Exercises
Exercise 1. For the following salts, if they are dissolved in water, indicate if the solution is acidic, basic, or neutral.
CO32- is the anion of the weak acid HCO3–. It will react with water to produce hydroxide ion, therefore, the solution will be basic. The sodium ion will not react to any extent with water.
b) KHSO4
Potassium ion is the cation of the strong base KOH. It will not react to any extent with water. HSO4– is a weak acid and is the anion of sulfuric acid. The hydrogen sulfate ion will react with water to produce hydronium ion, therefore, the solution will be acidic.
c) CH3NH3F
CH3NH3+ is the cation of the weak base CH3NH2. It will react with water to form hydronium ion. The solution will be acidic.
d) CsBr
Cs is the cation of the strong base, CsOH. Br is the anion of the strong acid HBr. Neither will react with water to any extent. The solution will be neutral.
e) RbNO2
The Rb+ ion is the cation of a strong base and will not react with water. The NO2– ion is the anion of the weak acid HNO2. It will react with water to produce hydroxide ion. The solution will be basic.
Exercise 2. What is the pH of a solution that is 2.5 M NaCN? Ka (HCN) = 4.9 x 10-10
NaCN will dissociate into sodium and cyanide ions. The cyanide ions will react with water to produce OH– ion.
CN– (aq) + H2O (l) ⇄ HCN (aq) + OH– (aq) Kb = 4.9 x 10-10
Set up an ICE table
Assume x is neglible and 2.5 M – x ≅ 2.5 M
\(\displaystyle \frac{x^2}{2.5}\;=\;4.9\times 10^{-10}\)
\(\displaystyle x\;=\;\sqrt{2.5\times\;(4.9\times 10^{-10})}\;=\;3.5\times 10^{-5}\;M\)
Check assumption.
\(\displaystyle \frac{3.5\times 10^{-5}\;M}{2.5\;M}\times\;100\;=\;0.0014%\)
0.0014% < 5% -- our assumption is valid
[OH–] = 3.5 x 10-5 M.
pH = 14.00 – (-log(3.5 x 10-5) = 9.54
Exercise 3. Calculate the pH of a solution that is 0.5 M Ca(ClO4)2.
This solution will be neutral, pH = 7.00. The calcium ion is the cation of a strong base, Ca(OH)2. The ClO4– is the anion of the strong acid, HClO4. Neither ion will react to a great extent with water.
Exercise 4. What is the pH of an aqueous solution that contains 0.35 M NH4NO3?
The nitrate ion is the anion of a strong acid. The ammonium ion is the cation of a weak base and will undergo hydrolysis in water.
NH4+ (aq) + H2O (l) ⇄ NH3 (aq) + H3O+ (aq) Ka = 5.6 x 10-10
Set up an ICE table.
Assume x is neglible and 0.35 M – x ≅ 0.35 M
\(\displaystyle \frac{x^2}{0.35}\;=\;5.6\times 10^{-10}\)
\(\displaystyle x\;=\;\sqrt{0.35\times\;(5.6\times 10^{-10})}\;=\;1.4\times 10^{-5}\;M\)
Check assumption.
\(\displaystyle \frac{1.4\times 10^{-5}\;M}{0.35\;M}\times\;100\;=\;0.004%\)
0.004% < 5% -- our assumption is valid
[H3O+] = 1.4 x 10-5 M.
pH = -log(1.4 x 10-5) = 4.85
Exercise 5. Calculate the pH of a solution that is 0.15 M NaHSO4.
The HSO4– ion is a weak acid with Ka2 = 1.0 x 10-2. It will undergo a hydrolysis reaction with water.
HSO4– (aq) + H2O (l) ⇄ SO42- (aq) + H3O+ (aq) Ka = 5.6 x 10-10
Set up an ICE table.
Assume x is neglible and 0.15 M – x ≅ 0.15 M
\(\displaystyle \frac{x^2}{0.15}\;=\;1.2\times 10^{-2}\)
\(\displaystyle x\;=\;\sqrt{0.15\times\;(1.2\times 10^{-2})}\;=\;0.042\;M\)
Check assumption.
\(\displaystyle \frac{0.042\;M}{0.15\;M}\times\;100\;=\;28%\)
28% > 5% — our assumption is not valid and we need to solve quadratic formula.
\(\displaystyle \frac{x^2}{0.15\;-\;x}\;=\;1.2\times 10^{-2}\)
Collect like terms and solve for x.
x2 + (1.2 x 10-2x) – 0.0018 = 0
\(\displaystyle \frac{-b\pm\;\sqrt{b^2\;-\;4ac}}{2a}\)
\(\displaystyle \frac{-(1.2\times 10^{-2})\pm\;\sqrt{(1.2\times 10^{-2})^2\;-\;4\times\;1\times\;-0.0018}}{2\times\;1}\)
\(\displaystyle \frac{-1.2\times 10^{-2}\pm\;0.085697}{2}\)
x = 0.0368 M, x = -0488 M
[H3O+] = 0.037 M
pH = -log(0.037) = 1.43