17.1 The common-ion effect 

Chapter 17: Additional Aspects of Aqueous  Equilibria 

17.1 The common-ion effect 

The extent of ionization of a weak electrolyte is ​decreased​ by adding to the solution a  strong electrolyte that has an ion in common with the weak electrolyte.   

CH​ _3​ COOH​(aq) ​+ H​ _2​ O ​(l) ​ ⇌ H​ _3​ O​ ^+​ ​(aq)​ + CH​ _3​ COO​ ^-​ (aq)  

If ​sodium acetate, NaCH​

COO​ was added to this solution, it would dissociate, the acetate  ion concentration would increase, how does Le Chatelier's principle say the equilibrium  will shift?  

What would happen to the pH of the solution?    

Sample Exercise 17.1 - What is the pH of a solution made by adding 0.30 mol of acetic acid  and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? 

What do we need to calculate the pH? 

We can find those quantities at equilibrium from an ice chart. 

Write the formula for the BCE   

Initial         

Change         

Equilibrium         

Use the K​

formula to solve for the Equilibrium row. K​

of acetic acid is 1.8 ×10​

Practice Exercise 1: For the generic equilibrium HA (​aq​) ⇌ H​

(​aq​) + A​

(​aq​), which of these  statements is true? 

(a)​ The equilibrium constant for this reaction changes as the pH changes. 

(b)​ If you add the soluble salt KA to a solution of HA that is at equilibrium, the  concentration of HA would decrease. 

(c)​ If you add the soluble salt KA to a solution of HA that is at equilibrium, the  concentration of A​

would decrease. 

(d)​ If you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would  increase 

Sample Exercise 17.2 Calculate the ​fluoride ion concentration​ and pH of a solution that is  0.20 ​M​ in HF and 0.10 ​M​ in HCl. 

What do we need to calculate the pH? 

We can find those quantities at equilibrium from an ice chart. 

Write the formula for the BCE   

Initial         

Change         

Equilibrium         

Use the K​

formula to solve for the Equilibrium row.  

17.2 Buffers 

Composed of a weak acid and its conjugate base (or weak base and its  conjugate acid) (SAME AS COMMON ION) 

Resists changes in pH when an acid or base is added.  

Strategy for approaching pH of a buffer solution is the SAME as pH of a Common Ion  Solution 

What is the pH of a buffer that is 0.12 M in lactic acid [CH​

CH(OH)COOH, or HC​

H​

O​

] and  0.10 M in sodium lactate [CH​

CH(OH)COONa or NaC​

H​

O​

]? For lactic acid, K​

= 1.4 × 10​

.   

What do we need to calculate the pH? 

We can find those quantities at equilibrium from an ice chart. 

Write the formula for the BCE   

Initial         

Change         

Equilibrium         

Use the K​

formula to solve for the Equilibrium row.  

How many moles of NH​

Cl must be added to 2.0 L of 0.10 M NH​

to form a buffer whose pH  is 9.00? (Assume that the addition of NH​

Cl does not change the volume of the solution.)   

Write the formula for the BCE   

Plug in what you know from the question   

Initial         

Change         

Equilibrium         

Use the K​

formula to solve for the Equilibrium row.  

Buffer Capacity and pH Range 

Buffer Capacity -    

Greater buffering capacity happens when… 

pH Range -    

Buffers have a usable range ± 1 pH unit of pKa.  

How do you choose the right acid/base to make a buffer solution?? 

Give it some thought...The K​

values for nitrous acid HNO​

and hypochlorous acid HClO are  4.5 x 10​

and 3.0 x 10​

respectively. Which one would be more suitable for use in a solution  buffered at pH = 7.0? What other substances would be needed to make the buffer?  

Addition of a Strong Acid or Base to a Buffer Solution   

When a strong acid or base is added to a buffer solution, we can assume ALL of it is  consumed by the weak base or acid present.  

