Question

$p{K}_b$ for $N{H}_{4}OH$ at certain temperature is 4.74 The pH of basic buffer containing equimolar concentration of $N{H}_{4}OH$ and $N{H}_{C}I$ will be:

Answer 1

$pOH=p{K}_b+log\frac{\left[N{H}_{4}I\right]}{\left[N{H}_{4}OH\right]}$ for basic buffer

Given that $p{K}_b=4.74$ $\left[N{H}_{4}I\right]=\left[N{H}_{4}CH\right]$

So $pOH=4.74+log1$

$pOH=4.74$

$pH=14-pOH=14-4.74=9.26$