SCH4U
Hydrolysis of Ions
1. Sodium phosphate, Na3PO4, is a common cleaning agent for floors and walls.
a) Will a solution of sodium phosphate be acidic, basic or neutral?
PO43-:
Kb = 1.0 x 10-14/4.2 x 10-13 = 0.0238 (Ka3 from HPO42-)
Kb is quite large so that the solution will be basic
b) Calculate the pH of 0.20M sodium phosphate.
PO43- + H2O HPO42- + OH[I] 0.20
[C] -x
[E] 0.20 – x
0
+x
x
0.0238 = x2/(0.20 – x)
0.0238(0.20 – x)= x2
X2 + 0.0238x - 0.00476 = 0
X = 0.058
[OH-] = 0.058M
pOH = 1.2
pH = 12.8
0
+x
x
2. Calculate the pH of a solution made by dissolving 8.20 g of NaCH3CO2 in
water to a final volume of 500.0 mL.
nNaCH3O2- = 8.20 g/82.034 = 0.10 mol
[CH3CO2-] = 0.10 mol/0.500L = 0.20M
CH3CO2- + H2O [I] 0.20
[C] -x
[E] 0.20 – x
CH3CO2H
0
+x
x
+ OH0
+x
x
Kb = 1.0 x 10-14/ 1.8 x 10-5 = 5.6 x 10-10
5.6 x 10-10 = x2/0.20
X2 = 1.12 x 10-10
X = 1.1 x 10-5
[OH] = 1.1 x 10-5 M
pOH = 5.0
pH = 9.0
*insignificant
3. How many grams of potassium nitrite, KNO2, must be dissolved in 1.0 L of
water to make a solution with a pH of 8.8?
Kb = 1.0 x 10-14/7.2 x 10-4 = 1.39 x 10-11
pH = 8.8 therefore pOH = 5.2 [OH-] = 6.3 x 10-6 M
NO2- +
H2O [I] 0.20
[C] -x
[E] 0.20 – 6.3 x 10-6
HNO2
+
0
+6.3 x 10-6
6.3 x 10-6
*insignificant
1.4 x 10-11 = (6.3 x 10-6)2/x
x = 2.8
Therefore [KNO2] = 2.8 mol/L
mKNO3 = 2.8 mol/L x 1.0L x 85.11 g/mol
= 238 g
OH0
+6.3 x 10-6
6.3 x 10-6
4. Calculate the pH of a 0.15M solution of a) NH4NO3
a) NH4+ / NO3Ka NH4+ = 10-14/(1.8 x 10-5) = 5.6 x 10-10
Kb NO3- = 10-14/ very large value = a value less than 10-14
therefore negligible
Since Ka for NH4+ > Kb for NO3- , NH4NO3 will be acidic
NH4+ [I] 0.15
[C] -x
[E] 0.15 – x
NH3
0
+x
x
5.6 x 10-10 = x2/0.15
X = 9.2 x 10-6 = [H+]
pH = 5.0
+ H+
0
+x
x
b) NaHC2O4
Na+/ HC2O4Na+ can’t act as an acid
HC2O4- is amphoteric – it can act as an acid or base – which
will it do to the greatest extent?
As an acid: HC2O4- H+ + C2O42- Ka = Ka2 = 5.4 x 10-5
As a base: HC2O4- + H2O H2C2O4 + OHKb = 10-14/Ka1 = 10-14/5.4 x 10-2 = 1.9 x 10-13
Ka >> Kb therefore it will act as an acid
HC2O4- H+ + C2O42[I]
0.15
0
0
[C]
-x
+x
+x
[E]
0.15 – x
x
x
-5
2
5.4 x 10 = x / 0.15
X = 2.8 x 10-3 = [H+]
pH = 2.5
Ka 5.4 x 10-5
5. 25.0 mL of 0.10M acetic acid is titrated with 25.0 mL of 0.10M NaOH.
a) Write the chemical equation for the titration reaction.
CH3CO2H + NaOH NaCH3CO2 + H2O
b) Calculate the concentration of the resulting solution of the salt, sodium
acetate.
n NaCH3CO2 = nNaOH = 0.10 mol/L x 0.025L = 0.0025 mol
Therefore nCH3CO2- = 0.0025 mol
(1:1)
CNaCH3CO2 = 0.0025 mol/0.050L = 0.050 M
c) Calculate the pH of the resulting solution.
CH3CO2- + H2O CH3CO2H + OH[I] 0.050
[C] -x
[E] 0.20 – x
5.6 x 10-10
0
+x
x
= x2/0.050
X = 5.3 x 10-6
0
+x
x
insignificant
[OH-] = 5.3 x 10-6 M
pOH = 5.3
pH = 8.7
Common Ion
6. A solution is made by dissolving 6.8 g of NH3 and 10.7 g of NH4Cl in water
with a final volume of 250 mL. The Kb for NH3 is 1.8 x 10-5. Calculate the OHconcentration of the solution.
The common ion will be NH4+ (from ionization of NH3 and from the salt NH4Cl)
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
nNH3 = 6.8 g/17.034 = 0.40 mol
[NH3] = 0.40mol/0.250L = 1.6M
nNH4Cl = 10.7g/53.49 = 0.20mol
[NH4+] = 0.20 mol/0.250L = 0.80M
[I]
[C]
[E]
NH3 + H2O 1.6
-x
1.6 – x
1.8 x 10-5
= 0.80x/1.6
X = 3.6 x 10-5
[OH-] =3.6 x 10-5 M
NH4+ + OH0.80
0
+x
+x
0.80 + x x
Kb = 1.8 x 10-5
insignificant
7. a) Calculate the pH of a 0.60M solution of acetic acid.
CH3CO2H CH3CO2- + H+
[I]
[C]
[E]
0.60
-x
0.60 – x
0
+x
x
0
+x
x
insignificant
1.8 x 10-5 = x2/0.60
X = 0.0033 = [H+]
pH = 2.5
b) Determine
the mass of sodium acetate that must be dissolved in 1.0 L of
0.60M acetic acid to bring the pH to 6.0.
pH = 6.0 [H+] = 10-6
CH3CO2H CH3CO2- + H+
[I]
[C]
0.60
- 10-6
x
+10-6
0
+10-6
[E]
0.60 - 10-6
x + 10-6
10-6
insignificant
1.8 x 10-5 = 10-5x /0.60
X = 10.8 = [CH3CO2-]
MCH3CO2- = 10.8 mol/L x 1.0L x 82.03 g/mol = 886g
8. How many grams of sodium formate, NaCHO2, would have to be dissolved in
1.0 L of 0.12M formic acid to make a solution with a pH of 3.80. Ka for formic
acid is 1.8 x 10-4.
[I]
[C]
[E]
HCHO2 0.12
- 1.6 x10-4
0.12 – 1.6 x 10-4
CH2O- + H+
x
0
-4
+1.6 x 10
+1.6 x 10-4
x +1.6 x 10-4 1.6 x 10-4
1.8 x 10-4 = x(1.6 x 10-4)/0.12
x = 0.14 mol/L = [CH2O-]
m = 0.14mol/L x 1.0 L x 68.01 g/mol = 9.5 g