Exercises
Exercise 1. What is the pH of aqueous 0.25 M nitrous acid, HNO2? Ka = 4.5 x 10-4.
First, write the equation and the Ka expression.
HNO2 (aq) + H2O (l) ⇄ NO2– + H3O+ (aq)
\(\displaystyle K_a\;=\;\frac{[\mathrm{NO_2^-}][\mathrm{H_3O^+}]}{[\mathrm{HNO_2}]}\;=\;4.5\times 10^{-4}\)
Set up an ICE table.
Put the equilibrium concentrations into the Ka expression.
\(\displaystyle \frac{(x)(x)}{0.25\;-\;x}\;=\;4.5\times 10^{-4}\)
Make the assumption x is very small. 0.25 M – x ≅ 0.25 M
\(\displaystyle \frac{x^2}{0.25}\;=\;4.5\times 10^{-4}\)
Solve for x.
\(\displaystyle x\;=\;\sqrt{0.25\times\;(4.5\times 10^{-4})}\;=\;0.0106\;M\)
Check the assumption using the 5% rule.
\(\displaystyle \frac{0.0106}{0.25}\times\;100\;=\;4.2\%\)
The assumption is valid: 4.2% < 5.0%.
The pH = -log(0.0106) = 1.97
Exercise 2. An aqueous hydrozoic acid, HN3, solution has a pH of 3.26. What was the initial concentration of the solution? Ka = 1.9 x 10-5.
First, we need the concentration of [H3O+].
[H3O+] = 10-pH = 10-3.26 = 5.5 x 10-4 M.
HN3 (aq) + H2O (l) ⇄ H3O+ (aq) + N3– (aq)
\(\displaystyle K_a\;=\;\frac{[\mathrm{N_3^-}][\mathrm{H_3O^+}]}{[\mathrm{HN_3}]}\;=\;1.9\times 10^{-5}\)
The concentration of H3O+ and N3– is 5.5 x 10-4 M. We need to solve Ka for [HN3].
\(\displaystyle [HN_3]\;=\;\frac{[\mathrm{N_3^-}][\mathrm{H_3O^+}]}{1.9\times 10^{-5}}\)
\(\displaystyle [HN_3]\;=\;\frac{(5.5\times 10^{-4})\times(5.5\times 10^{-4})}{1.9\times 10^{-5}}\;=\;\mathbf{0.016\;M}\)
Exercise 3. The percent ionization of ascorbic acid is 0.35%. What was the initial concentration of the ascorbic acid if the pH of the solution is 4.85? Ka = 1.8 x 10-5.
\(\displaystyle \%\;Ionization\;=\;\frac{x}{[\mathrm CH_3COOH]_{init}}\times\;100\)
The hydronium ion concentration is equal to x. Find the hydronium ion concentration using the pH.
[H3O+] = 10-pH = 10-4.85 = 1.4 x 10-5 M.
Now we can solve the equation for [CH3COOH]init
\(\displaystyle [CH_3COOH]_{init}\;=\;\frac{1.4\times 10^{-5}\;M}{0.35}\times\;100\;=\;\mathbf{4.0\times 10^{-3}\;M}\)
Exercise 4. Calculate the pH of an aqueous solution that is 0.025 M in formic acid, HCOOH. Ka = 1.8 x 10-4. What is the percent dissociation?
First, write the equation and the Ka expression.
HCOOH (aq) + H2O (l) ⇄ COOH– + H3O+ (aq)
\(\displaystyle K_a\;=\;\frac{[\mathrm{COOH^-}][\mathrm{H_3O^+}]}{[\mathrm{COOH}]}\;=\;1.8\times 10^{-4}\)
Set up an ICE table.
Put the equilibrium concentrations into Ka.
\(\displaystyle \frac{(x)(x)}{0.025\;-\;x}\;=\;1.8\times 10^{-4}\)
Make the assumption x is very small. 0.025 M – x ≅ 0.025 M
\(\displaystyle \frac{x^2}{0.025}\;=\;1.8\times 10^{-4}\)
Solve for x.
\(\displaystyle x\;=\;\sqrt{0.025\times\;(1.8\times 10^{-4})}\;=\;2.1\times 10^{-3}\;M\)
Check the assumption using the 5% rule.
\(\displaystyle \frac{2.1\times 10^{-3}}{0.025}\times\;100\;=\;8.4\%\)
The assumption is not valid: 8.4% > 5.0%, and we cannot neglect x. We have to use the quadratic formula. First, we rewrite our Ka expression with x.
