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Chapter 16: Problem 61

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{p} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the pH of a \(0.25 \mathrm{M}\) solution of this substance?

Short Answer

Expert verified
The pH of a 0.25 M solution of saccharin, a weak acid with a pKa of 2.32, is approximately 1.46.

Step by step solution

01

Write the ionization equilibrium equation and Ka expression

The ionization of saccharin in an aqueous solution can be represented by the following equation: \[ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) \] The equilibrium constant expression for this weak acid ionization is given by: \[K_a = \frac{[\mathrm{H}^+][\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]}{[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3]}\]
02

Determine the initial concentrations

The initial concentration of saccharin is given as 0.25 M. Since the solution is initially un-ionized, the initial concentrations of the other species (H+ and NC₇H₄SO₃⁻) are both 0: \[[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3]_0 = 0.25 \mathrm{M}\] \[[\mathrm{H}^+]_0 = 0\] \[[\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]_0 = 0\]
03

Set up an equilibrium table (ICE table)

An ICE (Initial, Change, Equilibrium) table helps us calculate the changes in concentration of the species in the ionization process: ``` [HNC₇H₄SO₃] [H⁺] [NC₇H₄SO₃⁻] Initial 0.25M 0 0 Change -x +x +x Equilibrium 0.25-x x x ```
04

Set up the Ka equation and solve

Substitute the equilibrium concentrations in the Ka expression: \[K_a = \frac{x \times x}{(0.25 - x)}\] Since the pKa is given as 2.32, the Ka can be calculated as: \[K_a = 10^{-\mathrm{p}K_a} = 10^{-2.32} = 4.82 \times 10^{-3}\] Now we can solve for "x": \[4.82 \times 10^{-3} = \frac{x^2}{(0.25 - x)}\] Simplifying further, we have \(x^2 \approx 4.82 \times 10^{-3} \times 0.25\): \[x \approx \sqrt{1.21 \times 10^{-3}}\] \[x \approx 0.0348\]
05

Calculate the pH

Since "x" is the concentration of H+ ions at equilibrium, we can now calculate the pH of the solution using the formula: \[\mathrm{pH} = -\log{[\mathrm{H}^+]}\] \[\mathrm{pH} = -\log{(0.0348)}\] \[\mathrm{pH} \approx 1.46\] The pH of the 0.25 M saccharin solution is approximately 1.46.

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