HCN(aq) + H20 (1) 3 H30+ (aq) + CN- (aq) Ka = H30+][CN"] – 62 x 10-10 [HCM] The equilibrium reaction shown above represents the partial ionization of the weak acid HCN (aq). A 0.200 M HCN (aq) solution has a pH ~ 4.95. If 0.05 g (0.010 mol) of NaCN (s) is added to 100 mL of 0.200 M HCN (aq), which of the following explains how and why the pH of the solution changes? a The pH will be higher than 4.95 because adding CN- will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of H3O+. c will disrupt the equilibrium, resulting in an increased production of HCN that decreases the concentration of ) The pH will be lower than 4.95 because adding CN H30+. c) The pH will be higher than 4.95 because CN is a strong base that can neutralize HCN. ( D ) The pH will remain close to 4.95 because the K, is so small that hardly any products form.