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Question

Chemistry Question on Equilibrium

A certain weak acid has a dissociation constant of 1.0×10−41.0 \times 10^{ - 4 }. The equilibrium constant for its reaction with a strong base is

A

1.0×10−41.0 \times 10^{ - 4}

B

1.0×10−101.0 \times 10^{ - 10}

C

1.0×10101.0 \times 10^{ 10}

D

1.0×10141.0 \times 10^{ 14}

Answer

1.0×10101.0 \times 10^{ 10}

Explanation

Solution

The reaction of HA with strong base is
HA+OH−⇌H2O+A−HA + OH^- \rightleftharpoons H_2O + A^-
K = [A−][HA] [OH−]×[H−][H−]=KaKw\frac{ [ A^- ]}{ [ HA ] \, [ OH^-] } \times \frac{ [ H^- ]}{ [ H^- ] } = \frac{ K_a }{ K_w }
= 10−410−14=1010\frac{ 10^{ - 4 }}{ 10^{ - 14 }} = 10^{ 10}