Chapter 18, Problem 18.152P

Interpretation Introduction

Interpretation:

A chemist makes four successive ten-fold dilutions of $\text{1}{\text{.0×10}}^{\text{-5}}\text{ M HCl}$.  The $\text{pH}$ of the original solution and of each diluted solution has to be calculated (through $\text{1}{\text{.0×10}}^{\text{-9}}\text{ M HCl}$).

Concept Introduction:

Molarity:

The number of moles of the substance per litre of liquid is said to be the molarity of the solution.

Molarity can be used to calculate the concentration of solutions.  The expression used is

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \\ \text{where,}{\text{M}}_{\text{1}}\text{ = molarity of initial solution}\\ \\ {\text{M}}_{\text{2}}\text{ = molarity of diluted solution}\\ \\ {\text{V}}_{\text{1}}\text{ = volume of initial solution}\\ \\ {\text{V}}_{\text{2}}\text{ = volume of diluted solution}\end{array}$

Explanation of Solution

As $\text{HCl}$ is a strong acid, it dissociates completely in water.  The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is same as the concentration of $\text{HCl}$.

  ${\text{[HCl] = [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

The $\text{pH}$ of the original solution can be calculated as

  $\begin{array}{l}\text{pH = -log (1}{\text{.0×10}}^{\text{-5}}\text{)}\\ \\ \text{pH}\text{=}\text{5}\text{.00}\end{array}$

Dilution-1:

The original solution of $\text{HCl}$ is diluted to $\text{10 mL}$.  The concentration can be calculated as follows:

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \\ \text{(1}{\text{.0×10}}^{\text{-5}}\text{ M)(1}\text{.0 mL) = x (10 mL)}\\ \\ {{\text{[H}}_{\text{3}}}_{{\text{O}}^{\text{+}}}\text{ = 1}{\text{.0×10}}^{\text{-6}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The $\text{pH}$ of the solution can be calculated as

  $\begin{array}{l}\text{pH = -log (1}{\text{.0×10}}^{\text{-6}}\text{)}\\ \\ \text{pH}\text{=}\text{6}\text{.00}\end{array}$

Dilution-2:

The diluted solution of $\text{HCl}$ is diluted to $\text{10 mL}$.  The concentration can be calculated as follows:

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \\ \text{(1}{\text{.0×10}}^{\text{-6}}\text{ M)(1}\text{.0 mL) = x (10 mL)}\\ \\ {{\text{[H}}_{\text{3}}}_{{\text{O}}^{\text{+}}}\text{ = 1}{\text{.0×10}}^{\text{-7}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The concentration of strong acid is close to the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in water ionization.  So, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the solution does not be the same as the initial concentration.

Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has to be calculated based on the ionization of water equilibrium.

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}{\text{K}}_{\text{w}}\text{=1}{\text{.0×10}}^{\text{-14}}\text{at}{\text{25}}^{\text{°}}\text{C}$

The dilution-2 gives the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ as $\text{1}{\text{.0×10}}^{\text{-7}}\text{M}$.  A reaction table is set up as follows to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}\text{Concentration(M)}{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial}-\text{1}{\text{.0×10}}^{\text{-7}}\text{0}\\ \\ \text{Change}-\text{+x}\text{+x}\end{array}}\\ \\ \text{Equilibrium}-\text{1}{\text{.0×10}}^{\text{-7}}\text{+x}\text{x}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ \text{1}{\text{.0×10}}^{\text{-14}}\text{ = (1}{\text{.0×10}}^{\text{-7}}\text{+ x)(x)}\end{array}$

A quadratic equation is given as

  $\begin{array}{l}{\text{x}}^{\text{2}}\text{ + 1}{\text{.0×10}}^{\text{-7}}\text{ x - 1}{\text{.0×10}}^{\text{-14}}\text{ = 0}\\ \\ \text{a=1}\text{b=1}{\text{.0×10}}^{\text{-7}}\text{c=-1}{\text{.0×10}}^{\text{-14}}\\ \\ \text{x = }\frac{\text{-1}{\text{.0×10}}^{\text{-7}}\text{±}\sqrt{{\text{(1}{\text{.0×10}}^{\text{-7}}\text{)}}^{\text{2}}\text{- 4(1)(-1}{\text{.0×10}}^{\text{-14}}\text{)}}}{\text{2(1)}}\\ \\ \text{x = 6}{\text{.18034×10}}^{\text{-8}}\end{array}$

Using the $\text{x}$ value, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is calculated as

  $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = (1}{\text{.0×10}}^{\text{-7}}\text{+ x) M}\\ \\ \text{= (1}{\text{.0×10}}^{\text{-7}}\text{+ 6}{\text{.18034×10}}^{\text{-8}}\text{)M}\\ \\ \text{= 1}{\text{.618034×10}}^{\text{-7}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The $\text{pH}$ of the solution can be calculated as

  $\begin{array}{l}{\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \\ \text{pH = -log (1}{\text{.618034×10}}^{\text{-7}}\text{)}\\ \\ \text{pH}\text{=}\text{6}\text{.79101 = 6}\text{.79}\end{array}$

Dilution-3:

The diluted solution of $\text{HCl}$ is diluted to $\text{10 mL}$.  The concentration can be calculated as follows:

