Chapter 15, Problem 15.157QP

Calculate the pH of a 0.20 M NaHCO₃ solution.(Hint: As an approximation, calculate hydrolysis and ionization separately first, followed by partial neutralization.)

Interpretation Introduction

Interpretation:

The $\text{0}\text{.20 }M$${\text{NaHCO}}_{\text{3}}$ solution $pH$ range has to be calculated

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\underset{}{\rightleftharpoons }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\underset{}{\rightleftharpoons }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Relationship between ${\text{K}}_{\text{a}}$and $K_b$

${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}$

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

$\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 15.157QP

The sodium bicarbonate $NaHC{O}_3$ $pH$ range is 9.82

Explanation of Solution

First we write the two separate equilibrium reactions with hydrogen carbonate functioning as basic as an acid in the other.

The hydrogen ions and hydroxide ions produced in the equilibria will then undergo partial neutralization.

The $K_b$ chemical equilibrium denoted by  

	${\text{HCO}}_3^{-}{}_{\text{(aq)}}+{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}$
Initial $(M)$	$0.20$ -x $0.20-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $k_b$ value $HC{O}_3=2.4\times {10}^{-8}$

The equilibrium expression is,

$\begin{array}{l}{K}_b=\frac{[H_{2}C{O}_3][O{H}^{-}]}{[HC{O}_3^{-}]}\\ 2.4\times {10}^{-8}=\frac{x^2}{(0.20-x)}\end{array}$

We assume that $x$ is small so, $(0.20-x)\approx 0.20$

$\begin{array}{l}2.4\times {10}^{-8}=\frac{x^2}{(0.20)}\\ x=[OH]=6.9\times {10}^{-5}M\end{array}$

The $K_a$ chemical equilibrium denoted by  

	${\text{HCO}}_3^{-}{}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}$
Initial $(M)$	$0.20$ -x $0.20-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $k_a$ value $HC{O}_3=4.8\times {10}^{-11}$

The equilibrium expression is,

$\begin{array}{l}{K}_a=\frac{[H^{+}][C{O}_3^{2-}]}{[HC{O}_3^{-}]}\\ 4.8\times {10}^{-11}=\frac{x^2}{(0.20-x)}\end{array}$

We assume that $x$ is small so, $(0.20-x)\approx 0.20$

$\begin{array}{l}4.8\times {10}^{-11}=\frac{x^2}{(0.20)}\\ x=[H^{+}]=3.1\times {10}^{-6}M\end{array}$

The hydroxide $O{H}^{-}$ produced in the first equilibrium will be partially neutralized by the $H^{+}$ produced in the second equilibrium.

	${\text{H}}^{\text{+}}{}_{\text{(aq)}}+{\text{OH}}^{\text{-}}{}_{(aq)}\underset{}{\overset{}{\rightleftharpoons }}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$
Initial $(M)$	$3.1\times {10}^{-6}$ $-3.1\times {10}^{-6}$      0	$6.9\times {10}^{-5}$	$0.00$
Change $(M)$		$3.1\times {10}^{-6}$	$3.1\times {10}^{-6}$
Equilibrium $(M)$		$+6.6\times {10}^{-5}$	$3.1\times {10}^{-6}$

The $K_b$ value of hydroxide is a $6.6\times {10}^{-5}M$

Calculation of $\text{pOH}$ value

$\begin{array}{l}\text{pOH}\text{=}\text{log}\text{(6}\text{.6}\times {\text{10}}^{\text{-5}}\text{)}\\ \text{pOH}=\underset{\_}{4.18}\\ \text{Calculation}\text{of}\text{pH}\text{value}\\ \text{pH=}\text{pOH}-4.18\\ \text{pH=}\text{14-4}\text{.18}\\ \text{pH=}\underset{\_}{9.82}\end{array}$

Therefore given ${\text{NaHCO}}_{\text{3}}$ solution $pOH=4.18$ and $pH=9.82$

The $\text{pOH}\text{and}\text{pH}$  result is only an approximation to the more rigorous treatment.