Question

A buffer solution with pH 9 is to be prepared by mixing $N{H}_{4}CI$ and $N{H}_{4}OH$. Calculate the number of moles of $N{H}_{4}CI$ that should be added to 1 liter of 1M $N{H}_{4}OH$. $K_b\left(N{H}_{4}OH\right)=1.8\ast {10}^{-5}$.

Answer 1

pOH of desired buffer is calculated by -

pOH = 14 - pH

pOH = 14 - 9

pOH = 5

By Hendersan-Hasselbach eqn,

$pOH=p{K}_b+log\frac{\left[salt\right]}{\left[base\right]}$

$pOH=-log\left(K_b\right)+log\frac{\left[N{H}_{4}Cl\right]}{\left[NH4OH\right]}$

Substituting values,

$5=-log(1.8\times {10}^{-5}+log\frac{\left[NH4Cl\right]}{1}$

$5-4.745=log\left[N{H}_{4}Cl\right]$

$0.255=log\left[N{H}_{4}Cl\right]$

Taking antilog,

$\left[N{H}_{4}Cl\right]=1.8M$

To prepare 1.8 M $N{H}_{4}Cl$ soln, 1.8 moles of $N{H}_{4}Cl$ are to be added in 1 L soln.