Question

The pH of a $\frac{M}{10}$ solution of a weak base, if its $K_b$ value at ${25}^{o}Cis2.5\times {10}^{-4},$ will be: $[log2=0.3]$

• A. 11.7
• B. 10.7
• C. 12.3
• D. 12.7

Answer 1

The correct option is A 11.7

$K_b=\frac{C{\alpha }^2}{1-\alpha }$
$\therefore \alpha =\sqrt{2}.5\times {10}^{-4}\times 10=5\times {10}^{-2}$
$\left[O{H}^{-}\right]=5\times {10}^{-3}$
$pOH=3-log5=2.3$
$pH=14-2.3=11.7$