Q19.113CP - Chemistry: Molecular Nature Of Matter And Change

Question

Phosphate systems form essential buffers in organisms. Calculate the pH of a buffer made by dissolving  $\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{ }\mathbf{mol}\mathbf{of}\mathbf{NaOH}\mathbf{in}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{Lof}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{M}\mathbf{ }\mathbf{ }{\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\mathbf{.}$

Step-by-Step Solution

Answer

The pH of the buffer solution is 7.38.

Concept introduction

Buffer solutions can withstand changes in pH after addition of acids and bases. In buffer solutions the conjugate acid is in equilibrium with the acid. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation. For an acid HA, the pH of the buffer solution formed is given by:

$\mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

First neutralization reaction

Volume of  ${\text{H}}_3{\text{PO}}_{\text{4}}=0.50 \text{L}$   

Concentration of  ${\text{H}}_3{\text{PO}}_{\text{4}}=\text{1 M}$ 

Number of moles of ${\text{H}}_3{\text{PO}}_{\text{4}}$ present $= 0.50 \mathrm{L}\times 1.0 \mathrm{mol}/\mathrm{L} = 0.5 \mathrm{mole}$

Number of moles of$\text{NaOH}$added  $= 0.8 \text{mole}$

We have to write the acid base reaction:

${\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}_2{\mathrm{PO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ 

Because ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{ and NaOH}$ react in a 1:1 ratio,

$\text{0.5} \text{mole}$ of  $\text{NaOH}$ react with $\text{0.5} \text{mole}$ of  ${\text{H}}_3{\text{PO}}_{\text{4}}$.

Number of moles of $\text{NaOH}$ left in the solution = $(0.8 \mathrm{mol}-0.5 \mathrm{mol}=0.3 \mathrm{mol}).$

Second neutralization reaction

$0.3 \mathrm{mol} \mathrm{NaOH}$The second neutralization reaction is:

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}{\text{(aq) + OH}}^{\text{ - }}\text{(aq)}\rightleftharpoons {\text{HPO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$ 

This$0.3 \mathrm{mol} \mathrm{NaOH}$reacts with: ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ 

Number of moles of  ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ left in the solution = $(0.5 \mathrm{mol}-0.3 \mathrm{mol}=0.2 \mathrm{mol}).$ 

Number of moles of  ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$ formed in the solution =$(0.5 \mathrm{mol}-0.3 \mathrm{mol}=0.2 \mathrm{mol}).$

Calculation of pH by using the Henderson-Hasselbalch equation

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=6.3\times {10}^{-8} \\ {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ {\mathrm{pK}}_{\mathrm{a}}=-\log 6.3\times {10}^{-8} \\ {\mathrm{pK}}_{\mathrm{a}}=7.20 \end{gathered}$$

The Henderson-Hasselbalch equation can now be used:

 $$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \mathrm{pH}=7.20+\log \frac{0.3}{0.2} \\ \mathrm{pH}=7.38 \end{gathered}$$

Therefore, the value of pH is 7.38.