Chapter 16, Problem 71PS

Sulphurous acid, H₂SO₃, is a weak acid capable of providing two H⁺ ions.

1. (a) What is the pH of a 0.45M solution of H₂SO₃?
2. (b) What is the equilibrium concentration of the sulphate ion SO₃²⁻, in the 0.45M solution of H₂SO₃?

Interpretation Introduction

Interpretation:

The $\text{pH}$ of a 0.45M solution of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$ has to be determined

Concept Introduction:

If a molecule donate only one hydrogen atom then the acid is called a monoprotic acid.

Example: ${\text{HCl, HBr and HClO}}_{\text{3}}$

If each molecule can donate two or more hydrogen ions, the acid is a polyprotic acid.

A diprotic acid, such as sulfuric acid, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has two acidic hydrogen atoms.

Some acids, such as phosphoric acid, ${\text{H}}_3{\text{PO}}_{\text{4}}$ are triprotic acids.

Polyprotic bases:

Can accept more than on proton ${\text{(H}}^{\text{+}}\text{)}$. So, it can react with water many times to produce many hydroxide ions ${\text{(OH}}^{-}\text{)}$.

Diprotic bases:

Can accept two protons.

Example: Sulphate ion ${\text{SO}}_{\text{4(aq)}}^{\text{2-}}$

Answer to Problem 71PS

pH of a 0.45M solution of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$ is 1.17

Explanation of Solution

Sulfurous acid is a diprotic acid.

$\begin{array}{l}\text{First ionization:}\\ {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{(aq) + H}}_{\text{2}}\text{O(aq)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + HSO}}_{\text{3}}^{\text{-}}\text{(aq)}\\ \text{Equilibrium expression:}\\ {\text{K}}_{\text{a1}}\text{ = }\frac{{\text{[HSO}}_{\text{3}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[H}}_{\text{2}}{\text{SO}}_{\text{3}}\text{]}}\\ {\text{K}}_{\text{a1}}\text{= 1}{\text{.2×10}}^{\text{-2}}\\ \text{Second ionization:}\\ {\text{HSO}}_{\text{3}}^{\text{-}}{\text{(aq) + H}}_{\text{2}}\text{O(aq)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + SO}}_{\text{3}}^{\text{2-}}\text{(aq)}\\ \text{Equilibrium expression:}\\ {\text{K}}_{\text{a2}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][SO}}_{\text{3}}^{\text{2-}}\text{]}}{{\text{[HSO}}_{\text{3}}^{\text{-}}\text{]}}\\ {\text{K}}_{\text{a2}}\text{= 6}{\text{.2×10}}^{\text{-8}}\end{array}$

From the ${\text{K}}_{\text{a1}}{\text{ and K}}_{\text{a2}}$ values, ${\text{ K}}_{\text{a2}}$ is smaller than the ${\text{ K}}_{\text{a1}}$.

Therefore, ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is almost produced entirely from ${\text{ K}}_{\text{a1}}$.

Let’s calculate the ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ from ${\text{ K}}_{\text{a1}}$.

$\begin{array}{l}\text{First ionization:}\\ {\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{(aq) + H}}_{\text{2}}\text{O(aq)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + HSO}}_{\text{3}}^{\text{-}}\text{(aq)}\\ \text{Equilibrium expression:}\\ {\text{K}}_{\text{a1}}\text{ = }\frac{{\text{[HSO}}_{\text{3}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[H}}_{\text{2}}{\text{SO}}_{\text{3}}\text{]}}\\ {\text{K}}_{\text{a1}}\text{= 1}{\text{.2×10}}^{\text{-2}}\end{array}$

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

$\begin{array}{l}{\text{ H}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{(aq) + H}}_{\text{2}}\text{O (aq)}\rightleftharpoons {\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + HSO}}_{\text{3}}^{\text{-}}\text{(aq)}\\ \text{I 0}\text{.45M -- -- --}\\ \text{C -x -- +x +x}\\ \text{E (0}\text{.45-x) -- x x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a1}}\text{ = }\frac{{\text{[HSO}}_{\text{3}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[H}}_{\text{2}}{\text{SO}}_{\text{3}}\text{]}}\\ {\text{K}}_{\text{a1}}\text{= 1}{\text{.2×10}}^{\text{-2}}\end{array}$

