Chapter 17: Acid-Base Equilibria and Solubility Equilibria
Chapter 17
Acid-Base Equilibria and Solubility Equilibria
17.5 / a. / This is a weak acid problem. Setting up the standard equilibrium table:CH3COOH(aq) / / H+(aq) / + CH3COO(aq)
Initial (M): / 0.40 / 0.00 / 0.00
Change (M): / x / +x / +x
Equilibrium (M): / (0.40 x) / x / x
x = [H+] = 2.7 103 M
pH = 2.57
b. / In addition to the acetate ion formed from the ionization of acetic acid, we also have acetate ion formed from the sodium acetate dissolving.
CH3COONa(aq) CH3COO(aq) + Na+(aq)
Dissolving 0.20 M sodium acetate initially produces 0.20 M CH3COO and 0.20 M Na+. The sodium ions are not involved in any further equilibrium (why?), but the acetate ions must be added to the equilibrium in part (a).
CH3COOH(aq) / / H+(aq) / + CH3COO(aq)
Initial (M): / 0.40 / 0.00 / 0.20
Change (M): / x / +x / +x
Equilibrium (M): / (0.40 x) / x / (0.20 x)
x = [H+] = 3.6 105 M
pH = 4.44
Think About It: / Could you have predicted whether the pH should have increased or decreased after the addition of the sodium acetate to the pure 0.40 M acetic acid in part (a)?
An alternate way to work part (b) of this problem is to use the Henderson-Hasselbalch equation.
17.6 / a. / This is a weak base problem.
/ / / /
Initial (M): / 0.20 / 0 / 0
Change (M): / x / +x / +x
Equilibrium (M): / 0.20 x / +x / +x
x = 1.9 103 M = [OH]
pOH = 2.72
pH = 11.28
b. / The initial concentration of NH is 0.30 M from the salt NH4Cl. We set up a table as in part (a).
/ / / /
Initial (M): / 0.20 / 0.30 / 0
Change (M): / x / +x / +x
Equilibrium (M): / 0.20 x / 0.30 + x / +x
x = 1.2 105 M = [OH]
pOH = 4.92
pH = 9.08
Alternatively, we could use the Henderson-Hasselbalch equation to solve this problem. We get the value of Ka for the ammonium ion using the Kb for ammonia (Table 16.7) and Equation 16.8. Substituting into the Henderson-Hasselbalch equation gives:
pH = 9.25 0.18 = 9.07
Think About It: / Is there any difference in the Henderson-Hasselbalch equation in the cases of a weak acid and its conjugate base and a weak base and its conjugate acid?
17.9 / Strategy: / What constitutes a buffer system? Which of the solutions described in the problem contains a weak acid and its salt (containing the weak conjugate base)? Which contains a weak base and its salt (containing the weak conjugate acid)? Why is the conjugate base of a strong acid not able to neutralize an added acid?
Solution: / The criteria for a buffer system are that we must have a weak acid and its salt (containing the weak conjugate base) or a weak base and its salt (containing the weak conjugate acid).
a. / HCl (hydrochloric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak base. Therefore, this is not a buffer system.
b. / H2SO4 (sulfuric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak base. Therefore, this is not a buffer system.
c. / This solution contains both a weak acid, H2PO and its conjugate base, HPO. Therefore, this is a buffer system.
d. / HNO2 (nitrous acid) is a weak acid, and its conjugate base, NO (nitrite ion, the anion of the salt KNO2), is a weak base. Therefore, this is a buffer system.
Only (c) and (d) are buffer systems.
17.10 / a. / HCN is a weak acid, and its conjugate base, CN, is a weak base. Therefore, this is a buffer system.
b. / HSO is a weak acid, and its conjugate base, SO is a weak base. Therefore, this is a buffer system.
c. / NH3 (ammonia) is a weak base, and its conjugate acid, NH is a weak acid. Therefore, this is a buffer system.
d. / Because HI is a strong acid, its conjugate base, I, is an extremely weak base. This means that the I ion will not combine with a H+ ion in solution to form HI. Thus, this system cannot act as a buffer system.
