH−] = 10-pOH
pH Scale: pH < 7: acidic pH = 7: neutral pH > 7: basic
[H+] scale 10-1 10-3 10-5 10-7 10-9 10-11 10-13
pH scale 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 strongly weakly acidic weakly basic strongly
acidic basic
contains H+ neutral contains OH− 1M HCl lemon juice pure eggs drain cleaner stomach acid vinegar water baking soda 1M NaOH
coffee NaCl(aq) liquid bleach
Calculate pH or [H+]: Recognize that pH = -log [H+] and [H+] = 10–pH To get pH if [H3O+] = 0.1M for a given solution, change [H+] given to 10–pH form.
→ so [H+] = 0.1M = 10–1M → pH = 1 and [H+]=0.0001M=10–4M → pH = 4
Calculate pOH or [OH–]: Recognize that pOH = -log [OH–] and [OH–] = 10–pOH To get pOH, change [OH–] given to 10–pOH form.
→ so [OH–] = 0.01M = 10–2M → pOH = 2
Converting between pH and pOH: pH + pOH = 14.00
Converting between [H+] and [OH–]: Kw = [H+][OH−] = 1.0×10-14
Complete the following table:
[H+] pH [OH–] pOH Classify
3.1×10-8M
2.6×10-9M 2.65
3.18 1.5×10-2M
16.5 CALCULATING THE pH of STRONG ACID SOLUTIONS
Ex. 1: Calculate the pH for the following solutions:
a. [HNO3]= 2.5×10-4M → [H+]= _____________ → pH = _____________
b. [HCl]= 1.5×10-3M → [H+]= _____________ → pH = _____________
c. [H2SO4]= 3.5×10-5M → [H+]= _____________ → pH = _____________
d. [NaOH]= 1.5×10-3M → [OH−]= _____________
→ pOH = _____________ → pH = _____________
e. [Ca(OH)2]= 2.0×10-5M pH = _____________
f. [Sr(OH)2]= 8.0×10-4M pH = _____________
15.7 NEUTRALIZATION REACTIONS
Let’s consider three different kinds of acid-base neutralization reactions:
• A strong acid reacts completely with a strong base to produce water and a salt.
• A strong acid reacts completely with a weak base to produce water and a salt.
• A strong base reacts completely with a weak acid to produce water and a salt.
In all three of these examples, we can determine the amount of acid (or base) present in a solution if we know the balanced chemical equation and how much base (or acid) is needed to completely neutralize it.
Ex 1: Find the molarity of a hydrochloric acid solution if 25.50 mL of HCl is required to neutralize 0.375 g of Na2CO3 as shown in the following equation:
2 HCl(aq) + Na2CO3(s) → 2 NaCl(aq) + H2O(l) + CO2(g)
Ex 2: A 25.0 mL sample of hydrochloric acid requires 37.55 mL of a 0.250M NaOH solution for complete neutralization. Calculate the molarity of the hydrochloric solution if the balanced equation for the reaction is:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Ex. 3: A 10.00 mL sample of calcium hydroxide requires 26.18 mL of a 0.5565M nitric acid solution for complete neutralization. Calculate the molarity of the calcium hydroxide.
a. Write the balanced equation for the reaction below.
b. Calculate the molarity of the calcium hydroxide solution.
Ex. 4: Citric acid (abbreviated H3X) is a triprotic acid—i.e. it has three H+ ions that can react to produce water—that contains a combination of carbons, hydrogens, and oxygens. If 36.10 mL of 0.223M NaOH is used to neutralize a 0.515 g sample of citric acid, calculate the molar mass of the acid.
H3X(aq) + 3 NaOH(aq) → Na3X(aq) + 3 H2O(l)