Question

What will be the ionisation constant of formic acid if its $0.01$M solution is $14.5\%$ ionised?

• A. $2.1\times {10}^{-4}$
• B. $14.5$
• C. $0.145$
• D. $1.45\times {10}^{-4}$

Answer 1

The correct option is A $2.1\times {10}^{-4}$

$HCOOH+H_{2}O\to H_3{O}^{+}+H{C}^{-}$

$C$

$C(1-\alpha )$ $C\alpha$ $C\alpha$

$K_a=\frac{c\alpha .c\alpha }{c(1-\alpha )}$

$=\frac{c{\alpha }^2}{1-{\alpha }^2}$

$\alpha =14.5\%=0.145$

$K_a=\frac{0.01(0.145{)}^2}{1-0.145}$

$K_a=2.1\times {10}^{-4}$