Q152CP - Chemistry: Molecular Nature Of Matter And Change

Question

A chemist makes four successive  dilutions of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{ }\mathbf{MHCl}$ . Calculate the pH of the original solution and of each diluted solution (through) $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{ }\mathbf{MHCl}$.

Step-by-Step Solution

Answer

The pH of the original solution is 5.00.

The pH of the first dilution is 6.00.

The pH of the second dilution is 6.70.

The pH of the third dilution is 6.96.

The pH of the fourth dilution is 7.00.

Step 1: Original solution

Use the ratio  _$(1:10)$as volume in the equation below.

${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}$

_$\text{HCl}$ is a strong acid and will dissociate completely. Therefore, $\left[{\text{H}}_3{\text{O}}^{+}\right]=\left[\text{HCl}\right]$.

The pH of the original solution is calculated below.

$\begin{array}{rcl}& & \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\\ & & \text{ =-log}\left(1.0\times {10}^{-5} \mathrm{M}\right)\\ & & \text{ =5.00}\end{array}$

Hence, the original solution is 5.00.

Step 2: First dilution

$$\begin{gathered} {\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{ = } {\text{M}}_{\text{2}}\text{V}\left(1.0\times {10}^{-5} \mathrm{M}\right) \\ {\text{ =M}}_{\text{2}}{\text{(10)M}}_{\text{2}} \\ =\frac{\left(1.0\times {10}^{-5}\mathrm{M}\right){\displaystyle \left(1\right)}}{10}{\text{M}}_{\text{2}} \\ =1.0\times {10}^{-6}\mathrm{M} \end{gathered}$$

Then solve for the pH as shown below.

$$\begin{gathered} \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = } \text{ - log}\left(1.0\times {10}^{-6}M\right) \\ \text{=6.00} \end{gathered}$$

Hence, the first dilution is 6.00.

Step 3: Second dilution

$$\begin{gathered} {\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = M}}_{\text{2}}{\text{V}}_{\text{2}}\left(1.0\times {10}^{-6} \mathrm{M}\right) \\ {\text{ = M}}_{\text{2}}{\text{(10)M}}_{\text{2}} \\ =\frac{\left(1.0\times {10}^{-6} \mathrm{M}\right){\displaystyle \left(1\right)}}{10} \\ {\text{M}}_{\text{2}}\text{ = }1.0\times {10}^{-7} \mathrm{M} \end{gathered}$$

Then solve for the pH.

$$\begin{gathered} \text{pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = - log}\left(1.0\times {10}^{-7}\mathrm{M}\right) \\ \text{=7.00} \end{gathered}$$

HCl is a strong acid so going beyond or having a $\text{pH} \text{ = } \text{7}$  is not ideal. In this case, the $\left[{\text{H}}_3{\text{O}}^{+}\right]$ from the auto ionization of water should also be added to the total $\left[{\text{H}}_3{\text{O}}^{+}\right]$ then solve for the total pH.

${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{total }}\text{ = } {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{HCl}}\text{ + }{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{{\text{H}}_{\text{2}}\text{O}}$

$$\begin{gathered} \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = } \text{ - log}\left(2.0\times {10}^{-7}\mathrm{M}\right) \\ =6.70 \end{gathered}$$

Hence, the second dilution is 6.70.

Step 4: Third dilution

$$\begin{gathered} {\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{ = } {\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\left(1.0\times {10}^{-7}\mathrm{M}\right)\left(1\right) \\ {\text{ =M}}_{\text{2}}{\text{(10)M}}_{\text{2}} \\ \text{ =}\frac{\left(1.0\times {10}^{-7}\mathrm{M}\right)}{10}{\text{M}}_{\text{2}} \\ =1.0\times {10}^{-8}\mathrm{M} \end{gathered}$$

Then solve for the pH

$$\begin{gathered} \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = } \text{ - log}\left(1.0\times {10}^{-8}\mathrm{M}\right) \\ \text{=8.00} \end{gathered}$$

Solve for the total pH.

$$\begin{gathered} {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{total }}={\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{HCl}}\text{ + }{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{{\text{H}}_{\text{2}}\text{O}} \\ {\text{=1.0×10}}^{-8}\text{M+}{\mathrm{1.0\times 10}}^{-7}M{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{total }} \\ ={\mathrm{1.0\times 10}}^{-7}M\text{pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = } \text{ - log}\left({\mathrm{1.0\times 10}}^{-7}M\right) \\ \text{ pH} \text{ = } \text{6.96.} \end{gathered}$$

Hence, the third dilution is 6.96.

Step 5: Fourth dilution

$$\begin{gathered} {\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{ = } {\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\left(1.0\times {10}^{-8}\mathrm{M}\right)\left(1\right) \\ \text{ = } {\text{M}}_{\text{2}}{\text{(10)M}}_{\text{2}} \\ =\frac{\left(1.0\times {10}^{-8}\mathrm{M}\right){\displaystyle \left(1\right)}}{10}{\text{M}}_{\text{2}} \\ =1.0\times {10}^{-9}\mathrm{M} \end{gathered}$$

Then solve for the pH.

$$\begin{gathered} \text{pH} \text{ = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = } \text{ - log}\left(1.0\times {10}^{-9}\mathrm{M}\right) \\ \text{=9.00} \end{gathered}$$

        

_HCl is a strong acid so going beyond or having a pH=7 is not ideal. In this case, the $\left[{\text{H}}_3{\text{O}}^{+}\right]$ from the auto ionization of water should also be added to the total $\left[{\text{H}}_3{\text{O}}^{+}\right]$  then solve for the total pH.

$$\begin{gathered} {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{total }}\text{ = }{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{HCl}}\text{ + }{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{{\text{H}}_{\text{2}}\text{O}} \\ {\text{=1.0×10}}^{-9}\text{M +}1{.0\times 10}^{-7}\mathrm{M}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}_{\text{total }} \\ =1{.01\times 10}^{-7}\mathrm{M} \\ \text{ pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = } \text{ - log}\left(1{.01\times 10}^{-7}\mathrm{M}\right) \\ \text{ =7.00} \end{gathered}$$

Hence, the fourth dilution is 7.00.