Question

$N{H}_{4}OH$ is titrated with $HCl$ solution. Find the value of $\log \left(\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right)$ when the volume of $HCl$ added is half of that of equivalence point.

• A. \N
• B. 0.5
• C. 0.75
• D. 1

Answer 1

The correct option is A \N

$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)$

At half equivalence point, $pOH=p{K}_b$

Since for basic buffer ,

$pOH=p{K}_b+log\left(\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right)$
When $pOH=p{K}_b$

$log\left(\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right)=0$