Question

$1M$ weak monoacidic base $\left(BOH\right)$ solution is dliuted by $100$ times. The $pH$ change of the solution is: $(K_b={10}^{-5})$

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

$1M$ $BOH$ contains $1$ mol of $BOH$ in $1L$ solution

Final volume= $100L$

$\therefore \left[O{H}^{-}\right]=\frac{1}{100}={10}^{-2}=C$

For weak base, $pH=\frac{1}{2}[p{K}_b-\log C]$

Initial $pH=\frac{1}{2}\left[p{K}_b\right]$

Final $pH=\frac{1}{2}[p{K}_b-\log {10}^{-2}]=\frac{p{K}_b+2}{2}$

$\therefore$ Change in $pH=1$