Q18.12P - Chemistry: Molecular Nature Of Matter And Change

Question

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

(a) ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{ - }}}$

(b) ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{2}}$

(c) ${\mathbf{\text{HSO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{ - }}}$

Step-by-Step Solution

Answer

(a) The $K_a$ expression for ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ is ${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{][HPO}}_{\text{4}}^{\text{2 - }}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}\text{]}}$.

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

(c) The $K_a$ expression for ${\text{HSO}}_{\text{4}}^{\text{ - }}$ is ${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{][SO}}_{\text{4}}^{\text{2 - }}\text{]}}{{\text{[HSO}}_{\text{4}}^{\text{ - }}\text{]}}$ .

Step 1: Concept Introduction

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

Step 2: (a) K a expression for H 2 PO 4  - 

To write the $K_a$ expression of ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ in water, we should write the balanced reaction first –

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\rightleftharpoons {\text{ H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{(aq) + HPO}}_{\text{4}}^{\text{2 - }}\text{(aq)}$

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

$$\begin{gathered} {\text{K}}_{\text{a}}\text{ = }\frac{\text{[product]}}{\text{[reactant]}} \\ {\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{][HPO}}_{\text{4}}^{\text{2 - }}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}\text{]}} \end{gathered}$$

 

Therefore, the expression is obtained as ${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{][HPO}}_{\text{4}}^{\text{2 - }}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}\text{]}}$.

Step 3: (b) K a expression for H 3 PO 2

To write $K_a$ the expression of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{2}}$ in water, we should write the balanced reaction first –

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

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

$$\begin{gathered} {\text{K}}_{\text{a}}\text{ = }\frac{\text{[product]}}{\text{[reactant]}} \\ {\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{][H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{ - }}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{2}}\text{]}} \end{gathered}$$

 

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

Step 4: (c) K a expression for HSO 4  - 

To write the $K_a$ expression of ${\text{HSO}}_{\text{4}}^{\text{ - }}$ in water, we should write the balanced reaction first –

${\text{HSO}}_{\text{4}}^{\text{ - }}{\text{(aq) + H}}_{\text{2}}\text{O(l) }\rightleftharpoons {\text{ H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 - }}\text{(aq)}$

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

$$\begin{gathered} {\text{K}}_{\text{a}}\text{ = }\frac{\text{[product]}}{\text{[reactant]}} \\ {\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{][SO}}_{\text{4}}^{\text{2 - }}\text{]}}{{\text{[HSO}}_{\text{4}}^{\text{ - }}\text{]}} \end{gathered}$$

 

Therefore, the expression is obtained as ${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{][SO}}_{\text{4}}^{\text{2 - }}\text{]}}{{\text{[HSO}}_{\text{4}}^{\text{ - }}\text{]}}$ .