Question

How many gram moles of $HCl$ will be required to prepare one litre of a buffer solution (containing $NaCN$ and $HCN$) of $pH8.5$ using $0.01gram$ formula weight of $NaCN$ ?

$K_a\left(HCN\right)$ = $4.1\times {10}^{-10}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

One of the very first things to check is if this is an acidic or basic buffer. $K_{a}HCN$ is given; so is the pH. $HCN$ and $NaCN$ is a combination of weak acid and its conjugate base

$pH$ = $p{K}_a$ + $log\frac{\left[salt\right]}{\left[Acid\right]}$

$NaCN+HCl\to NaCl+HCN$

At Equilibrium 0.1 - x x x

At equb. NaCN + HCl $\to$ NaCl + HCN

0.01 - x x x

Applying the knows and the unknowns into the henderson hasselbach equation, we have

$8.5$ = $(10-log4.1)+log\frac{(0.01-x)}{x}$

Solving, $x$ = $8.85\times {10}^{-3}M$

Amount of HCl required = $8.85\times {10}^{-3}M$