Question

What is the \(\mathrm{pH}\) of a \(0.1500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution if (a) the ionization of \(\mathrm{HSO}_{4}^{-}\) is ignored? (b) the ionization of \(\mathrm{HSO}_{4}^{-}\) is taken into account? \(\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HSO}_{4}^{-}\) is \(\left.1.1 \times 10^{-2} .\right)\)

Short answer

Answer: The pH values for the given H2SO4 solution are approximately (a) 0.82 and (b) 0.79.

Step-by-step solution

Find the pH ignoring ionization of HSO4-

Calculate the pH by evaluating the expression: \(pH \approx 0.82\) This calculation is for part (a) of the exercise, where the ionization of HSO4- is ignored.

Calculate pH taking ionization of HSO4- into account

First, write the equilibrium expression for the ionization of HSO4-: \(\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-} \) Now, set up the equilibrium expression using the given Ka value: \(K_a = \frac{[\text{H}^+][\text{SO}_4^{2-}]}{[\text{HSO}_4^-]}\)

Use the ICE method

Set up the ICE (Initial, Change, Equilibrium) table to find the equilibrium concentration of the ions: Initial: \([\text{H}^+] = 0.1500\text{ M}, [\text{SO}_4^{2-}] = 0\text{ M}, [\text{HSO}_4^-] = 0.1500\text{ M}\) Change: \([\text{H}^+] = +x, [\text{SO}_4^{2-}] = +x, [\text{HSO}_4^-] = -x\) Equilibrium: \([\text{H}^+] = 0.1500 + x, [\text{SO}_4^{2-}] = x, [\text{HSO}_4^-] = 0.1500 - x\) Substitute equilibrium concentrations into the Ka expression: \(1.1 \times 10^{-2} = \frac{(0.1500 + x)x}{(0.1500 - x)}\)

Solve for x

x is a relatively small value compared to 0.1500, so the equation can be approximated by: \(1.1 \times 10^{-2} = \frac{(0.1500)x}{0.1500}\) Now, solve for x: \(x = 1.1 \times 10^{-2}\) This value corresponds to the concentration of additional H+ ions from the second ionization of HSO4-.

Calculate the total H+ concentration

To find the total concentration of H+ ions, add the contribution from both ionizations: \([\text{H}^+]_{total} = [\text{H}^+]_{\text{first ionization}} + [\text{H}^+]_{\text{second ionization}}\) \([\text{H}^+]_{total} = 0.1500 + 1.1 \times 10^{-2}\) \([\text{H}^+]_{total} \approx 0.1610\text{ M}\)

Calculate the pH

Now, calculate the pH using the total H+ concentration: \(pH = -\log_{10} (0.1610)\) \(pH \approx 0.79\) Thus, when taking the ionization of HSO4- into account, the pH of the solution is approximately 0.79. So, the pH values for parts (a) and (b) of the exercise are approximately 0.82 and 0.79, respectively.