Question

$p{K}_b$ of $N{H}_{4}OH$ is $4.74$. If the ratio of the buffer mixture $\left[N{H}_{4}Cl\right]:\left[N{H}_{4}OH\right]$ is $1:10$, then the $pH$ of that buffer is:

• A. $4.74$
• B. $10.26$
• C. $9.26$
• D. $3.74$

Answer 1

The correct option is B $10.26$

The expression for the $pOH$ of the basic buffer solution is as given below:

$pOH=p{K}_b+log\frac{\left[Salt\right]}{\left[Base\right]}$ $...\left(i\right)$

Given that the $p{K}_b$ of $N{H}_{4}OH$ is $4.74$.

Also, $\left[N{H}_{4}Cl\right]:\left[N{H}_{4}OH\right]=1:10$ $=\left[Salt\right]:\left[Base\right]$

Substituting values in the equation $\left(i\right)$, we get

$pOH=4.74+log\frac{1}{10}=3.74$

$pH=14-pOH=14-3.74=10.26$

Option B is correct.