Q19.22P - Chemistry: Molecular Nature Of Matter And Change

Question

A buffer consists of 0.50 M ${\mathbf{\text{NaH}}}_{\mathbf{\text{2}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}$ and 0.40 M ${\mathbf{\text{Na}}}_{\mathbf{\text{2}}}{\mathbf{\text{HPO}}}_{\mathbf{\text{4}}}$. Phosphoric acid is a triprotic acid (${\mathbf{\text{K}}}_{\mathbf{\text{a1}}}{\mathbf{\text{ = 7.2×10}}}^{\mathbf{-}\mathbf{3}}{\mathbf{\text{,K}}}_{\mathbf{\text{a2}}}{\mathbf{\text{ = 6.3×10}}}^{\mathbf{-}\mathbf{8}}$ and ${\mathbf{\text{K}}}_{\mathbf{\text{a3}}}{\mathbf{\text{ = 4.2×10}}}^{\mathbf{-}\mathbf{1}}$ ).

(a) Which ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$ value is most important to this buffer? 

(b) What is the buffer pH?

Step-by-Step Solution

Answer

(a) The value important to this buffer is${\text{K}}_{\text{a2}}$.

(b) The buffer pH obtained as is 7.10.

Step 1: Phosphoric acid buffer

The basicity of a phosphoric acid is three. The mixture of mono sodium phosphate and disodium phosphate forms a buffer mixture.

Step 2: Subpart (a)

First, write the phosphoric acid dissociation process.

$$\begin{gathered} {\text{ H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }} \\ {\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{HPO}}_{\text{4}}^{\text{2 - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }} \\ {\text{HPO}}_{\text{4}}^{\text{2 - }}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{PO}}_{\text{4}}^{\text{3 - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }} \end{gathered}$$

Write the sodium salt's dissolution.

$$\begin{gathered} {\text{NaH}}_{\text{2}}{\text{PO}}_{\text{4}}\to {\text{Na}}^{\text{ + }}{\text{ + H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }} \\ {\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}\to {\text{2Na}}^{\text{ + }}{\text{ + HPO}}_{\text{4}}^{\text{2 - }} \end{gathered}$$

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}$and ${\text{HPO}}_{\text{4}}^{\text{ - }}$were both generated by salts. Both sides of the phosphoric acid's second dissociation include these chemical entities. 

 Therefore, for this buffer solution, we'll utilize ${\text{K}}_{\text{a2}}$.

 

Step 3: Subpart (b)

The given dihydrogen phosphate and hydrogen phosphate concentrations are:

$$\begin{gathered} {\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}\text{]=0.50M} \\ {\text{[HPO}}_{\text{4}}^{\text{2 - }}\text{]=0.40M} \end{gathered}$$

To find the pH, we may utilize the Henderson-Hasselbalch equation. First, use${\text{K}}_{\text{a2}}$ to solve for the ${\text{pK}}_{\text{a}}$ .

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(6.3×10}}^{-8}\text{)} \\ =7.20 \\ {\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\text{[base]}}{\text{[ acid ]}} \\ {\text{ = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2 - }}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{ - }}\right]} \\ \text{ = 7.20 + log}\frac{\text{0.40M}}{\text{0.50M}} \\ \text{ = 7.10.} \end{gathered}$$

Therefore, the value of the buffer pH is 7.10.