17.11 / Strategy: / The pH of a buffer system can be calculated using the Henderson-Hasselbalch equation (Equation 17.3). The Ka of a conjugate acid such as NH is calculated using the Kb of its weak base (in this case, NH3) and Equation 16.8.
Solution: / NH(aq) NH3(aq) + H+(aq)
Ka = Kw /1.8 105 = 5.56 1010
pKa = log Ka = 9.26
pH = 8.89
17.12 / a. / We summarize the concentrations of the species at equilibrium as follows:
CH3COOH(aq) / / H+(aq) / + CH3COO(aq)
Initial (M): / 2.0 / 0 / 2.0
Change (M): / x / +x / +x
Equilibrium (M): / 2.0 x / x / 2.0 + x
Ka = [H+]
Taking the log of both sides,
pKa = pH
Thus, for a buffer system in which the [weak acid] = [weak base],
pH = pKa
pH = log(1.8 105) = 4.74
b. / Similar to part (a),
pH = pKa = 4.74
Buffer (a) will be a more effective buffer because the concentrations of acid and base components are ten times higher than those in (b). Thus, buffer (a) can neutralize 10 times more added acid or base compared to buffer (b).
17.13 / H2CO3(aq) HCO(aq) + H+(aq)
17.14 / Step 1:Write the equilibrium that occurs between H2PO and HPO. Set up a table relating the initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations.
/ / /
Initial (M): / 0.15 / 0 / 0.10
Change (M): / x / +x / +x
Equilibrium (M): / 0.15 x / x / 0.10 + x
Step 2:Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (Ka), solve for x.
You can look up the Ka value for dihydrogen phosphate (Ka2 for phosphoric acid) in Table 16.8 of your text.
x = [H+] = 9.3 108 M
Step 3:Having solved for the [H+], calculate the pH of the solution.
pH = log[H+] = log(9.3 108) = 7.03
17.15 / Using the HendersonHasselbalch equation:
Thus,
17.16 / We can use the Henderson-Hasselbalch equation to calculate the ratio [HCO3]/[H2CO3]. The Henderson-Hasselbalch equation is:
For the buffer system of interest, HCO3 is the conjugate base of the acid, H2CO3. We can write:
The [conjugate base]/[acid] ratio is:
The buffer should be more effective against an added acid because ten times more base is present compared to acid. Note that a pH of 7.40 is only a two significant figure number (Why?); the final result should only have two significant figures.
17.17 / For the first part we use Ka for ammonium ion. (Why?) The HendersonHasselbalch equation is
For the second part, the acidbase reaction is
NH3(g) + H+(aq) NH(aq)
We find the number of moles of HCl added
The number of moles of NH3 and NH4+ originally present are
Using the acid-base reaction, we find the number of moles of NH3 and NH after addition of the HCl.
/ / /
Initial (mol): / 0.013 / 0.0010 / 0.013
Change (mol): / 0.0010 / 0.0010 / +0.0010
Final (mol): / 0.012 / 0 / 0.014
We find the new pH:
17.18 / As calculated in Problem 17.12, the pH of this buffer system is equal to pKa.
pH = pKa = log(1.8 105) = 4.74
a. / The added NaOH will react completely with the acid component of the buffer, CH3COOH. NaOH ionizes completely; therefore, 0.080 mol of OH are added to the buffer.
Step 1:The neutralization reaction is:
CH3COOH(aq) / + OH(aq) / → / CH3COO(aq) / + H2O(l)
Initial (mol): / 1.00 / 0.080 / 1.00
Change (mol): / 0.080 / 0.080 / +0.080
Final (mol): / 0.92 / 0 / 1.08
Step 2:Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration.
CH3COOH(aq) / / H+(aq) / + CH3COO(aq)
Initial (M): / 0.92 / 0 / 1.08
Change (M): / x / +x / +x
Equilibrium (M): / 0.92 x / x / 1.08 + x
Write the Ka expression, then solve for x.
x = [H+] = 1.5 105 M
Step 3:Having solved for the [H+], calculate the pH of the solution.
pH = log[H+] = log(1.5 105) = 4.82
The pH of the buffer increased from 4.74 to 4.82 upon addition of 0.080 mol of strong base.
b. / The added acid will react completely with the base component of the buffer, CH3COO. HCl ionizes completely; therefore, 0.12 mol of H+ ion are added to the buffer
Step 1:The neutralization reaction is:
CH3COO(aq) / + H+(aq) / → / CH3COOH(aq)
Initial (mol): / 1.00 / 0.12 / 1.00
Change (mol): / 0.12 / 0.12 / +0.12
Final (mol): / 0.88 / 0 / 1.12
Step 2:Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration.
