What will be the exact pH of a aqueous solution of ammonium formate () at ? Given the dissociation c — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Ammonium formate is a salt of a weak acid ($HCOOH$) and a weak base ($NH_3$). The pH is given by $pH = 7 + \frac{1}{2}(pK_a - pK_b)$. First, $pK_a = -\log(10^{-4}) = 4$ and $pK_b = -\log(10^{-5}) = 5$. Thus, $pH = 7 + \frac{1}{2}(4 - 5) = 7 - 0.5 = 6.5$. Note that the concentration (1.0 M) does not affect the pH.