Question 8.E.B: (Without activities), calculate the pH of (a) 1.0 × 10^−8 M ...

(a) Charge balance: [H^{+}] = [OH^{−}] + [Br^{−}]

Mass balance:     [Br^{−}] = 1.0 × 10^{−8} M

Equilibrium:     [H^{+}][OH^{−}] = K_{w}

Setting [H^{+}] = x and [Br^{−}] = 1.0 × 10^{−8} M, the charge balance tells us that [OH^{−}] = x  −  1.0 × 10^{−8}. Putting this into the K_{w} equilibrium gives (x)(x − 1.0 × 10^{−8}) =  1.0 × 10^{−14} ⇒ x = 1.0_{5} × 10^{−7} M ⇒ pH = 6.98.
(b) Charge balance:     [H^{+}] = [OH^{−}] + 2[SO_{4}^{2−}]

Mass balance:     [SO_{4}^{2−}] = 1.0 × 10^{−8} M

Equilibrium:     [H^{+}][OH^{−}] = K_{w}

As before, writing [H^{+}] = x and [SO_{4}^{2−}] = 1.0 × 10^{−8} M gives [OH^{−}] = x − 2.0 × 10^{−8} M and  [H^{+}][OH^{−}] = (x)[x   −  (2.0 × 10^{−8})] = 1.0 × 10^{−14}

⇒ x = 1.10 × 10^{−7} M ⇒ pH = 6.96.