Two parts to these calculations 

1. Limiting reactant stoichiometry problem (BCA)  2. Equilibrium (ICE) 

Example Problem: 1.00 L buffer solution with .140 mol cyanic acid (HCNO) and .110 mol  potassium cyanate (KCNO) K​

= 3.5 x10​

What is the pH ​before​ any acid or base is added (this is just normal common ion problem)   

Initial         

Change         

Equilibrium         

Same buffer, with the addition of .015 mol HNO​

1. Stoichiometric (show the acid (H​

) being consumed (reacting with) by the conjugate  base (CNO​

) 

DO THIS IN MOLES 

Before (MOL)       

Change (MOL)       

2. Use your answers from the stoichiometric calculations to do normal ICE chart. ICE  charts are done in MOLARITY so you may have to divide by the ​total volume​. 

Initial (M)         

Change (M)         

Equilibrium (M)         

Use your equilibrium concentrations in the K​

equation and solve for H​

and then pH   

Same buffer, with the addition of .015 mol KOH   

1. Stoichiometric (show the base (OH​

) being consumed by the acid (HCNO)  DO THIS IN MOLES 

Before (MOL)     

Change (MOL)     

After (MOL)     

2. Use your answers from the stoichiometric calculations to do normal ICE chart. ICE  charts are done in MOLARITY so you may have to divide by the ​total volume. 

Initial (M)         

Change (M)         

Equilibrium (M)         

Use your equilibrium concentrations in the K​

equation and solve for H​

and then pH   

Sample Exercise 17.6: A buffer is made by adding 0.300 mol CH​

COOH and 0.300 mol  CH​

COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74  (Sample Exercise 17.1).  

(a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) solution is added.  

(b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M  NaOH(aq) solution to 1.000 L of pure water. 

17.3 Acid-Base Titrations 

Solution containing a known concentration of an base is slowly added to an acid. (or vice  versa) 

Acid-base indicators are used to signal the ​equivalence point (# of mole of acid and  base are equal).  

STRONG ACID/STRONG BASE TITRATION 

Write the balanced molecular and net ionic equations for each of the following  neutralization reactions 

A. aqueous acetic acid is neutralized by aqueous barium hydroxide   

B. Solid chromium (III) hydroxide reacts with nitrous acid   

For neutralization calculations, remember that the number of moles of H​

(or H​

O​

) equals  the moles of OH​

.   

1.  Find mole of H​

or OH​

of the known substance  2.  State moles of H​

=mol OH​

3.  Use the mole of H​

or OH​

= to find the needed quantity of the unknown substance.  

A 25.0 ml solution of 0.500 M NaOH is titrated until neutralized into a 50.0 ml sample of  HCl. What was the concentration of the HCl? 

What volume of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)​

?    

For titration problems, treat these as a limiting reactant stoichiometry problem, or set up a  BCA chart.  

1. Write the BCE, for strong acid/strong base  H​

+ OH​

→ H​

O  2. Calculate moles of each and set up BCA chart 

3. Subtract, find moles of leftover species 

4. Divide by final volume to find new concentration, find pH.  

Sample Exercise 17.6 - Calculate the pH when ​(a)​ 49.0 mL and ​(b)​ 51.0 mL of 0.100 ​M​ NaOH  solution have been added to 50.0 mL of 0.100 ​M​ HCl solution. 

Acid Base Titration with a Weak acid/Strong base Combo 

4 Types of Calculations 

  Initial pH – no titrant has been added 

1. Do regular ICE chart with the acid being an acid (donating a proton to water) and  K​

from table to solve for [H​

] 

Between initial pH and equivalence point​ – some titrant has been added but not enough to  reach the equivalence point 

1. Do stoichiometric calculation with # of moles in BCA chart and the neutralization  reaction 

HA + OH- → A​

+ H​

O 

2. Take the # of moles of HA and A​

leftover and change back to molarity using the  new volume (original volume + volume of titrant added) 

3. Do regular ICE chart with acid being an acid (donating a proton to water) and K​

from table to solve for [H​

] 

pH at the equivalence point​ – number of moles of acid and base are equal 

1. Calculate the number of moles of A​

formed. (use molarity and volume of acid)  2. Calculate new molarity using the new volume (original volume + volume of titrant  added) 

3. Do ICE chart with the reaction of the A​

and water: A​

+ H​

O ⇄ HA + OH​

. Use K​

  and K​

to get K​

and solve for OH​

then pH. 

  pH after the equivalence point​ – more titrant is added than is needed to neutralize the acid. 

