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 is the concentration ratio y : x has $pH=(pH{)}_2$. If $(pH{)}_2-(pH{)}_1=$ 1 unit and $(pH{)}_1+(pH{)}_2=9.5units$, 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)

The correct options are
A $p{K}_a=4.75$
C $\frac{x}{y}=3.162$

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

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

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

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

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

$\Rightarrow \frac{x}{y}=3.162$

$(pH{)}_2+(pH{)}_1=2p{K}_a+log\left(\frac{y}{x}\right)+log\left(\frac{x}{y}\right)$

$2p{K}_a=9.5$

So, $p{K}_a=\frac{9.5}{2}=4.75$

Options A and C are correct.