Question

$1MN{H}_{4}OH$ and $1MHCl$ are mixed to make total volume of $300$mL. If pH of the mixture is $9.26$ and $pKa\left(N{H}_4^{+}\right)=9.26$ then volume ratio of $N{H}_{4}OH$ and $HCl$ will be:

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $3:1$

Answer 1

The correct option is C $2:1$

The buffer should contain ${NH}_{4}OH$ (weak base) and its salt ${NH}_{4}Cl$.

now, $pOH={pK}_b+log\frac{\left[salt\right]}{\left[base\right]}$

$\Rightarrow 14-pH=4.74+log\frac{\left[salt\right]}{\left[base\right]}$

$\Rightarrow 14-9.26=4.74+log\frac{\left[salt\right]}{\left[base\right]}$

$\Rightarrow log\frac{\left[salt\right]}{\left[base\right]}=0$

$\therefore$ $\frac{\left[salt\right]}{\left[base\right]}=\frac{\left[{NH}_{4}Cl\right]}{\left[{NH}_{4}OH\right]}=1$

$1$ mole of $HCl$ neutralises $1$ mole of ${NH}_{4}OH$ to form $1$ mole of ${NH}_{4}Cl$.

Therefore $200$ ml of $1M$ ${NH}_{4}OH$ and $100$ ml of $1M$ $HCl$ are mixed to make total volume of $300$ ml.

$\therefore$ Volume ratio of ${NH}_{4}OH$ and $HCl$ will be $2:1$.