Saccharin, a sugar substitute, has the formula $\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}$ and is a weak acid with \(K_{\mathrm{a}}=2.0 \times 10^{-12} .\) If 100.0 \(\mathrm{g}\) of saccharin is dissolved in enough water to make 340 \(\mathrm{mL}\) of solution, calculate the \(\mathrm{pH}\) of the resulting solution.

First, we need to convert the given weight of saccharin into moles, and then find its molar concentration. The molecular formula of saccharin is \(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}\). Using the molar masses of the different elements in the molecular formula (H: 1 g/mol, C: 12 g/mol, N: 14 g/mol, S: 32 g/mol, O: 16 g/mol), we can determine the molar mass of saccharin as follows: Molar Mass of Saccharin = (1)(1) + (7)(12) + (4)(1) + (1)(14) + (1)(32) + (3)(16) = 183 g/mol Now, we can find the number of moles of saccharin by dividing the given mass of saccharin by its molar mass: Moles of Saccharin = \(\frac{100.0 \, \text{g}}{183 \, \text{g/mol}} = 0.546 \, \text{mol}\) The molar concentration of the solution can be calculated by dividing the moles of saccharin by the volume of the solution in liters: Molar Concentration = \(\frac{0.546\, \text{mol}}{0.340\, \text{L}} = 1.61\, \text{M}\)

Saccharin, being a weak acid, will dissociate as follows: $\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3} \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}^-$ For simplicity purposes, let's denote saccharin as HA and the saccharin ion as A-. The equilibrium expression for the dissociation reaction is: \(K_{a}=\frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}\) At the start of the reaction, [H3O+] = [A-] = 0 M while [HA] = 1.61 M.

Let x be the change in initial concentrations at equilibrium. After dissociation, the equilibrium concentrations will be: [H3O+] = x M [A-] = x M [HA] = (1.61 - x) M Substituting these values in the Ka expression: \(2.0\times 10^{-12}=\frac{(x)(x)}{(1.61-x)}\) We can approximate that (1.61-x) ≈ 1.61 since x << 1.61 \(2.0\times 10^{-12}=\frac{x^{2}}{1.61}\) Now, we solve for x. \(x^{2}=2.0\times 10^{-12} \times 1.61 = 3.22\times 10^{-12}\) \(x=\sqrt{3.22\times 10^{-12}} = 5.67 \times 10^{-7}\) Since x ≈ [H3O+], we found the concentration of hydronium ions in the solution.

Now, we can calculate the pH of the solution using the formula: pH = -log[H3O+] pH = -log[5.67 × 10^{-7}] pH ≈ 6.25 The pH of the saccharin solution is approximately 6.25.