Question

Write the ${\text{K}}_{\text{a}}$expression for each of the following in water:

(a)$\text{HCN}$

(b)${\text{HCO}}_{\text{3}}^{\text{-}}$

(c)$\text{HCOOH}$

Short answer

(a) The $K_a$expression for is ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CN}}^{\text{-}}\text{]}}{\text{[HCN]}}$.

(b) The$K_a$ expression for is ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CO}}_{\text{3}}^{\text{2-}}\text{]}}{{\text{[HCO}}_{\text{3}}^{\text{-}}\text{]}}$.

(c) The$K_a$ expression for is ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{\text{[HCOOH]}}$.

Step-by-step solution

Concept Introduction

An acid dissociation constant$\left(K_a\right)$is a numerical representation of an acid's strength in solution. The dissociation constant${\text{K}}_{\text{a}}\text{=}\frac{{\text{[A}}^{\text{-}}{\text{][H}}^{\text{+}}\text{]}}{\text{[HA]}}$is commonly stated as a quotient of the equilibrium concentrations (in$mol/L$).

 Step 2:  Kaexpression for HCN

(a)

To write the $K_a$expression of $\text{HCN}$in water, we should write the balanced reaction first –

${\text{HCN(aq)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)+CN}}^{\text{-}}\text{(aq)}$

Then write the $K_a$expression using the formula below. Note that it only includes aqueous species.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CN}}^{\text{-}}\text{]}}{\text{[HCN]}}\end{array}$

Therefore, the expression is obtained as ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CN}}^{\text{-}}\text{]}}{\text{[HCN]}}$.

Step 3:expression for HCO3-

(b)

To write the expression of${\text{HCO}}_{\text{3}}^{\text{-}}$ in water, we should write the balanced reaction first –

${\text{HCO}}_{\text{3}}^{\text{-}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)+CO}}_{\text{3}}^{\text{2-}}\text{(aq)}$

Then write the $K_a$expression using the formula below. Note that it only includes aqueous species.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CO}}_{\text{3}}^{\text{2-}}\text{]}}{{\text{[HCO}}_{\text{3}}^{\text{-}}\text{]}}\end{array}$

Therefore, the expression is obtained as .${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CO}}_{\text{3}}^{\text{2-}}\text{]}}{{\text{[HCO}}_{\text{3}}^{\text{-}}\text{]}}$

Step 4: Kaexpression for HCOOH

(c)

To write the $K_a$expression of$\text{HCOOH}$ in water, we should write the balanced reaction first –

${\text{HCOOH(aq)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)+HCOO}}^{\text{-}}\text{(aq)}$

Then write the $K_a$expression using the formula below. Note that it only includes aqueous species.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{\text{[HCOOH]}}\end{array}$

Therefore, the expression is obtained as .${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{\text{[HCOOH]}}$