Chapter U4.20, Problem 3E

Interpretation Introduction

Interpretation:

The given table needs to be completed by calculating missing hydrogen ion concentrations, pH and hydroxide ion concentrations.

Concept Introduction:

The pH is used to determine the concentration of hydroniumion. The formula is represented as follows:

  $\text{pH = - log }\left[{\text{H}}^{\text{+}}\right]$

Also,

  $\left[{\text{OH}}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]\text{ = Kw}\text{=}\text{1×}{\text{10}}^{\text{-14}}$

Thus,

  $\left[{\text{OH}}^{\text{-}}\right]\text{ = }\frac{\text{1×}{\text{10}}^{\text{-14}}}{\left[{\text{H}}^{\text{+}}\right]}$

Answer to Problem 3E

The complete table is as follows:

Explanation of Solution

• 0.010 M HCl

Concentration of ${\text{H}}^{+}$ = 0.010M

        $=1\times {10}^{-2}\text{M}$

The calculation of pH is shown below:

  $\begin{array}{l}\text{pH = -log}\left[{\text{H}}^{\text{+}}\right]\hfill \end{array}$

  $\begin{array}{c}\text{ = -log }\left[\text{0}\text{.010}\right]\hfill \end{array}$

  $\begin{array}{c}\text{ = 2}\hfill \end{array}$

The calculation of hydroxide ion is shown below:

  $\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{ = 1× 10}}^{\text{-14}}\hfill \end{array}$

  $\begin{array}{c}\text{0}\text{.010 }\left[{\text{OH}}^{\text{-}}\right]{\text{ = 1× 10}}^{\text{-14}}\hfill \end{array}$

  $\begin{array}{c}\left[{\text{OH}}^{\text{-}}\right]{\text{ = 1× 10}}^{\text{-12}}\text{M}\hfill \end{array}$

• 0.0010 M HCl

Concentration of ${\text{H}}^{+}$ = 0.0010 M

        $=1\times {10}^{-3}\text{M}$

The calculation of pH is shown below:

  $\begin{array}{l}\text{pH = -log}\left[{\text{H}}^{\text{+}}\right]\hfill \end{array}$

  $\begin{array}{c}\text{ = -log }\left[\text{0}\text{.0010}\right]\hfill \end{array}$

  $\begin{array}{c}\text{ = 3}\hfill \end{array}$

The calculation of hydroxide ion is shown below:

  $\begin{array}{l}{\text{[H}}^{\text{+}}\left]\right[{\text{OH}}^{\text{-}}{\text{] = 1× 10}}^{\text{-14}}\hfill \end{array}$

  $\begin{array}{c}\text{0}\text{.0010 }\left[{\text{OH}}^{\text{-}}\right]{\text{ = 1× 10}}^{\text{-14}}\hfill \end{array}$

     $\begin{array}{c}\left[{\text{OH}}^{\text{-}}\right]{\text{ = 1× 10}}^{\text{-11}}\text{M}\hfill \end{array}$

• 0.00000010 M HCl

Concentration of HCl = 0.00000010M

         $=1\times {10}^{-7}\text{M}$

The pH of the solution will be7 which means,a neutral range of pH but it is not true. While calculating the pH of the solution, the ${\text{H}}^{+}$ ions which are produced by the dissociation of water because the concentration of hydrogen ion is low and comparable to that produced from water.

To calculate the ${\text{H}}^{+}$ concentration from water, consider the following equilibrium in the presence of HCl.The dissociation of H₂O is suppressed by HCl due to the common ion effect.