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \\ \text{(1}{\text{.0×10}}^{\text{-7}}\text{ M)(1}\text{.0 mL) = x (10 mL)}\\ \\ {{\text{[H}}_{\text{3}}}_{{\text{O}}^{\text{+}}}\text{ = 1}{\text{.0×10}}^{\text{-8}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The dilution-3 gives the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ as $\text{1}{\text{.0×10}}^{\text{-8}}\text{M}$.  A reaction table is set up as follows to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}\text{Concentration(M)}{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial}-\text{1}{\text{.0×10}}^{\text{-8}}\text{0}\\ \\ \text{Change}-\text{+x}\text{+x}\end{array}}\\ \\ \text{Equilibrium}-\text{1}{\text{.0×10}}^{\text{-8}}\text{+x}\text{x}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ \text{1}{\text{.0×10}}^{\text{-14}}\text{ = (1}{\text{.0×10}}^{\text{-8}}\text{+ x)(x)}\end{array}$

A quadratic equation is given as

  $\begin{array}{l}{\text{x}}^{\text{2}}\text{ + 1}{\text{.0×10}}^{\text{-8}}\text{ x - 1}{\text{.0×10}}^{\text{-14}}\text{ = 0}\\ \\ \text{a=1}\text{b=1}{\text{.0×10}}^{\text{-8}}\text{c=-1}{\text{.0×10}}^{\text{-14}}\\ \\ \text{x = }\frac{\text{-1}{\text{.0×10}}^{\text{-8}}\text{±}\sqrt{{\text{(1}{\text{.0×10}}^{\text{-8}}\text{)}}^{\text{2}}\text{- 4(1)(-1}{\text{.0×10}}^{\text{-14}}\text{)}}}{\text{2(1)}}\\ \\ \text{x = 9}{\text{.51249×10}}^{\text{-8}}\end{array}$

Using the $\text{x}$ value, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is calculated as

  $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = (1}{\text{.0×10}}^{\text{-8}}\text{+ x) M}\\ \\ \text{= (1}{\text{.0×10}}^{\text{-8}}\text{+ 9}{\text{.51249×10}}^{\text{-8}}\text{)M}\\ \\ \text{= 1}{\text{.051249×10}}^{\text{-7}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The $\text{pH}$ of the solution can be calculated as

  $\begin{array}{l}{\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \\ \text{pH = -log (1}{\text{.051249×10}}^{\text{-7}}\text{)}\\ \\ \text{pH}\text{=}\text{6}\text{.97829 = 6}\text{.98}\end{array}$

Dilution-4:

The diluted solution of $\text{HCl}$ is diluted to $\text{10 mL}$.  The concentration can be calculated as follows:

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \\ \text{(1}{\text{.0×10}}^{\text{-8}}\text{ M)(1}\text{.0 mL) = x (10 mL)}\\ \\ {{\text{[H}}_{\text{3}}}_{{\text{O}}^{\text{+}}}\text{ = 1}{\text{.0×10}}^{\text{-9}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The dilution-4 gives the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ as $\text{1}{\text{.0×10}}^{\text{-9}}\text{M}$.  A reaction table is set up as follows to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}\text{Concentration(M)}{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial}-\text{1}{\text{.0×10}}^{\text{-9}}\text{0}\\ \\ \text{Change}-\text{+x}\text{+x}\end{array}}\\ \\ \text{Equilibrium}-\text{1}{\text{.0×10}}^{\text{-9}}\text{+x}\text{x}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \\ \text{1}{\text{.0×10}}^{\text{-14}}\text{ = (1}{\text{.0×10}}^{\text{-9}}\text{+ x)(x)}\end{array}$

A quadratic equation is given as

  $\begin{array}{l}{\text{x}}^{\text{2}}\text{ + 1}{\text{.0×10}}^{\text{-9}}\text{ x - 1}{\text{.0×10}}^{\text{-14}}\text{ = 0}\\ \\ \text{a=1}\text{b=1}{\text{.0×10}}^{\text{-9}}\text{c=-1}{\text{.0×10}}^{\text{-14}}\\ \\ \text{x = }\frac{\text{-1}{\text{.0×10}}^{\text{-9}}\text{±}\sqrt{{\text{(1}{\text{.0×10}}^{\text{-9}}\text{)}}^{\text{2}}\text{- 4(1)(-1}{\text{.0×10}}^{\text{-14}}\text{)}}}{\text{2(1)}}\\ \\ \text{x = 9}{\text{.95012×10}}^{\text{-8}}\end{array}$

Using the $\text{x}$ value, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is calculated as

  $\begin{array}{l}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = (1}{\text{.0×10}}^{\text{-9}}\text{+ x) M}\\ \\ \text{= (1}{\text{.0×10}}^{\text{-9}}\text{+ 9}{\text{.95012×10}}^{\text{-8}}\text{)M}\\ \\ \text{= 1}{\text{.005012×10}}^{\text{-7}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The $\text{pH}$ of the solution can be calculated as

  $\begin{array}{l}{\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \\ \text{pH = -log (1}{\text{.005012×10}}^{\text{-7}}\text{)}\\ \\ \text{pH}\text{=}\text{6}\text{.9978 = 7}\text{.00}\end{array}$

The $\text{pH}$ of the original and diluted solution are

$\text{pH}$ of the original solution: $\text{5}\text{.00}$

$\text{pH}$ of the dilution-1: $\text{6}\text{.00}$

$\text{pH}$ of the dilution-2: $\text{6}\text{.79}$

$\text{pH}$ of the dilution-3: $\text{6}\text{.98}$

$\text{pH}$ of the dilution-4: $\text{7}\text{.00}$