Question

Find the $pH$ of $1L$ of a buffer solution containing $0.01MN{H}_{4}Cl$ and $0.05MN{H}_{4}OH$ at ${25}^{\circ }C$
The base dissociation constant for $N{H}_{4}OH$ is $1.8\times {10}^{-5}$
Take $log\left(1.8\right)=0.26$

• A. 9.96
• B. 10.36
• C. 11.56
• D. 12.56

Answer 1

The correct option is A 9.96

This is a basic buffer formed by a weak base $\left(N{H}_{4}OH\right)$ and its salt $\left(N{H}_{4}Cl\right)$ with strong acid $\left(HCl\right)$
$p{K}_b=-log\left(K_b\right)p{K}_b=-log(1.8\times {10}^{-5})p{K}_b=(5-0.26)=4.74$
$pOH=p{K}_b+log\frac{\left[salt\right]}{\left[base\right]}pOH=p{K}_b+log\left(\frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_{4}OH\right]}\right)pOH=4.74+log\left(\frac{0.01}{0.05}\right)pOH=4.74+log\left(0.2\right)=(4.74-0.7)pOH=4.04pH+pOH=14pH=14-pOH=14-4.04=9.96$