Chemistry - Ionic Equilibrium Question with Solution | TestHub

The Henderson-Hasselbalch equation is:

$$pH=p{K}_a+\log \left(\frac{[A^{-}]}{[HA]}\right)$$

For buffer A: $p{H}_1=p{K}_a+\log \left(\frac{y}{x}\right)$

For buffer B: $p{H}_2=p{K}_a+\log \left(\frac{x}{y}\right)$

Given:

$p{H}_2-p{H}_1=1$

$p{H}_1+p{H}_2=9.5$

Adding the two equations: $2p{H}_2=10.5⟹p{H}_2=5.25$

Subtracting the two equations: $2p{H}_1=8.5⟹p{H}_1=4.25$

Substitute $p{H}_1$ and $p{H}_2$ into the Henderson-Hasselbalch equations:

$4.25=p{K}_a+\log \left(\frac{y}{x}\right)$ (1)

$5.25=p{K}_a+\log \left(\frac{x}{y}\right)$ (2)

Subtract (1) from (2):

$5.25-4.25=\log \left(\frac{x}{y}\right)-\log \left(\frac{y}{x}\right)$

$1=\log \left(\frac{x}{y}\right)+\log \left(\frac{x}{y}\right)$

$1=2\log \left(\frac{x}{y}\right)$

$\log \left(\frac{x}{y}\right)=0.5$

$\frac{x}{y}=10^{0.5}=\sqrt{10}\approx 3.162$

Substitute $\log \left(\frac{x}{y}\right)=0.5$ into equation (2):

$5.25=p{K}_a+0.5$

$p{K}_a=5.25-0.5=4.75$

Thus, $\frac{x}{y}=3.162$ and $p{K}_a=4.75$.

The final answer is $\frac{x}{y}=3.162,p{K}_a=4.75$.