$\begin{array}{l}\text{At equilibrium,}\\ {\text{x = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = [HSO}}_{\text{3}}^{\text{-}}\text{]}\\ \text{1}{\text{.2×10}}^{\text{-2}}\text{ = }\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.45-x)}}\\ {\text{x}}^{\text{2}}\text{ + 1}{\text{.2×10}}^{\text{-2}}\text{ - 5}{\text{.4×10}}^{\text{-3}}\text{ = 0}\\ \text{x = }\frac{\text{-1}{\text{.2×10}}^{\text{-2}}\text{±}\sqrt{{\text{(1}{\text{.2×10}}^{\text{-2}}\text{)}}^{\text{2}}\text{-4(1)(-5}{\text{.4×10}}^{\text{-3}}\text{)}}}{\text{2(1)}}\\ \text{x = 6}{\text{.77×10}}^{\text{-2}}\\ \text{Therefore,}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] = [HSO}}_{\text{3}}^{\text{-}}\text{] = 6}{\text{.77×10}}^{\text{-2}}\\ {\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{pH = -log(6}{\text{.77×10}}^{\text{-2}}\text{)}\\ \text{ =1}\text{.17}\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium concentration has to be determined for the sulphate ion ${\text{SO}}_{\text{3}}^{\text{2-}}$ in the 0.45M solution of ${\mathrm{H}}_{2}S{O}_3$.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$ is base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

If a molecule donate only one hydrogen atom then the acid is called a monoprotic acid.

Example: ${\text{HCl, HBr and HClO}}_{\text{3}}$

If each molecule can donate two or more hydrogen ions, the acid is a polyprotic acid.

A diprotic acid, such as sulfuric acid, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has two acidic hydrogen atoms.

Some acids, such as phosphoric acid, ${\text{H}}_3{\text{PO}}_{\text{4}}$ are triprotic acids.

Polyprotic bases:

Can accept more than on proton ${\text{(H}}^{\text{+}}\text{)}$. So, it can react with water many times to produce many hydroxide ions ${\text{(OH}}^{-}\text{)}$.

Diprotic bases:

Can accept two protons.

Example: Sulphate ion ${\text{SO}}_{\text{4(aq)}}^{\text{2-}}$

Answer to Problem 71PS

The equilibrium concentration of the sulphate ion ${\text{SO}}_{\text{3}}^{\text{2-}}$ in the 0.45M solution of ${\mathrm{H}}_{2}S{O}_3$ is $\text{6}{\text{.28×10}}^{\text{-8}}\text{M}$

Explanation of Solution

$\begin{array}{l}\text{Second ionization:}\\ {\text{HSO}}_{\text{3}}^{\text{-}}{\text{(aq) + H}}_{\text{2}}\text{O(aq)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + SO}}_{\text{3}}^{\text{2-}}\text{(aq)}\\ \text{Equilibrium expression:}\\ {\text{K}}_{\text{a2}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][SO}}_{\text{3}}^{\text{2-}}\text{]}}{{\text{[HSO}}_{\text{3}}^{\text{-}}\text{]}}\\ {\text{K}}_{\text{a2}}\text{= 6}{\text{.2×10}}^{\text{-8}}\end{array}$

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

$\begin{array}{l}{\text{ HSO}}^{-}{}_{\text{3}}{\text{(aq) + H}}_{\text{2}}\text{O (aq)}\rightleftharpoons {\text{ SO}}_3^{2-}{\text{(aq) + H}}_3{\text{O}}^{+}\text{(aq)}\\ \text{I 6}\text{.77}\times {\text{10}}^{-2}\text{M -- -- 6}\text{.77}\times {\text{10}}^{-2}M\\ \text{C -y -- +y +y}\\ \text{E (6}\text{.77}\times {\text{10}}^{-2}\text{-x) -- y (6}\text{.77}\times {\text{10}}^{-2}M)\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a2}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][SO}}_{\text{3}}^{\text{2-}}\text{]}}{{\text{[HSO}}_{\text{3}}^{\text{-}}\text{]}}\\ {\text{K}}_{\text{a2}}\text{= 6}{\text{.2×10}}^{\text{-8}}\end{array}$

$\begin{array}{l}\text{6}{\text{.2×10}}^{\text{-8}}\text{ = }\frac{\text{y (6}{\text{.77×10}}^{\text{-2}}\text{ + y)}}{\text{(6}{\text{.77×10}}^{\text{-2}}\text{- y)}}\\ \text{Y is so small compared to 6}{\text{.77×10}}^{\text{-2}}\\ \text{So, all values are almost equal to 6}{\text{.77×10}}^{\text{-2}}\\ \text{6}{\text{.2×10}}^{\text{-8}}\text{ = }\frac{\text{y(6}{\text{.77×10}}^{\text{-2}}\text{)}}{\text{(6}{\text{.77×10}}^{\text{-2}}\text{)}}\\ \text{ y = 6}{\text{.2×10}}^{\text{-8}}\end{array}$