Step 2:Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration.
CH3COOH(aq) / / H+(aq) / + CH3COO(aq)
Initial (M): / 1.12 / 0 / 0.88
Change (M): / x / +x / +x
Equilibrium (M): / 1.12 x / x / 0.88 + x
Write the Ka expression, then solve for x.
x = [H+] = 2.3 105 M
Step 3:Having solved for the [H+], calculate the pH of the solution.
pH = log[H+] = log(2.3 105) = 4.64
The pH of the buffer decreased from 4.74 to 4.64 upon addition of 0.12 mol of strong acid.
17.19 / In order for the buffer solution to function effectively, the pKa of the acid component must be close to the desired pH. We write
Therefore, the proper buffer system is Na2A/NaHA.
17.20 / For a buffer to function effectively, the concentration of the acid component must be roughly equal to the conjugate base component. According to Equation 17.3 of the text, this occurs when the desired pH is close to the pKa of the acid, that is, when pH pKa,
or
To prepare a solution of a desired pH, we should choose a weak acid with a pKa value close to the desired pH. Calculating the pKa for each acid:
For HA,pKa = log(2.7 103) = 2.57
For HB,pKa = log(4.4 106) = 5.36
For HC,pKa = log(2.6 109) = 8.59
The buffer solution with a pKa closest to the desired pH is HC. Thus, HC is the best choice to prepare a buffer solution with pH = 8.60.
17.21 / 1. / In order to function as a buffer, a solution must contain species that will consume both acid and base. Solution (a) contains HA, which can consume either acid or base:
HA + H+ H2A
HA + OH A2 + H2O
and A2, which can consume acid:
A2 + H+ HA
Solution (b) contains HA, which can consume either acid or base, and H2A, which can consume base:
H2A + OH HA + H2O
Solution (c) contains only H2A and consequently can consume base but not acid. Solution (c) cannot function as a buffer.
Solution (d) contains HA and A2.
Solutions (a), (b), and (c) can function as buffers.
2. / Solution (a) should be the most effective buffer because it has the highest concentrations of acid- and base-consuming species.
17.22 / 1. / Solution (d) has the lowest pH because it has the highest ratio of HA to A (6:4). Solution (a) has the highest pH because it has the lowest ratio of HA to A (3:5).
2. / After addition of two H+ ions to solution (a), there will be two species present: HA and A.
3. / After addition of two OH ions to solution (b), there will be two species present: HA and A.
17.27 / Since the acid is monoprotic, the number of moles of KOH is equal to the number of moles of acid.
17.28 / We want to calculate the molar mass of the diprotic acid. The mass of the acid is given in the problem, so we need to find moles of acid in order to calculate its molar mass.
The neutralization reaction is:
2KOH(aq) + H2A(aq) K2A(aq) + 2H2O(l)
From the volume and molarity of the base needed to neutralize the acid, we can calculate the number of moles of H2A reacted.
We know that 0.500 g of the diprotic acid were reacted (1/10 of the 250 mL was tested). Divide the number of grams by the number of moles to calculate the molar mass.
17.29 / The neutralization reaction is:
H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)
Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write:
17.30 / We want to calculate the molarity of the Ba(OH)2 solution. The volume of the solution is given (19.3 mL), so we need to find the moles of Ba(OH)2 to calculate the molarity.
The neutralization reaction is:
2HCOOH + Ba(OH)2 (HCOO)2Ba + 2H2O
From the volume and molarity of HCOOH needed to neutralize Ba(OH)2, we can determine the moles of Ba(OH)2 reacted.
The molarity of the Ba(OH)2 solution is:
17.31 / a. / Since the acid is monoprotic, the moles of acid equals the moles of base added.
