Question

A \(50.0-\mathrm{mL}\) sample of \(\mathrm{NaHSO}_{3}\) is titrated with \(22.94 \mathrm{~mL}\) of \(0.238 \mathrm{M} \mathrm{KOH}\). (a) Write a balanced net ionic equation for the reaction. (b) What is \(\left[\mathrm{HSO}_{3}^{-}\right]\) before the titration? (c) Find \(\left[\mathrm{HSO}_{3}^{-}\right],\left[\mathrm{SO}_{3}^{2-}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{K}^{+}\right],\) and \(\left[\mathrm{Na}^{+}\right]\) at the equivalence point. (d) What is the \(\mathrm{pH}\) at the equivalence point?

Short answer

(a) The balanced net ionic equation for the reaction between KOH and NaHSO3 is: \(HSO_{3}^{-}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l) + SO_{3}^{2-}(aq)\) (b) Before titration, the initial concentration of HSO3- is 0.109186 M. (c) At the equivalence point, the concentration of the following species are: - [HSO3-] = 0 - [SO3^2-] = 0.109186 M - [OH-] = 0 - [K+] = 0.07486 M - [Na+] = 0.07486 M (d) The pH at the equivalence point is 7.955.

Step-by-step solution

(a) Balanced Net Ionic Equation

The balanced net ionic equation involves the major species that react with each other. In this case, it will be: \(HSO_{3}^{-}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l) + SO_{3}^{2-}(aq)\) The spectator ions, K+ and Na+, do not participate in the reaction and are not included in the net ionic equation.

(b) Initial Concentration of HSO3-

Since KOH and NaHSO3 react with each other in a 1:1 ratio, we first need to calculate the initial moles of NaHSO3: moles of KOH: \(n(KOH) = volume \times molarity = 22.94\mathrm{mL} \times 0.238\mathrm{M} = 5.45932\mathrm{mmol}\) Since moles of NaHSO3 and KOH are the same at the equivalence point, we have 5.45932 mmol of NaHSO3 initially. Now, calculate the initial concentration of HSO3-: \(\left[HSO_{3}^{-}\right]_{initial} = \frac{n(NaHSO_{3})}{volume} = \frac{5.45932\mathrm{mmol}}{50.0\mathrm{mL}} = 0.109186\mathrm{M}\)

(c) Concentrations at the Equivalence Point

At the equivalence point, all moles of HSO3- have reacted with OH-. Therefore: \([\mathrm{HSO}_{3}^{-}] = 0\) Since all moles of HSO3- have reacted to form SO3^2-, their concentrations are equal: \([\mathrm{SO}_{3}^{2-}] = 0.109186\mathrm{M}\) Similarly, all OH- ions have reacted with HSO3- ions, and thus: \([\mathrm{OH}^{-}] = 0\) Potassium and sodium ions are spectator ions, their concentrations can be found based on the initial moles and the total volume at the equivalence point: At the equivalence point, total volume = 50.0 mL (initial) + 22.94 mL (KOH added) = 72.94 mL \([\mathrm{K}^{+}] = \frac{n(KOH)}{72.94\mathrm{mL}} = \frac{5.45932\mathrm{mmol}}{72.94\mathrm{mL}} = 0.07486\mathrm{M}\) \([\mathrm{Na}^{+}] = \frac{n(NaHSO_{3})}{72.94\mathrm{mL}} = \frac{5.45932\mathrm{mmol}}{72.94\mathrm{mL}} = 0.07486\mathrm{M}\)

(d) pH at the Equivalence Point

At the equivalence point, the pH is determined by the remaining basic ion, which is in this case SO3^2-. We first need to find the concentration of the remaining acidic ion HSO4-, which is produced by the hydrolysis of SO3^2-. The hydrolysis reaction of SO3^2- can be described as: \(SO_{3}^{2-}(aq) + H_{2}O(l) \rightleftharpoons HSO_{4}^{-}(aq) + OH^{-}(aq)\) The equilibrium expression for the reaction is: \(K_{b} = \frac{\left[\mathrm{HSO}_{4}^{-}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{SO}_{3}^{2-}\right]}\) \(K_{b}\) for \(HSO_{4}^{-}\) can be calculated as: \(K_{b} = K_{w} / K_{a} = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-2}} = 8.33 \times 10^{-13}\) Let x be the concentrations of HSO4- and OH- ions at equilibrium, then: \(8.33 \times 10^{-13} = \frac{x^{2}}{0.109186-x}\) Assuming x <<< 0.109186, the equation simplifies to: \(x^{2} = 8.33 \times 10^{-13} \times 0.109186\) \(x = \sqrt{8.33 \times 10^{-13} \times 0.109186} = 9.021 \times 10^{-7} \mathrm{M}\) The concentrations of \(\mathrm{HSO}_{4}^{-}\) and OH- are now calculated. The \(\mathrm{pOH}\) can be calculated as: \(\mathrm{pOH}=-log(9.021\times 10^{-7})=6.045\) Finally, obtain the pH at the equivalence point: \(\mathrm{pH}=14-\mathrm{pOH}=14-6.045=7.955\)