1. Calculate number of moles of acid and number of moles of base (from volume and  molarity) 

2. Subtract to find the number of excess moles of base and divide by new volume to  find new molarity. 

Use new molarity of base to find pH. 

  A 35.0 mL sample of 0.150 M acetic acid CH​

COOH is titrated with .150 M NaOH solution. 

Calculate the pH after the following volumes of titrant have been added.   

  1. 0 mL 

2. 17.5 mL (HALF EQUIVALENCE POINT)  3. 34.5 mL 

4. 35.0 mL  5. 35.5 mL  6. 50.0 mL 

  Which ones of these are each of the 4 types of calculations??? 

1. 0 mL     

  Between initial pH and equivalence point​ – some titrant has been added but not enough to  reach the equivalence point 

A 35.0 mL sample of 0.150 M acetic acid CH​

COOH is titrated with .150 M NaOH solution. 

Calculate the pH after the following volumes of titrant have been added.   

  2. 17.5 mL (HALF EQUIVALENCE POINT)     

3. 34.5 mL     

pH at the equivalence point​ – number of moles of acid and base are equal 

1. Calculate the number of moles of A​

formed. (use molarity and volume of acid)  2. Calculate new molarity using the new volume (original volume + volume of titrant  added) 

3. Do ICE chart with the reaction of the A​

and water: A- + H​

O ⇄ HA + OH​

. Use K​

  and K​

to get K​

and solve for OH​

then pH. 

  A 35.0 mL sample of 0.150 M acetic acid CH​

COOH is titrated with .150 M NaOH solution. 

Calculate the pH after the following volumes of titrant have been added.   

4.   35.0 mL     

pH after the equivalence point​ – more titrant is added than is needed to neutralize the acid. 

1. Calculate number of moles of acid and number of moles of base (from volume and  molarity) 

2. Subtract to find the number of excess moles of base and divide by new volume to  find new molarity. 

Use new molarity of base to find pH. 

  A 35.0 mL sample of 0.150 M acetic acid CH​

COOH is titrated with .150 M NaOH solution. 

Calculate the pH after the following volumes of titrant have been added. 

  5. 35.5 mL     

6. 50.0 mL       

  WEAK BASE WITH STRONG ACID 

  Consider the titration of 30.0 mL of .100 M NH​

with .150 M HCl. Calculate the pH after the  following volumes of titrant have been added. The K​

of NH​

is 1.8 × 10​

  1.  0 mL 

2. 10.0 mL (HALF EQUIVALENCE POINT)  3. 19.5 mL 

4. 20.0 mL  5. 20.5 mL  6. 30.0 mL 

  Initial pH – no titrant has been added    1.  0.0 mL 

Between initial pH and equivalence point​ – some titrant has been added but not enough to  reach the equivalence point

  2. 10.0 mL (HALF EQUIVALENCE POINT)     

3. 19.5 mL     

pH at the equivalence point​ – number of moles of acid and base are equal

4. 20.0 mL 

pH after the equivalence point​ – more titrant is added than is needed to neutralize the base.

5. 20.5 mL       

  6. 30.0 mL     

Titrations curves and acid strength     What is the pH at the half equivalence  point?  

Does the pH change much around the  half equivalence point? Why?  

Titrations with an Acid-Base Indicator   

Choose an indicator with a pKa similar to the pH at the equivalence point of  your titration 

Titration of Polyprotic acids     

17.4 Solubility Equilibria 

  Solubility rules are a ​qualitative​ description of how some ionic compound dissolve in water. 

Now we get to ​quantify​ it. 

  The solubility product constant, K​

Saturated solution of an ionic cpd dissolving is in equilibrium    BaSO​

(s)​ Ba​

​(aq)​ + SO​

​(aq) 

  K​

=      

  Sample Exercise 17.10 - Write the expression for the solubility-product constant for CaF​

,  and look up the corresponding K​

value in Appendix D. 

We can go back and forth between solubility and K​

Solubility (M) to K​

Sample Exercise 17.11 - Solid silver chromate is added to pure water at 25 °C, and some of  the solid remains undissolved. The mixture is stirred for several days to ensure that  equilibrium is achieved between the undissolved Ag​

CrO​

(s)​ and the solution. Analysis of  the equilibrated solution shows that its silver ion concentration is 1.3 × 10​

M. Assuming  that the Ag​

CrO​

solution is saturated and that there are no other important equilibria  involving Ag​

or CrO​

ions in the solution, calculate K​

for this compound. 

