Question

The pH of buffer of $N{H}_{4}OH+{NH}_{4}Cl$ type is given by:

• A. $pH-{pK}_b$
• B. $pH=\frac{1}{2}{K}_b-\frac{1}{2}log\frac{\left[Salt\right]}{\left[base\right]}$
• C. $PH=14-{pK}_b-log\frac{\left[salt\right]}{\left[base\right]}$
• D. $pH=pOH-{pK}_b+log\frac{\left[salt\right]}{\left[base\right]}$

Answer 1

The correct option is C $PH=14-{pK}_b-log\frac{\left[salt\right]}{\left[base\right]}$

Solution:-

Buffer of $N{H}_{4}OH+N{H}_{4}Cl$

This buffer solution composed of weak base $\left(N{H}_{4}OH\right)$ and its salt $\left(N{H}_{4}Cl\right)$.

To calculate the pH of the above solution:-

$\left[{OH}^{-}\right]=K_b.\frac{\left[base\right]}{\left[salt\right]}$

Taking $\log$ both sides, we have

$\log \left[{OH}^{-}\right]=\log (K_b.\frac{\left[base\right]}{\left[salt\right]})$

$\log \left[{OH}^{-}\right]=\log K_b+\log \frac{\left[base\right]}{\left[salt\right]}$

$-\log \left[{OH}^{-}\right]=-\log K_b-\log \frac{\left[base\right]}{\left[salt\right]}$

$\Rightarrow pOH=p{K}_b+\log \frac{\left[salt\right]}{\left[base\right]}..........\left(1\right)$

As we know that,

$pH+pOH=p{K}_w=14$

$\Rightarrow pH=14-pOH$

$\therefore pH=14-p{K}_b-\log \frac{\left[salt\right]}{\left[base\right]}(\because From\left(1\right))$

Hence, the correct answer is-

$pH=14-p{K}_b-\log \frac{\left[salt\right]}{\left[base\right]}$