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Chemistry Question on Ionic Equilibrium In Solution

The degree of dissociation of a 0.01 M0.01\, M weak acid is 10−310^{-3} . Its pOHpOH is

A

55

B

33

C

99

D

1111

Answer

99

Explanation

Solution

[H+]=Ka⋅C\left[ H ^{+}\right] =\sqrt{K_{a} \cdot C}

or =α2⋅C2=\sqrt{\alpha^{2} \cdot C^{2}}
[∴Ka=α2C]\left[\therefore K_{a}=\alpha^{2} C\right]
=(10−3)2⋅(0.01)2=\sqrt{\left(10^{-3}\right)^{2} \cdot(0.01)^{2}}
=10−5=10^{-5}
pH=−log⁡[H+]=−log⁡10−5pH =-\log \left[ H ^{+}\right] =-\log 10^{-5}
=5=5
pOH=14−5=9pOH =14-5= 9