Saccharin, a sugar substitute, is a weak acid with pKa = 2.32 at 25 °C. It ionizes in aqueous solution as follows: HNC7H4SO31(aq) ⇌ H+(aq) + NC7H4SO3-(aq). What is the pH of a 0.10 M solution of this substance?

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1. Identify the given information: The concentration of saccharin is 0.10 M, and the pKa is 2.32.
2. Use the relationship between pKa and Ka: \( \text{pKa} = -\log_{10}(K_a) \). Calculate \( K_a \) by rearranging the formula to \( K_a = 10^{-\text{pKa}} \).
3. Set up the expression for the ionization of saccharin: \( K_a = \frac{[H^+][NC_7H_4SO_3^-]}{[HNC_7H_4SO_3]} \). Assume \([H^+] = [NC_7H_4SO_3^-] = x\) and \([HNC_7H_4SO_3] = 0.10 - x\).
4. Substitute the expressions into the \( K_a \) equation: \( K_a = \frac{x^2}{0.10 - x} \). Since \( K_a \) is small, assume \( x \ll 0.10 \), simplifying to \( K_a \approx \frac{x^2}{0.10} \).
5. Solve for \( x \) (which is \([H^+]\)) using the simplified equation: \( x = \sqrt{K_a \times 0.10} \). Finally, calculate the pH using \( \text{pH} = -\log_{10}(x) \).