Chapter 6, Problem 6H.31E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{50}\text{.0}\text{mL}$ solution made by $\text{0}\text{.122}\text{g}$ of phosphorous acid dissolved in water has to be estimated.

Concept Introduction:

Buffer:

An aqueous solution of mixture of weak acid and its salt, or weak base and its conjugate acid solution is called “buffer”. It is resistant to change in $p^H$.

Acid Buffer:

A mixture of weak acid and its corresponding salt solution is called “Acidic Buffer”.

Base Buffer:

A mixture weak base and its corresponding salt solution forms a buffer is called “Basic Buffer”.

Explanation of Solution

The Henderson-Hassel batch equation for buffer is given by,

  ${\text{pH=pK}}_{\text{a}}\text{+log}\frac{{\left[\text{Base}\right]}_{\text{initial}}}{{\left[\text{Acid}\right]}_{\text{initial}}}$

In the case of acidic buffer,

${\left[\text{Base}\right]}_{\text{initial}}$= initial concentration of conjugate base

${\left[\text{Acid}\right]}_{\text{initial}}$= initial concentration of weak acid

Therefore, the above equation becomes,

  $\begin{array}{l}{\text{pH=pK}}_{\text{a}}\text{+log1}\\ \\ {\text{pH=pK}}_{\text{a}}\end{array}$

The dissociation reaction for the given Phosphorous acid is given by,

  ${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}{}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

Concentration of starting phosphorous acid solution,

  $\text{concentration=}\frac{\text{weight}}{\text{MW}\text{.}\text{}\text{of}\text{}{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{×volume}}$

  $=\frac{0.122g}{81.99g.mo{l}^{-1}\times 0.050L}$

  $=0.0298mol.L^{-1}$

The equilibrium constant and its values as follows:

  ${\text{K}}_{\text{a}}{\text{(H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{)=}\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\right]}$

  ${\text{K}}_{\text{a}}{\text{(H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{)=1}{\text{.0×10}}^{\text{-2}}$

  $\begin{array}{l}1.0\times {10}^{-2}=\frac{x\times x}{0.0298-x}\\ \\ (0.0298-x)\times 0.01=x^2\\ \\ x^2+0.01x-0.000298=0\end{array}$

Finally, $x=0.013$

Calculation for pH:

  $\begin{array}{l}\text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{=-log(0}\text{.013)}\\ \\ \text{=1}\text{.89}\end{array}$

Therefore, the pH of the initial solution is 1.89

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{5}\text{.00}\text{mL}$ solution made by $\text{0}\text{.175}\text{m}$ of $\text{NaOH}$ added to phosphoric acid solution has to be estimated.

Explanation of Solution

The reaction between phosphorous acid and NaOH is as follows:

  ${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The total number of moles of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}$ and ${\text{OH}}^{\text{-}}$ can be calculated as

  $\begin{array}{l}\text{=0}\text{.0298mol}{\text{.L}}^{\text{-1}}\text{×0}\text{.0500L}\\ \\ \text{=1}{\text{.49×10}}^{\text{-3}}\text{mol}\end{array}$

The ${\text{OH}}^{\text{-}}$ will be

  $\begin{array}{l}=0.175mol.L^{-1}\times 0.00500L\\ \\ =8.76\times {10}^{-4}mol\end{array}$

From the above calculations, the remaining number of moles of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}$ in solution is $6.14\times {10}^{-4}mol$.

Total volume of the solution is 0.055L

Calculation for remaining concentration of${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}$:

  $\begin{array}{l}=\frac{6.14\times {10}^{-4}mol}{0.055L}\\ \\ =0.011mol.L^{-1}\end{array}$

Calculation for remaining concentration of${\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}^{\text{-}}$:

  $\begin{array}{l}=\frac{8.76\times {10}^{-4}mol}{0.055L}\\ \\ =0.016mol.L^{-1}\end{array}$

Number of moles of Arsenic acid in the starting solution can be calculated as follows,

Number of moles = $Molarity\times volume$

  $\begin{array}{l}=0.010mol.L^{-1}\times 0.0050L\\ \\ =5.0\times {10}^{-5}mol\end{array}$

The expression for equilibrium constant and its value as follows,

    ${\text{K}}_{\text{a}}{\text{(H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{)=}\frac{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\right]}$

    ${\text{K}}_{\text{a}}{\text{(H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{)=1}{\text{.0×10}}^{\text{-2}}$

    $\begin{array}{l}1.0\times {10}^{-2}=\frac{(0.016+x)\times x}{0.011-x}\\ \\ (0.011-x)\times 0.01=0.016+x^2\\ \\ x^2+0.026x-0.00011=0\end{array}$

    $x=1.08\times {10}^{-2}$

Thus, ‘x’ is the equilibrium concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $1.08\times {10}^{-2}mol.L^{-1}$.

Calculation for pH:

    $\begin{array}{l}\text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{=-log(1}{\text{.08×10}}^{\text{-2}}\text{)}\\ \\ \text{=1}\text{.96}\end{array}$

Therefore, pH of the solution is 1.96

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{5}\text{.00}\text{mL}$ solution made by $\text{0}\text{.175}\text{m}$ of $\text{NaOH}$ added to phosphoric acid solution in addition to the solution in part (b) has to be estimated.

Explanation of Solution

Total number of moles of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}$ is $1.49\times {10}^{-3}mol$

After addition of NaOH, the total number of moles of ${\text{OH}}^{\text{-}}$ will be:

    $\begin{array}{l}\text{=8}{\text{.76×10}}^{\text{-4}}\text{mol+8}{\text{.76×10}}^{\text{-4}}\text{mol}\\ \\ \text{=1}{\text{.75×10}}^{\text{-3}}\text{mol}\end{array}$

Calculation for remaining moles of ${\text{OH}}^{\text{-}}$ and its concentration:

    $\begin{array}{l}\text{Moles=1}{\text{.75×10}}^{\text{-3}}\text{mol-1}{\text{.49×10}}^{\text{-4}}\text{mol}\\ \\ \text{=2}{\text{.62×10}}^{\text{-4}}\text{mol}\end{array}$

Total volume of the solution is 0.060L. The concentration will be

    $\begin{array}{l}=\frac{1.49\times {10}^{-3}mol}{0.060L}\\ \\ =2.48\times {10}^{-2}mol.L^{-1}\end{array}$

Thus the ${\text{OH}}^{\text{-}}$ concentration,

    $\begin{array}{l}=\frac{2.62\times {10}^{-4}mol}{0.060L}\\ \\ =4.37\times {10}^{-3}mol.L^{-1}\end{array}$

The dissociation concentration of phosphorous acid is small, so the second dissociation is ignored. Calculate the ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ concentration in solution,

    $\begin{array}{l}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]{\text{=10}}^{\text{-4}}\\ \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{{\text{10}}^{\text{-4}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \\ \text{=}\frac{{\text{10}}^{\text{-4}}}{\left[\text{4}{\text{.37×10}}^{\text{-3}}\right]}\end{array}$

    $=2.29\times {10}^{-12}$

Calculation for pH:

    $\begin{array}{l}\text{pH=-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{=-log(2}{\text{.29×10}}^{\text{-12}}\text{)}\\ \\ \text{=11}\text{.6}\end{array}$

Therefore, pH of the solution is 11.6

Conclusion:

The $\text{pH}$ of $\text{5}\text{.00}\text{mL}$ solution made by $\text{0}\text{.175}\text{m}$ of $\text{NaOH}$ added to phosphoric acid solution were estimated as above.