Chapter 14, Problem 112QRT

(a)

Interpretation Introduction

Interpretation:

The $pH$ of a $0.0100M$ trichloroacetic acid solution has to be calculated.

Answer to Problem 112QRT

The $pH$ of a $0.0100M$ trichloroacetic acid solution is $2.01.$

Explanation of Solution

The equation for the equilibrium can be written as given below.

  $CC{l}_{3}COOH\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+CC{l}_{3}CO{O}^{-}\left(aq\right)$

The acid ionization constant for the above equation can be written as given below.

  $K_a=\frac{\left[H_3{O}^{+}\right]\left[CC{l}_{3}CO{O}^{-}\right]}{\left[CC{l}_{3}COOH\right]}$

A table can be set up as given below.

$\begin{array}{cccc}& CC{l}_{3}COOH\left(aq\right)& H_3{O}^{+}\left(aq\right)& CC{l}_{3}CO{O}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.010& 1.0\times {10}^{-7}& 0\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.010-x& x& x\end{array}$

The concentration of $H_3{O}^{+}$ comes from its auto ionization that is $1.0\times {10}^{-7}.$  It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of $x$ can be calculated.

  $\begin{array}{c}{K}_a=\frac{\left(x\right)\left(x\right)}{\left(0.010-x\right)}=3.0\times {10}^{-1}\\ \\ x^2=\left(3.0\times {10}^{-1}\right)\left(0.010-x\right)\\ \\ x^2+3.0\times {10}^{-1}x-3.0\times {10}^{-5}=0\\ \\ x=9.7\times {10}^{-3}M=\left[H_3{O}^{+}\right]\end{array}$

The $pH$ of the solution can be calculated as follows,

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(9.7\times {10}^{-3}\right)=\mathrm{2.01.}$

Therefore, the $pH$ of a $0.0100M$ trichloroacetic acid solution is $2.01.$

(b)

Interpretation Introduction

Interpretation:

It has to be decided how many times $H_3{O}^{+}$ concentration in $0.0100M$ trichloroacetic acid solution is greater than in a $0.0100M$ acetic acid solution.

Answer to Problem 112QRT

The concentration of $H_3{O}^{+}$ ions in acetic acid is $23$ times greater than that of trichloroacetic acid solution.

Explanation of Solution

The equation for the equilibrium can be written as given below.

  $C{H}_{3}COOH\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{H}_{3}CO{O}^{-}\left(aq\right)$

The acid ionization constant for the above equation can be written as given below.

  $K_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$

A table can be set up as given below.

$\begin{array}{cccc}& C{H}_{3}COOH\left(aq\right)& H_3{O}^{+}\left(aq\right)& C{H}_{3}CO{O}^{-}\left(aq\right)\\ Initialconc.\left(M\right)& 0.010& 1.0\times {10}^{-7}& 0\\ Changeinconc.\left(M\right)& -x& +x& +x\\ Equilibriumconc.\left(M\right)& 0.010-x& x& x\end{array}$

The concentration of $H_3{O}^{+}$ comes from its auto ionization that is $1.0\times {10}^{-7}.$  It can be assumed to be small compared to the change.

By substituting all the values in the above equation, the value of $x$ can be calculated.

  $K_a=\frac{\left(x\right)\left(x\right)}{\left(0.010-x\right)}=1.8\times {10}^{-5}$

Assuming x is very small, it can be written as $0.010-x\cong \mathrm{0.010.}$

Then,

  $\begin{array}{c}1.8\times {10}^{-5}=\frac{x^2}{0.010}\\ \\ x^2=\left(1.8\times {10}^{-5}\right)\left(0.010\right)\\ \\ x=4.2\times {10}^{-4}M=\left[H_3{O}^{+}\right]\end{array}$

So, the concentration of $H_3{O}^{+}$ in $0.0100M$ trichloroacetic acid and $0.0100M$ acetic acid is $9.7\times {10}^{-3}Mand4.2\times {10}^{-4}M$ respectively.

The number of times more $H_3{O}^{+}$ ions in the trichloroacetic acid than in acetic acid can be calculated as shown below.

  $\frac{9.7\times {10}^{-3}M}{4.2\times {10}^{-4}M}=23$

Therefore, the concentration of $H_3{O}^{+}$ ions in acetic acid is $23$ times greater than that of trichloroacetic acid solution.