Question

Buffer solution $A$ of a weak monoprotic acid and its sodium salt in the concentration ratio $x:y$ has $pH=(pH{)}_1$. Buffer solution $B$ of the same acid and its sodium salt in the concentration ratio $y:x$ has $pH=(pH{)}_2$. If $(pH{)}_2-(pH{)}_1=1$ unit and $(pH{)}_1+(pH{)}_2=9.5$ units, then:

• A. $p{K}_a=4.75$
• B. $\frac{x}{y}=2.36$
• C. $\frac{x}{y}=3.162$
• D. $p{K}_a=5.25$

Answer 1

Answer: (A), (C)

$p{K}_a=4.75$
$\frac{x}{y}=3.162$

The expression for the $pH$ of the acidic buffer solution is as given below:
$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$

For buffer solution A, $\frac{\left[salt\right]}{\left[acid\right]}=\frac{y}{x}$

$(pH{)}_1=p{K}_a+log(y/x)$

For buffer solution B, $\frac{\left[salt\right]}{\left[acid\right]}=\frac{x}{y}$

$(pH{)}_2=p{K}_a+log(x/y)$

Hence, $(pH{)}_2-(pH{)}_1=log\left(\frac{x}{y}\right)-log\left(\frac{y}{x}\right)=1$.

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

or $\frac{x^2}{y^2}=10$

$\frac{x}{y}=3.162$
Also $(pH{)}_2+(pH{)}_1=2p{K}_a+log\left(\frac{y}{x}\right)+log\left(\frac{x}{y}\right)=2p{K}_a=9.5$.
Thus, $p{K}_a=4.75$.