HA(aq) + NaOH(aq) NaA(aq) + H2O(l)
We know the mass of the unknown acid in grams and the number of moles of the unknown acid.
b. / The number of moles of NaOH in 10.0 mL of solution is
The neutralization reaction is:
HA(aq) / + NaOH(aq) / → / NaA(aq) / + H2O(l)
Initial (mol): / 0.00116 / 6.33 104 / 0
Change (mol): / 6.33 104 / 6.33 104 / +6.33 104
Final (mol): / 5.3 104 / 0 / 6.33 104
Now, the weak acid equilibrium will be reestablished. The total volume of solution is 35.0 mL.
We can calculate the [H+] from the pH.
[H+] = 10pH = 105.87 = 1.35 106 M
HA(aq) / / H+(aq) + / A(aq)
Initial (M): / 0.015 / 0 / 0.0181
Change (M): / 1.35 106 / +1.35 106 / +1.35 106
Equilibrium (M): / 0.015 / 1.35 106 / 0.0181
Substitute the equilibrium concentrations into the equilibrium constant expression to solve for Ka.
17.32 / The resulting solution is not a buffer system. There is excess NaOH and the neutralization is well past the equivalence point.
CH3COOH(aq) / + NaOH(aq) / → / CH3COONa(aq) / + H2O(l)
Initial (mol): / 0.0500 / 0.0835 / 0
Change (mol): / 0.0500 / 0.0500 / +0.0500
Final (mol): / 0 / 0.0335 / 0.0500
The volume of the resulting solution is 1.00 L (500 mL + 500 mL = 1000 mL).
CH3COO(aq) / + H2O(l) / / CH3COOH(aq) / + OH(aq)
Initial (M): / 0.0500 / 0 / 0.0335
Change (M): / x / +x / +x
Equilibrium (M): / 0.0500 x / x / 0.0335 + x
x = [CH3COOH] = 8.4 1010 M
17.33 / HCl(aq) + CH3NH2(aq) CH3NH(aq) + Cl(aq)
Since the concentrations of acid and base are equal, equal volumes of each solution will need to be added to reach the equivalence point. Therefore, the solution volume is doubled at the equivalence point, and the concentration of the conjugate acid from the salt, CH3NH3+, is:
The conjugate acid undergoes hydrolysis.
/ / / /
Initial (M): / 0.10 / 0 / 0
Change (M): / x / +x / +x
Equilibrium (M): / 0.10 x / x / x
Assuming that, 0.10 x 0.10
x = [H3O+] = 1.5 106 M
pH = 5.82
17.34 / Let's assume we react 1 L of HCOOH with 1 L of NaOH.
HCOOH(aq) / + NaOH(aq) / / HCOONa(aq) / + H2O(l)
Initial (mol): / 0.10 / 0.10 / 0
Change (mol): / 0.10 / 0.10 / +0.10
Final (mol): / 0 / 0 / 0.10
The solution volume has doubled (1 L + 1 L = 2 L). The concentration of HCOONa is:
HCOO(aq) is a weak base. The hydrolysis is:
HCOO(aq) / + H2O(l) / → / HCOOH(aq) / + OH(aq)
Initial (M): / 0.050 / 0 / 0
Change (M): / x / +x / +x
Equilibrium (M): / 0.050 x / x / x
x = 1.7 106 M = [OH]
pOH = 5.77
pH = 8.23
17.35 / The reaction between CH3COOH and KOH is:
CH3COOH(aq) + KOH(aq) CH3COOK(aq) + H2O(l)
We see that 1 mole CH3COOH is stoichiometrically equivalent to 1 mol KOH. Therefore, at every stage of titration, we can calculate the number of moles of acid reacting with base, and the pH of the solution is determined by the excess acid or base left over. At the equivalence point, however, the neutralization is complete, and the pH of the solution will depend on the extent of the hydrolysis of the salt formed, which is CH3COOK.
a. / No KOH has been added. This is a weak acid calculation.
CH3COOH(aq) / + H2O(l) / / H3O+(aq) / + CH3COO(aq)
Initial (M): / 0.100 / 0 / 0
Change (M): / x / +x / +x
Equilibrium (M): / 0.100 x / x / x
x = 1.34 103 M = [H3O+]
pH = 2.87
b. / The number of moles of CH3COOH originally present in 25.0 mL of solution is:
The number of moles of KOH in 5.0 mL is:
We work with moles at this point because when two solutions are mixed, the solution volume increases. As the solution volume increases, molarity will change, but the number of moles will remain the same. The changes in number of moles are summarized.
