91Ó°ÊÓ

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:

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

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

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

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

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

We have to write the acid base reaction:

role="math" localid="1663671130867" ${\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)⇌{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

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 = $(0.8\mathrm{mol}-0.5\mathrm{mol}=0.3\mathrm{mol}).$

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

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

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 =role="math" localid="1663672569283" $(0.5\mathrm{mol}-0.3\mathrm{mol}=0.2\mathrm{mol}).$

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

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

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.