Question

A weak base 50 ml was titrated with $0.1M$ $HCl$. The $pH$ of the solution after the addition of acid $10.0$ ml and $25.0$ ml were found to be $9.84$ and $9.24$, respectively. Calculate $K_b$ of the base and $pH$ at the equivalence point?

• A. $1.73\times {10}^{-5}$, $5.27$
• B. $1.73\times {10}^{-6}$, $4.27$
• C. $1.73\times {10}^{-8}$, $3.27$
• D. $1.73\times {10}^{-4}$, $6.27$

Answer 1

The correct option is A $1.73\times {10}^{-5}$, $5.27$

Given weak base of $50ml$ is titrated against $0.1M$ $HCl$

Which is nothing but

$BOH+HCl⟶BCl+H_{2}O$

$0ml$ $0.1$ $0$

$10ml$ $1$ $x-1$

$25ml$ $2.5$ $x-2.5$

We know that$pOH=14-pH=p{K}_b+\log \frac{\left[Salt\right]}{\left[Acid\right]}$

For $10ml$

$14-9.84=p{K}_b+\log \frac{1}{(x-1)}$ ........$\left(1\right)$

For $25ml$

$14-9.24=p{K}_b+\log \frac{2.5}{(x-2.5)}$..........$\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$ we get $x=5$

Substituting $x$ value in $\left(2\right)$ or $\left(1\right)$ we get,

$K_b$ $=1.73\times {10}^{-5}$

Now we know that $K_b{K}_a=K_w{K}_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.73\times {10}^{-5}}\left[H^{+}\right]=\sqrt{K_{a}C}=\sqrt{\frac{{0.05\times 10}^{-14}}{1.73\times {10}^{-5}}}=5.37\times {10}^{-6}pH=-\log \left[H^{+}\right]=-\log (5.37\times {10}^{-6})=5.27$