CH3COOH(aq) / + KOH(aq) / → / CH3COOK(aq) / + H2O(l)
Initial (mol): / 2.50 103 / 1.00 103 / 0
Change (mol): / 1.00 103 / 1.00 103 / +1.00 103
Final (mol): / 1.50 103 / 0 / 1.00 103
At this stage, we have a buffer system made up of CH3COOH and CH3COO (from the salt, CH3COOK). We use the Henderson-Hasselbalch equation to calculate the pH.
pH = 4.56
c. / This part is solved similarly to part (b).
The number of moles of KOH in 10.0 mL is:
The changes in number of moles are summarized.
CH3COOH(aq) / + KOH(aq) / → / CH3COOK(aq) / + H2O(l)
Initial (mol): / 2.50 103 / 2.00 103 / 0
Change (mol): / 2.00 103 / 2.00 103 / +2.00 103
Final (mol): / 0.50 103 / 0 / 2.00 103
At this stage, we have a buffer system made up of CH3COOH and CH3COO (from the salt, CH3COOK). We use the Henderson-Hasselbalch equation to calculate the pH.
pH = 5.34
d. / We have reached the equivalence point of the titration. 2.50 103 mole of CH3COOH reacts with
2.50 103 mole KOH to produce 2.50 103 mole of CH3COOK. The only major species present in solution at the equivalence point is the salt, CH3COOK, which contains the conjugate base, CH3COO. Let's calculate the molarity of CH3COO. The volume of the solution is: (25.0 mL + 12.5 mL = 37.5 mL = 0.0375 L).
We set up the hydrolysis of CH3COO, which is a weak base.
CH3COO(aq) / + H2O(l) / / CH3COOH(aq) / + OH(aq)
Initial (M): / 0.0667 / 0 / 0
Change (M): / x / +x / +x
Equilibrium (M): / 0.0667 x / x / x
x = 6.09 106 M = [OH]
pOH = 5.22
pH = 8.78
e. / We have passed the equivalence point of the titration. The excess strong base, KOH, will determine the pH at this point. The moles of KOH in 15.0 mL are:
The changes in number of moles are summarized.
CH3COOH(aq) / + KOH(aq) / → / CH3COOK(aq) / + H2O(l)
Initial (mol): / 2.50 103 / 3.00 103 / 0
Change (mol): / 2.50 103 / 2.50 103 / +2.50 103
Final (mol): / 0 / 0.50 103 / 2.50 103
Let's calculate the molarity of the KOH in solution. The volume of the solution is now 40.0 mL =
0.0400 L.
KOH is a strong base. The pOH is:
pOH = log(0.0125) = 1.90
pH = 12.10
17.36 / The reaction between NH3 and HCl is:
NH3(aq) + HCl(aq) NH4Cl(aq)
We see that 1 mole NH3 is stoichiometrically equivalent to 1 mol HCl. Therefore, at every stage of titration, we can calculate the number of moles of base reacting with acid, and the pH of the solution is determined by the excess base or acid left over. At the equivalence point, however, the neutralization is complete, and the pH of the solution will depend on the extent of the hydrolysis of the salt formed, which is NH4Cl.
a. / No HCl has been added. This is a weak base problem.
x = 2.3 103 M = [OH]
pOH = 2.64
pH = 11.36
b. / The number of moles of NH3 originally present in 10.0 mL of solution is:
The number of moles of HCl in 10.0 mL is:
We work with moles at this point because when two solutions are mixed, the solution volume increases. As the solution volume increases, molarity will change, but the number of moles will remain the same. The changes in number of moles are summarized.
NH3(aq) + / HCl(aq) / → / NH4Cl(aq)
Initial (mol): / 3.00 103 / 1.00 103 / 0
Change (mol): / 1.00 103 / 1.00 103 / +1.00 103
Final (mol): / 2.00 103 / 0 / 1.00 103
NH3(aq) + HCl(aq) NH4Cl(aq)
At this stage, we have a buffer system made up of NH3 and NH (from the salt, NH4Cl). We use the Henderson-Hasselbalch equation to calculate the pH.
pH = 9.55
c. / This part is solved similarly to part (b).
