Q18.9P - Chemistry: Molecular Nature Of Matter And Change

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) $\text{HCOOH}$

Step-by-Step Solution

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 1: 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: K a expression for H C N

(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 HCO 3 -

(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: K a expression 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 .${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{\text{[HCOOH]}}$