  Practice Exercise 2 

A saturated solution of Mg(OH)​

in contact with undissolved Mg(OH)​

(s) is prepared at 25 

°C. The pH of the solution is found to be 10.17. Assuming that there are no other  simultaneous equilibria involving the Mg​

or 

OH​

ions, calculate K​

for this compound. 

K​

to Solubility   

The K​

for CaF​

is 3.9 × 10​

at 25 °C. Assuming equilibrium is established between solid  and dissolved CaF​

, and that there are no other important equilibria affecting its solubility,  calculate the solubility of CaF​

in grams per liter. 

Practice Exercise 1 

Of the five salts listed below, which has the highest concentration of its cation in water? 

Assume that all salt solutions are saturated and that the ions do not undergo any additional  reactions in water. 

(a) lead (II) chromate, K​

= 2.8 × 10​

(b) cobalt(II) hydroxide, K​

= 1.3 × 10​

(c) cobalt(II) sulfide, K​

= 5 × 10​

(d) chromium(III) hydroxide, K​

= 1.6 × 10​

(e) silver sulfide, K​

= 6 × 10​

Practice Exercise 2 

The K​

for LaF​

is 2 × 10​

. What is the solubility of LaF​

in water in moles per liter? 

17.5 Factors that Affect Solubility 

  Common Ion Effect - just like before with Le Chatelier's Principle    CaF​

(aq) ⇌ Ca​

(aq) + 2F​

(aq)  

Sample Exercise 17.3 - Calculate the molar solubility of CaF​

at 25 °C in a solution that is (a)    0.010 M in Ca(NO​

)​

and (b) 0.010 M in NaF. 

Solubility and pH 

  Solubility is affected when the anion is basic or cation is acidic 

  If a substance has a basic anion (anion of a weak acid) , it will be more soluble in an acidic  solution overall 

  PbF​

(s) ⇌ Pb 2​

(aq) + 2F​

(aq)   F​

(aq) + H​

(aq) ⇌ HF (aq)    

If a substance has an acidic cation, it will be more soluble in a basic solution   

Which of these substances are more soluble in acidic solution than in basic solution: 

(a) Ni(OH)​

(s)    

(b) CaCO​

(s)   

(c) BaF​

(s)    

(d) AgCl(s)   

Practice Exercise 1 

Which of the following actions will increase the solubility of AgBr in water? (a) increasing  the pH, (b) decreasing the pH, (c) adding NaBr, (d) adding NaNO​

, (e) none of the above. 

  Formation of Complex Ions 

  Metals (esp transition metal ions) can act as lewis acids 

  Water molecules, and other small electron rich molecules (NH​

, CN​

), can act as lewis bases    THE FORMATION OF A COMPLEX ION CAN INCREASE THE SOLUBILITY OF A METAL 

CATION - YOU ONLY NEED TO KNOW THIS QUALITATIVELY 

  17.6 Precipitation and Separation of Ions 

  If Q = K​

, the system is at equilibrium which means that the solution is saturated, this is  the highest concentration the solution can have without precipitation 

  If Q< K​

, the reaction will proceed to the right towards the soluble ions; no precipitate will  form.  

  If Q> K​

, the reaction will proceed to the left, towards the solid; precipitate will form    Sample Exercise 17.16 - Does a precipitate form when 0.10 L of 8.0 × 10​

M Pb(NO​

)​

is  added to 0.40 L of 5.0 × 10​

M Na​

SO​

? 

Selective precipitation of Ions 

  Look at the figure below. Sulfide salts are often used to separate ions since their solubility  varies greatly. CuS has a K​

of 6 × 10​

while ZnS has a K​

of 2 × 10​

Sample Exercise 17.17 - A solution contains 1.0 × 10​

M Ag​

and 2.0 × 10​

M Pb​

. When Cl​

  is added, both AgCl(K​

= 1.8 × 10​

) and PbCl​

(K​

= 1.7 × 10​

) can precipitate. What 

concentration of Cl​

is necessary to begin the precipitation of each salt? Which salt  precipitates first? 

17.7 Qualitative Analysis for Metallic 

Elements