The number of moles of HCl in 20.0 mL is:
The changes in number of moles are summarized.
NH3(aq) + / HCl(aq) / → / NH4Cl(aq)
Initial (mol): / 3.00 103 / 2.00 103 / 0
Change (mol): / 2.00 103 / 2.00 103 / +2.00 103
Final (mol): / 1.00 103 / 0 / 2.00 103
At this stage, we have a buffer system made up of NH3 and NH (from the salt, NH4Cl). We use the Henderson-Hasselbalch equation to calculate the pH.
pH = 8.95
d. / We have reached the equivalence point of the titration. 3.00 103 mole of NH3 reacts with
3.00 103 mole HCl to produce 3.00 103 mole of NH4Cl. The only major species present in solution at the equivalence point is the salt, NH4Cl, which contains the conjugate acid, NH. Let's calculate the molarity of NH. The volume of the solution is: (10.0 mL + 30.0 mL = 40.0 mL = 0.0400 L).
We set up the hydrolysis of NH, which is a weak acid.
/ / / /
Initial (M): / 0.0750 / 0 / 0
Change (M): / x / +x / +x
Equilibrium (M): / 0.0750 x / x / x
x = 6.5 106 M = [H3O+]
pH = 5.19
e. / We have passed the equivalence point of the titration. The excess strong acid, HCl, will determine the pH at this point. The moles of HCl in 40.0 mL are:
The changes in number of moles are summarized.
NH3(aq) + / HCl(aq) / → / NH4Cl(aq)
Initial (mol): / 3.00 103 / 4.00 103 / 0
Change (mol): / 3.00 103 / 3.00 103 / +3.00 103
Final (mol): / 0 / 1.00 103 / 3.00 103
Let's calculate the molarity of the HCl in solution. The volume of the solution is now 50.0 mL = 0.0500 L.
HCl is a strong acid. The pH is:
pH = log(0.0200) = 1.70
17.37 / a. / HCOOH is a weak acid and NaOH is a strong base. Suitable indicators are cresol red and phenolphthalein.
b. / HCl is a strong acid and KOH is a strong base. Most of the indicators in Table 17.3 are suitable for a strong acid-strong base titration. Exceptions are thymol blue and, to a lesser extent, bromophenol blue and methyl orange.
c. / HNO3 is a strong acid and CH3NH2 is a weak base. Suitable indicators are bromophenol blue, methyl orange, methyl red, and chlorophenol blue.
17.38 / CO2 in the air dissolves in the solution:
CO2 + H2O H2CO3
The carbonic acid neutralizes the NaOH, lowering the pH. When the pH is lowered, the phenolphthalein indicator returns to its colorless form.
17.39 / The weak acid equilibrium is
HIn(aq) H+(aq) + In(aq)
We can write a Ka expression for this equilibrium.
Rearranging,
From the pH, we can calculate the H+ concentration.
[H+] = 10pH = 104 = 1.0 104 M
Since the concentration of HIn is 100 times greater than the concentration of In, the color of the solution will be that of HIn, the nonionized formed. The color of the solution will be red.
17.40 / When [HIn] [In] the indicator color is a mixture of the colors of HIn and In. In other words, the indicator color changes at this point. When [HIn] [In] we can write:
[H+] = Ka = 2.0 106
pH = 5.70
17.41 / 1. / Diagram (c)
2. / Diagram (b)
3. / Diagram (d)
4. / Diagram (a). The pH at the equivalence point is below 7 (acidic) because the species in solution at the equivalence point is the conjugate acid of a weak base.
17.42 / 1. / Diagram (b)
2. / Diagram (d)
3. / Diagram (a)
4. / Diagram (c). The pH at the equivalence point is above 7 (basic) because the species in solution at the equivalence point is the conjugate base of a weak acid.
17.49 / a. / The solubility equilibrium is given by the equation
AgI(s) Ag+(aq) + I(aq)
The expression for Ksp is given by
Ksp = [Ag+][I]
The value of Ksp can be found in Table 17.4 of the text. If the equilibrium concentration of silver ion is the value given, the concentration of iodide ion must be