Question

A weak base $BOH$ was titrated against a strong acid. The $pH$ at $\frac{1}{4}$th equivalence point was $9.24$. Enough strong base was now added ($6$ meq.) to completely convert the salt. The total volume was $50mL$. Find the $pH$ at this point.

• A. $pH=2.72$
• B. $pH=14.7$
• C. $pH=9.28$
• D. $pH=11.22$

Answer 1

The correct option is D $pH=11.22$

Initially, let $x$ moles of $BOH$ be present.

$BOH+HCl⟶BCl+H_{2}O$
$x$
$\frac{3x}{4}$ $\frac{x}{4}$

$⟹pOH=p{k}_b+\log \frac{1}{3}$

$⟹14-9.24=p{k}_b-\log 3$

$⟹p{k}_b=5.237$

Hence, $k_b=5.8\times {10}^{-6}$

Now,
$BCl+NaOH⟶BOH$
$\frac{x}{4}$ $6$
$-$ $-$ $x$
$\frac{x}{4}=6$ $\Rightarrow$ $x=24$

$\left[BOH\right]=\frac{24}{50}=0.48$

$\left[{OH}^{-}\right]=\sqrt{k_b\times C}=1.668\times {10}^{-3}$

$pOH=2.77$

As we know, $pH=14-pOH$

$pH=11.22$

Hence, the correct option is $\text{D}$