  $\text{HCl }=\left[{\text{H}}^{+}\right]={10}^{-7}$

Total $\left[{\text{H}}^{+}\right]$ in the solution = $\left[x+{10}^{-7}\right]$

       $\begin{array}{c}\text{Kw = }\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\end{array}$

         $\begin{array}{c}\text{=}\left[\text{x}\text{+}{\text{10}}^{\text{-7}}\right]\left[\text{x}\right]\end{array}$

       $\begin{array}{c}{\text{10}}^{\text{-14}}\text{=}{\text{x}}^{\text{2}}\text{+}{\text{10}}^{\text{-7}}\text{x}\end{array}$

  $\begin{array}{c}{\text{x}}^{\text{2}}\text{+}{\text{10}}^{\text{-7}}\text{x}\text{-}{\text{10}}^{\text{-14}}\text{=}\text{0}\end{array}$

         $\begin{array}{c}\text{x}\text{=}\text{0}{\text{.62×10}}^{\text{-7}}\end{array}$

Total $\left[{\text{H}}^{+}\right]$ in the solution will be:

  $=0.62\times {10}^{-7}+{10}^{-7}$

Total $\left[{\text{H}}^{+}\right]$ in the solution = $1.62\times {10}^{-7}$

The calculation of pH is shown below:

  $\begin{array}{c}\text{pH of the solution = - log}\left[\text{1}{\text{.62 ×10}}^{\text{-7}}\right]\end{array}$

          $\begin{array}{c}\text{ =6}\text{.79}\end{array}$

The calculation of hydroxide ion is shown below:

  $\begin{array}{l}{\text{ [H}}^{\text{+}}\left]\right[{\text{OH}}^{\text{-}}{\text{] = 1× 10}}^{\text{-14}}\hfill \end{array}$

  $\begin{array}{c}\text{1}{\text{.62 ×10}}^{\text{-7}}\left[{\text{OH}}^{\text{-}}\right]{\text{ = 1× 10}}^{\text{-14}}\hfill \end{array}$

  $\begin{array}{c}\left[{\text{OH}}^{\text{-}}\right]\text{ = 6}{\text{.2× 10}}^{\text{-8}}\text{M}\hfill \end{array}$

• Concentration of HCl = 0.000000010M

$\left[{\text{H}}^{+}\right] =1\times {10}^{-8}\text{M}$ .The pH of acid cannot be more than 7 thus, again here contribution of H+ from water takes place.

Total [H⁺] in the solution = $\left[x+{10}^{-8}\right]$

  $\begin{array}{l}\text{ Kw = }\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\end{array}$

  $\begin{array}{c}\text{ =}\left[\text{x}\text{+}{\text{10}}^{\text{-8}}\right]\left[\text{x}\right]\end{array}$

  $\begin{array}{c}{\text{ 10}}^{\text{-14}}\text{=}{\text{x}}^{\text{2}}\text{+}{\text{10}}^{\text{-8}}\text{x}\end{array}$

  $\begin{array}{c}{\text{x}}^{\text{2}}\text{+}{\text{10}}^{\text{-8}}\text{x}\text{-}{\text{10}}^{\text{-14}}\text{=}\text{0}\end{array}$

  $\begin{array}{c}\text{ x}\text{=9}{\text{.5×10}}^{\text{-8}}\end{array}$

The concentration of H+ will be $\text{1}{\text{.05×10}}^{\text{-7}}$ that is approximately $\text{1}{\text{.0×10}}^{\text{-7}}$ .

The calculation of pH is shown below:

  $\begin{array}{l}\text{pH = -log}\left[{\text{H}}^{\text{+}}\right]\hfill \end{array}$

  $\begin{array}{c}\text{ = -log [ 1}\text{.05}\times {\text{10}}^{\text{-7}}\text{]}\hfill \end{array}$

  $\begin{array}{c}\begin{array}{l}\text{ = -0}\text{.0212+7}\text{.00}\end{array}\hfill \end{array}$

  $\begin{array}{c}\text{ =6}\text{.98}\end{array}$

The value is approximately 7.

