Chapter 15, Problem 15.152QP

Calculate the pH of a solution that is 1.00 M HCN and 1.00 M HF. Compare the concentration (in molarity) of the CN⁻ ion in this solution with that in a 1.00 M HCN solution. Comment on the difference.

Interpretation Introduction

Interpretation:

The $\text{1}\text{.00 }M$$\text{HCN}\text{and}\text{HF}$ solution $pH$ range and molar concentration has to be calculated under given concentration of both solutions.

Concept Information:

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.

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

Molarity:

Molarity is defined as the ratio of number of moles of solute to the number of liters of solution. The concentration of solute in solution is measured using a concentration unit called molarity.

The unit of molarity is moles per liter.

The equation for molarity can be given as,

$\text{Molarity (M)}\text{= }\frac{numberofmolesofsolute}{litresofsolution}$

Answer to Problem 15.152QP

The hydrogen cyanide $[HF]$ $pH=1.57$ and $1.00MHCN$ molar concentration is $\underset{\_}{2.2\times {10}^{-5}M}$

Explanation of Solution

The hydrogen fluoride $HF$ is much stronger acid than $HCN$ therefore, let us assume that $pH$ is determined by the ionization of $HF$

Hence we can write the equation corresponding to the reaction that takes place along with the equilibrium expression.

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

Construct an equilibrium table and determine, in terms of the unknown $x$ the equilibrium concentrations of the species in the equilibrium expression denoted by,

	${\text{HF}}_{\text{(aq)}}+{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{F}}^{\text{-}}{}_{\text{(aq)}}$
Initial $(M)$	$1.00$ -x $1.00-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The $k_a$ value $HF=7.1\times {10}^{-4}$

The equilibrium expression is,

$\begin{array}{l}{K}_a=\frac{[H_3{O}^{+}][F^{-}]}{[HF]}\\ 7.1\times {10}^{-4}=\frac{x^2}{(1.00-x)}\end{array}$

We assume that $x$ is small so, $(1.00-x)\approx 1.00$

$\begin{array}{l}7.1\times {10}^{-4}=\frac{x^2}{(1.00)}\\ x^2=(7.1\times {10}^{-4})(1.00)\\ x^2=7.1\times {10}^{-4}\\ Takingforsquarerootbothside\\ x=0.027M=[H_3{O}^{+}](\because 2.6645\times {10}^{-3})\\ pH=-\log (0.027)\\ pH=1.57\end{array}$

Next we calculate the equilibrium $[HCN]\approx 1.00M$

The $k_a$ value $HCN=4.9\times {10}^{-10}$

The equilibrium expression is,

$\begin{array}{l}{K}_a=\frac{[H_3{O}^{+}][C{N}^{-}]}{[HCN]}\\ 4.9\times {10}^{-10}=\frac{(0.027)[C{N}^{-}]}{1.00}[\because H_3{O}^{+}Molarconcentration=0.027]\\ C{N}^{-}=\frac{(4.9\times {10}^{-10})}{(0.027)}\\ C{N}^{-}=1.8\times {10}^{-}{}^{8}M\end{array}$

Hydrogen cyanide $HCN$ is a very weak acid, so at equilibrium $[HCN]\approx 1.00M$

The equilibrium expression denoted by

	${\text{HCN}}_{\text{(aq)}}+{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{CN}}^{\text{-}}{}_{\text{(aq)}}$
Initial $(M)$	$1.00$ -x $1.00-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

The equilibrium expression is,

$\begin{array}{l}{K}_a=\frac{[H_3{O}^{+}][C{N}^{-}]}{[HCN]}\\ 4.9\times {10}^{-10}=\frac{x^2}{(1.00-x)}\end{array}$

The $k_a$ value $HCN=4.9\times {10}^{-10}$

The  $x$ is small so, $(1.00-x)\approx 1.00$

$\begin{array}{l}4.9\times {10}^{-10}=\frac{x^2}{(1.00)}\\ x^2=(4.9\times {10}^{-10})(1.00)\\ x^2=4.9\times {10}^{-10}\\ Takingforsquarerootbothside\\ x=2.2\times {10}^{-5}M=[C{N}^{-}]\end{array}$

Conclusion

Finally we conclude that cyanide $[C{N}^{-}]$ is greater in the $1.00MHCN$ solution compared to the $1.00MHCN/1.00MHF$ solution.