Additional Problems: Answers
1. For each of the following aqueous solutions, calculate the pH:
a. Nitric acid (HNO3), 0.25M
For strong acids, assume essentially complete dissociation:
HNO3 (aq) + H2O(l)  H3O+ (aq) + NO3- (aq)
[H3O+] = 0.25 M ;
pH=-log[H3O+]= -log(0.25M) = 0.6
b. 0.01M NaOH
For strong bases, assume essentially complete dissociation:
NaOH(aq)  HO-(aq) + Na+(aq)
pH +pOH =14; pOH = -log[OH] = -log (0.01M) = 2; pH=14-2 = 12.
c. 0.1M HCl
strong acid, pH= -log (0.1M) = 1
d. 0.02 M ClCH2CO2H which is 10% dissociated
weak acid:
ClCH2CO2H (aq) + H2O(l)  H3O+ (aq) + ClCH2CO2-(aq)
Initial: 0.02M
1x10-7M
0M
Final: 0.02-0.002
0.002
0.002
pH=-log (0.002) = 2.69
e. 2M LiOH: strong base
pH +pOH =14; pOH = -log[OH] = -log (2M) = -0.3; pH=14-(-.3) = 14.3
2. Write ionization constant expressions for the following acids and bases:
a. NH4+
b.-CN
c. NO2d. HCOOH
+
a. NH4 (aq) + H2O (l) H3O+(aq) + NH3(aq)
Ka = [H3O+]eq[NH3]eq / [NH4+]eq
b. CN-(aq) + H2O (l)  HCN (aq) + HO-(aq)
Kb = [HCN]eq[HO-]eq / [CN-]eq
c. NO2-(aq) + H2O (l)  HNO2 (aq) + HO- (aq)
Kb = [HNO2]eq[HO-]eq/ [NO2-]eq
d. HCOOH (aq) + H2O (l)  H3O+ (aq) + HCOO- (aq)
Ka = [H3O+]eq[HCOO-]eq/[HCOOH]eq
3. Calculate the [H3O+] and [-OH] and the pH at equilibrium that results when 0.01
mol of each of the following acids is added to sufficient water to produce a 1L
solution (treat this problem as you would any typical equilibrium problem):
(a) HOCl (Ka= 3.7x10-8)
First, write out the dissolution equation:
HOCl(aq) + H2O(l)  H3O+(aq) + ClO-(aq)
Initial
0.01 M
1x10-7M
0M
-7
Equilibrium: 0.01 M-x
1x10 M+ x x
Since Ka = [H3O+]eq[ClO-]eq / [HOCl]eq = 3.7 x10-8 = (1x10-7 + x)(x) / (0.01M-x)
then we have:
3.7 x10-8 = (1x10-7x + x2) / (0.01 mol-x); x2 + 1.37x10-7x – 3.7 x10-10 = 0 ; use
quadratic
x=1.92 x10-5 mol
For 1 L, [HOCl] =0.01 mol-x = 0.0099 M
[H3O+] = 1x10-7 + x = 1.93 x 10-5 M; pH = 4.71; pOH = 9.28;
[OH]= 5.2 x10-10 M
[ClO-] = x = 1.92 x 10-5 M
(b) HCN (Ka = 7.2 x 10-10)
First, write out the dissolution equation:
HCN(aq) + H2O(l)  H3O+(aq) + CN-(aq)
Initial:
0.01 M
1x10-7M
0M
Equilibrium: 0.01 M-x
1x10-7 M +x x
Since: Ka = [H3O+]eq[CN-]eq / [HCN]eq = 7.2 x 10-10 = (1x10-7 + x)(x) / (0.01 M-x)
Since Ka<<1x10-7 , we cannot assume x>> 1x10-7, thus:
7.2 x10-12 – 7.2x10-10x = 1x10-7x + x2; x2 + 1x10-7x –7.2 x10-12 = 0
using quadratic, x=2.63x10-6 M
thus [H3O+] =1x10-7 mol+x = 2.73 x10-6 M; pH=5.56; [OH-]= 3.6x10-9 M
[CN-] = x = 2.63 x10-6 M
[HCN] = 0.01 mol-x = .0099 M
4. Calculate the [H3O+] and [-OH] and the pH at equilibrium that results when 0.15
mol of each of the following bases is added to sufficient water to produce a 1L
solution (treat this problem as you would any typical equilibrium problem):
(a) NH3 (Kb= 1.8x10-5)
NH3 (aq) + H2O(l)  NH4+ (aq) + HO- (aq)
Initial:
0.15M
0
1x10-7 M
Equilibrium: 0.15-x
x
1x10-7 +x
Kb = [NH4+]eq[HO-]eq/[NH3]eq
Kb= 1.8 x10-5 = (x)(1x10-7+x) / (0.15 –x)
2.7x10-6 – 1.8x10-5x = 1x10-7x + x2; x2 + 1.81x10-5 x – 2.7 x10-6 = 0
quadratic: x= .0016 M
[NH4+] = x = 0.0016 M
[HO-] = 1x10-7 +x = 0.0016M; [H3O+] = 6.25 x 10-12M; pH= 11.2
[NH3] = 0.15-x = 0.1484 M
(b) (CH3)3N (Kb = 6.3 x 10-5)
5. The dissociation constant for formic acid (HCOOH) at 25°C is 1.8 x 10-4
(a) For a 0.1 M solution, calculate the percentage of the acid that is dissociated
into ions.
Assume 1L of solution:
HCOOH (aq) + H2O(l)  H3O+(aq) + HCOO-(aq)
Initial:
0.1M
1x10-7M
0M
Eq.
0.1 – x
1x10-7 +x
x
-4
-7
Ka= 1.8 x 10 = (x)(1x10 +x) / (0.1 – x )
Assume x>>1x10-7mol, then (x)2/(0.1 –x) = 1.8 x10-4; x2 = 1.8x10-5 – 1.8x10-4x
x2 + 1.8x10-4x – 1.8x10-5 = 0 ; quadratic: x=.00415
percentage of acid dissociated = x/0.1 *100% = 4.15%
(b) Calculate the pH of a water sample containing 6.32 g of formic acid per liter
Molar mass of HCOOH = 46 g/mol; 6.32 g / 46g/mol = 0.137 mol
HCOOH (aq) + H2O(l)  H3O+(aq) + HCOO-(aq)
Initial:
0.137 M
1x10-7M
0M
Eq.:
0.137-x
x
x
-4
2
Ka= 1.8 x10 = x / (0.137-x); 2.466x10-5 – 1.8x10-4x = x2
x2 + 1.8x10-4x – 2.46 x10-5 = 0. quadratic: x=0.0048
[H3O+] = x = 0.0048 M; pH = 2.3
6. Calculate the dissociation constant at 25°C for nitrous acid (HNO2) in a 0.1M
solution if it is known to be 6.3% dissociated.
Init:
Eq.:
6.3% of 0.1M = 0.1M x 0.063 = 0.0063M
HNO2 (aq) + H2O(l)  H3O+ (aq) + NO2-(aq)
0.1M
1x10-7 M
0M
0.0937M
.0063 M
.0063M
Ka = [H3O+]eq[ NO2-]eq / [HNO2]eq = (0.0063M)2 / (0.0937M) = 4.2 x10-4M