Chapter 15, Problem 15.124QP

Interpretation Introduction

Interpretation:

The $\text{pH}$ of given $\text{1}{\text{.00×10}}^{\text{-7}}\text{M}\text{HCl}$ solution has to be determined.

Concept introduction:

$\text{pH}$ is an logarithmic expression to express a solution is acidic, basic or neutral.  $\text{pH}$ scale lies values between 1-14 and on which 7 is neutral, below 7 values are more acidic in nature and above 7 values are more basic in nature.

The $\text{pH}$ is equal to $\text{pH = -log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$

Answer to Problem 15.124QP

The $\text{pH}$ of given $\text{1}{\text{.00×10}}^{\text{-7}}\text{M}\text{HCl}$ solution is $\text{6}\text{.79}$

Explanation of Solution

The equilibrium of the given reaction can be represented as follows.

$\begin{array}{l}\underset{\_}{\text{Conc}{\text{. (M) H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}}\\ \text{from}\text{HCl}\text{1}{\text{.00×10}}^{\text{-7}}\text{0}\\ \text{from}\text{water}\text{1}{\text{.00×10}}^{\text{-7}}\text{1}{\text{.00×10}}^{\text{-7}}\\ \text{change}{}^{\text{-}}\text{x}{}^{\text{-}}\text{x}\\ \text{equlibrium}\text{2}{\text{.00×10}}^{\text{-7}}{}^{\text{-}}\text{x}\text{1}{\text{.00×10}}^{\text{-7}}{}^{\text{-}}\text{x}\end{array}$

${\text{K}}_{\text{w}}\text{=}\text{1}{\text{.00×10}}^{\text{-14}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{] = (2}{\text{.00×10}}^{\text{-7}}{}^{\text{-}}\text{x)(1}{\text{.00×10}}^{\text{-7}}{}^{\text{-}}\text{x)}$

Now, rearrange the above equation into standard quadratic form to solve for x.

${\text{x}}^{\text{2}}\text{- 3}\text{.00}\times {\text{10}}^{\text{-7}}\text{x + 1}\text{.00}\times {10}^{-14}=0$

$\text{x = }\frac{\text{-}\left(\text{-3}{\text{.00×10}}^{\text{-7}}\right)\text{ ± }\sqrt{{\left(\text{-3}{\text{.00×10}}^{\text{-7}}\right)}^{\text{2}}\text{-}\left(\text{4}\right)\left(\text{1}{\text{.00×10}}^{\text{-14}}\right)}}{\text{2}\left(\text{1}\right)}\text{ = }\frac{\text{3}{\text{.00×10}}^{\text{-7}}\text{± 2}{\text{.236×10}}^{\text{-7}}}{\text{2}}$

The positive root is $2.618\times {10}^{-7}$ and which is not allowed.  So, we must use negative root and which can be equal to $0.382\times {10}^{-7}.$

Hence, the concentration of hydronium ion and $\text{pH}$ are as follows

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = 2}{\text{.00 ×10}}^{\text{-7}}\text{- 0}{\text{.382×10}}^{\text{-7}}\text{=1}{\text{.618×10}}^{\text{-7}}\text{= 1}{\text{.62×10}}^{\text{-7}}\text{M}$

$\text{pH = -log }\left(1.618\times {10}^{-7}\right)=6.791=6.79$

Conclusion

From the given information the $\text{pH}$ of the given solution was determined as $\text{6}\text{.79}\text{.}$