Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is

A

1.0X10−31.0 X 10^{-3}

B

1.0X10−61.0 X 10^{-6}

C

1.0X10−81.0 X 10^{-8}

D

1.0X10−111.0 X 10^{-11}

Answer

1.0X10−111.0 X 10^{-11}

Explanation

Solution

Ionisation constant, Ka=a2×c1−aK_{a}=\frac{a^{2} \times c}{1-a} α\alpha= degree of dissociation =0.001%= 0.001 \% =0.001100=1×10−5=\frac{0.001}{100}=1\times10^{-5} CC= concentration =0.1 M= 0.1 \,M when α\alpha is very small K=α2×CK=\alpha^{2} \times C =(1×10−5)2×0.1=1×10−11=(1 \times 10^{-5})^{2} \times 0.1=1\times 10^{-11}