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  4. When a 20 mL of 0.08 M weak base BOH is titrated with 0.08 HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 mL of 0.04 M NaOH is added to the resulting solution ? [Given: log 2 = 0.30 and log 3 = 0.48]

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Q. When a $20\, mL$ of $0.08\, M$ weak base $BOH$ is titrated with $0.08\, HCl$, the pH of the solution at the end point is 5. What will be the $pOH$ if $10\, mL$ of $0.04\, M\, NaOH$ is added to the resulting solution ? [Given : $\log 2 = 0.30$ and $\log 3 = 0.48$]

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

image
Total volume $=40\, mL$
$[ BCl ]=1.6 / 40=0.04\, M$
$pH =\frac{1}{2}\left[ pK _{ w }- pK _{ b }-\log C\right]$
$5=\frac{1}{2}\left[14- pK _{ b }-\log 0.04\right]$
$pK _{ b }=5.4$
image
$p OH = pK _{ b }+\log \frac{\left[ B ^{+}\right]}{[ BOH ]}$
$=5.4+\log \frac{1.2}{0.4}$
$=5.4+\log 3$
$=5.4+0.48$
$=5.88$