The calculation of hydroxide ion is shown below:

  $\begin{array}{l}{\text{ K}}_{\text{w}}{\text{= [H}}^{\text{+}}{\text{]×[OH}}^{\text{-}}\text{]}\end{array}$

  $\begin{array}{c}{\text{[OH}}^{\text{-}}\text{] =}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}^{\text{+}}\text{]}}\end{array}$

  $\begin{array}{c}{\text{[OH}}^{\text{-}}\text{] =}\frac{{\text{10}}^{\text{-14}}}{\text{1}\text{.05}\times {\text{10}}^{\text{-7}}}\end{array}$

  $\begin{array}{c}{\text{[OH}}^{\text{-}}\text{] =}9.5\times {\text{10}}^{\text{-8}}\end{array}$

This is also approximately $1.0\times {\text{10}}^{\text{-7}}$ .

• NaCl is a strong electrolyte that is not dissociated into water. Therefore, the concentration of H+ and OH- is 1× 10-⁷M.

pH = 7

The concentration of OH- = 1× 10-7

The calculation of hydronium ion is shown below:

  $\begin{array}{l}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{ = 1× 10}}^{\text{-14}}\hfill \end{array}$

  $\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]\left({\text{1× 10}}^{\text{-7}}\right){\text{= 1× 10}}^{\text{-14}}\hfill \end{array}$

  $\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]{\text{= 1× 10}}^{\text{-7}}\text{M}\hfill \end{array}$

The calculation of pH is shown below:

  $\begin{array}{l}\text{pH = -log}\left[{\text{H}}^{\text{+}}\right]\hfill \end{array}$

  $\begin{array}{c}{\text{ = -log [1× 10}}^{\text{-7}}\text{]}\hfill \end{array}$

  $\begin{array}{c}\text{ = 7}\hfill \end{array}$

But the pH of the base should be greater than 7 thus, the contribution of hydroxide ion should also be considered. The total hydroxide solution will be:

  $\left[{\text{OH}}^{-}\right]=1\times {10}^{-7}+y$

Now,

  $\begin{array}{l}\text{ Kw = }\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\end{array}$

  $\begin{array}{c}\text{ =}\left[\text{y}\right]\left[\text{y}\text{+}{\text{10}}^{\text{-7}}\right]\end{array}$

  $\begin{array}{c}{\text{ 10}}^{\text{-14}}\text{=}{\text{y}}^{\text{2}}\text{+}{\text{10}}^{\text{-7}}\text{y}\end{array}$

  $\begin{array}{c}{\text{y}}^{\text{2}}\text{+}{\text{10}}^{\text{-7}}\text{y-}{\text{10}}^{\text{-14}}\text{=}\text{0}\end{array}$

   $\begin{array}{c}\text{ y}\text{=}\text{0}{\text{.62×10}}^{\text{-7}}\end{array}$

Thus, the concentration of hydroxide ion is as follows:

  $\left[{\text{OH}}^{-}\right]=1\times {10}^{-7}+0.62\times {10}^{-7}=1.62\times {10}^{-7}$

Thus,

  $\left[{\text{H}}^{+}\right]=\frac{{10}^{-14}}{1.62\times {10}^{-7}}=6.2\times {10}^{-8}$

Therefore, the pH of the solution is:

  $pH=-\log \left[H^{+}\right]=-\log \left(6.2\times {10}^{-8}\right)=8-0.79=7.21$

• The molarity of NaOH is 0.0000010 M thus, $\left[{\text{OH}}^{-}\right]={10}^{-6}\text{ M}$ .

Now, hydrogen ion concentration is calculated as follows:

  $\left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{{10}^{-6}}={10}^{-8}$

Thus, pH can be calculated as follows:

  $pH=-\log \left[H^{+}\right]=-\log \left({10}^{-8}\right)=8$

• The molarity of NaOH is 0.000010 M thus, $\left[{\text{OH}}^{-}\right]=1\times {10}^{-5}\text{ M}$ .

Now, hydrogen ion concentration is calculated as follows:

  $\left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{{10}^{-5}}=1\times {10}^{-9}$

Thus, pH can be calculated as follows:

  $pH=-\log \left[H^{+}\right]=-\log \left({10}^{-9}\right)=9$