What is the pH of 0.16 M solution of \( \mathrm{H}_{3} \mathrm{PO}_{4} \) with \( \mathrm{K}_{1}=7.5 \times 10^{-3}, \mathrm{~K}_{2}=6.2 \times 10^{-8} \), and \( \mathrm{K}_{3}=4.2 \times 10^{-13} \) ? 1.6 3.2 1.5 4.1

**Step 1. Write the first dissociation reaction and equilibrium expression** For phosphoric acid, the first dissociation is \[ \mathrm{H}_3\mathrm{PO}_4 \rightleftharpoons \mathrm{H}_2\mathrm{PO}_4^- + \mathrm{H}^+ \] with equilibrium constant \[ K_1 = \frac{[\mathrm{H}_2\mathrm{PO}_4^-][\mathrm{H}^+]}{[\mathrm{H}_3\mathrm{PO}_4]} = 7.5 \times 10^{-3}. \] Let the concentration of \(\mathrm{H}^+\) produced from the first dissociation be \(x\). The initial concentration of \(\mathrm{H}_3\mathrm{PO}_4\) is \(0.16\, \mathrm{M}\), so at equilibrium we have: - \([\mathrm{H}_3\mathrm{PO}_4] = 0.16 - x\) - \([\mathrm{H}_2\mathrm{PO}_4^-] = x\) - \([\mathrm{H}^+] = x\) Thus, the equilibrium expression becomes \[ K_1 = \frac{x \cdot x}{0.16-x} = \frac{x^2}{0.16-x}. \] **Step 2. Solve for \(x\) using the approximation** We have: \[ \frac{x^2}{0.16-x} = 7.5 \times 10^{-3}. \] This leads to the quadratic equation: \[ x^2 + (7.5 \times 10^{-3})x - (7.5 \times 10^{-3})(0.16) = 0. \] Compute the last term: \[ (7.5 \times 10^{-3})(0.16) = 0.0012. \] So the quadratic is: \[ x^2 + 0.0075\,x - 0.0012 = 0. \] The quadratic formula gives: \[ x = \frac{-0.0075 \pm \sqrt{(0.0075)^2 + 4 \times 0.0012}}{2}. \] Calculate the discriminant: \[ (0.0075)^2 = 0.00005625, \] \[ 4 \times 0.0012 = 0.0048, \] \[ \text{Discriminant} = 0.00005625 + 0.0048 \approx 0.00485625. \] Taking the square root: \[ \sqrt{0.00485625} \approx 0.0697. \] So, the positive solution is: \[ x = \frac{-0.0075 + 0.0697}{2} \approx \frac{0.0622}{2} \approx 0.0311\, \mathrm{M}. \] **Step 3. Calculate the pH from the first dissociation** Since the concentration of \(\mathrm{H}^+\) is approximately \(x = 0.0311\, \mathrm{M}\), the pH is: \[ \mathrm{pH} = -\log (0.0311) \approx 1.51. \] **Step 4. Consider the contributions of further dissociations** The second dissociation is: \[ \mathrm{H}_2\mathrm{PO}_4^- \rightleftharpoons \mathrm{HPO}_4^{2-} + \mathrm{H}^+, \] with \(K_2 = 6.2 \times 10^{-8}\). Because \(K_2\) is many orders of magnitude smaller than \(K_1\), its contribution to the \(\mathrm{H}^+\) concentration is negligible compared to the first dissociation. The third dissociation has an even smaller \(K_3\). Thus, the pH is essentially determined by the first dissociation. **Final Answer** The pH of the \(0.16\, \mathrm{M}\) phosphoric acid solution is approximately \(\boxed{1.5}\).