Gather all of the like terms and set the equation to zero.
x2 + 1.8 x 10-4x – 4.5 x 10-6 = 0
a = 1, b = 1.8 x 10-4, and c = – 4.5 x 10-6
Use the quadratic formula.
\(\displaystyle \frac{-b\;\pm\;\sqrt{b^2\;-\;4ac}}{2a}\)
\(\displaystyle \frac{-1.8\times 10^{-4}\;\pm\;\sqrt{(1.8\times 10^{-4})^2\;-\;4\times 1\times(-4.5\times 10^{-6})}}{2\times\;1}\)
\(\displaystyle \frac{-1.8\times 10^{-4}\;\pm 0.004246}{2}\)
There are two solutions for x. x = 0.00203, x = -0.00221. We cannot have a negative concentration, so x = 0.00203 M.
pH = -log(0.00203) = 2.69
% Dissociation = \(\displaystyle\frac{0.00203\;M}{0.025\;M}\times\;100\;=\;\mathbf{8.12\%}\)
Exercise 5. Calculate the pH, pOH, and the concentrations of all species in 0.45 M HF. Ka = 3.5 x 10-4.
Write a balanced equation and the Ka expression.
HF (aq) + H2O (l) ⇄ F– (aq) + H3O+ (aq)
\(\displaystyle K_a\;=\;\frac{[H_3O^+][F^-]}{[HF]}\;=\;3.5\times 10^{-4}\)
Set up an ICE Table
Make the assumption that 0.45 – x ≅ 0.45 M.
Plug the equilibrium concentrations into Ka and solve for x.
\(\displaystyle \frac{x^2}{0.45}\;=\;3.5\times 10^{-4}\)
\(\displaystyle x\;=\;\sqrt{0.45\times\;(3.5\times 10^{-4})}\;=\;0.0125\;M\)
Check the assumption using the 5% rule.
\(\displaystyle \frac{0.0125\;M}{0.45\;}\times 100\;=\;2.8\%\)
2.8% < 5%, therefore, the assumption is valid.
At equilibrium, the concentrations of [HF], [F–], and [H3O+] are:
[HF]eq = 0.45 M – 0.0125 M = 0.44 M
[F–]eq = [H3O+]eq = 0.013 M
The pH = -log(0.0125) = 1.90
The pOH = 14.00 – 1.90 = 12.10
Exercise 6. A benzoic acid, C6H5COOH, solution has a pH of 2.68. What is the percent dissociation of this solution? Ka = 6.5 x 10-5
C6H5COOH (aq) + H2O (l) ⇄ C6H5COO– (aq) + H3O+ (aq)
\(\displaystyle K_a\;=\;\frac{[C_6H_5COO^-][H_2O]}{C_6H_5COOH}\;=\;6.5\times 10^{-5}\)
[H3O+] = 10-2.68 = 2.09 x 10-3 M. [C6H5COO–] is also equal to 2.09 x 10sup>-3 M.
Solve Ka for [C6H5COOH]init
\(\displaystyle [C_6H_5COOH]\;=\;\frac{(2.09 \times 10^{-3})^2}{6.5\times 10^{-5}}\;=\;0.067\;M\)
\(\displaystyle \%\;dissociation\;=\;\frac{2.09\times 10^{-3}\;M}{0.067\;M}\times 100\;=\;\mathbf{3.1\;\%}\)
Exercise 7. Vinegar contains 5.0% by mass of acetic acid (CH3COOH) in water. If the pH of the solution is 2.50 and Ka = 1.8 x 10-5, what is the percent dissociation of CH3COOH in vinegar? The density of vinegar is 1.00 g/mL.
We need to calculate the molarity of the acetic acid. The solution is 5.0% by mass which means we have
\(\displaystyle\frac{5.0\;g\;CH_3COOH}{100\;g\;\mathrm{solution}}\).
Molarity is
The molar mass of acetic acid is 60.052 g/mol.
\(\displaystyle 5.00\;g\times\frac{1\;mol}{60.052\;g}\;=\;0.0833\;mol\;CH_3COOH\)
Use the density to convert g of solution to mL of solution. The mass of the solution is 100 g.
\(\displaystyle 100\;g\frac{1\;mL}{1.00\;g}\;=\;100\;mL\;=\;0.100\;L\)
\(M\;=\;\frac{0.0833\;mol}{0.100\;L}\;=\;0.833\;M\)
Calculate the percent dissociation with x = 10-2.5 = 3.2 x 10-3