To repress the ionization of formic acid, HCOOH(aq), which of the following should be added to the solution? (a) \(\mathrm{NaCl} ;\) (b) \(\mathrm{NaOH}\); (c) \(\mathrm{NaHCOO}\); (d) \(\mathrm{NaNO}_{3}\)

Now, we need to evaluate the effects of each compound in the given options.\n\n(a) \(NaCl\) - This compound will dissociate into \(Na^+\) and \(Cl^-\) ions. Neither of these ions participates in the reaction, so adding this salt would have no effect on the ionization of formic acid.\n\n(b) \(NaOH\) - When \(NaOH\) is added to the solution, it will dissociate into \(Na^+\) and \(OH^-\) ions. The \(OH^-\) ions will combine with \(H^+\) ions from the ionization of formic acid to form water. This effectively reduces the amount of \(H^+\) present, and according to Le Chatelier’s principle, the reaction will shift to the right to compensate for the reduction in \(H^+\) ions, leading to increased ionization of formic acid, which is not what we want.\n\n(c) \(NaHCOO\) - Adding this compound adds \(HCOO^-\) ions to the solution, which is a product of the formic acid ionization. Again, by Le Chatelier's principle, the reaction will shift to the left in an attempt to reduce the increase in product concentration. In this case, it has the desired effect of repressing the ionization of formic acid.\n\n(d) \(NaNO_{3}\) - This ionic compound will dissociate into \(Na^+\) and \(NO_{3}^-\). Neither of these ions is involved in the ionization of formic acid, so there would be no effect.

Having looked at the effects of all the given substances, it can be concluded that \(NaHCOO\) is the correct answer since it directly adds a product to the ionization reaction of formic acid. Le Chatelier's principle predicts that the reaction will shift to the left to counteract this, causing suppression of ionization of formic acid.