Question

The dissociation constants for $$HCOOH$$ and $${CH}_{3}COOH$$ are $$2.1\times { 10 }^{ -4 }$$ a.nd $$1.8\times { 10 }^{ -5 }$$ respectively. Calculate the relative strengths of the acids.

Solution

The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are$$\frac { { (Acidic\ Strength })_{ HCOOH } }{ { (Acidic\ Strength })_{ CH_{ 2 }COOH } } =\frac { \sqrt { { (Dissociation\ constant })_{ HCOOH } } }{ \sqrt { { (Dissociation\ constant })_{ CH_{ 2 }COOH } } }$$Relative Acidic Strength= $$\frac { { (Acidic\ Strength })_{ HCOOH } }{ { (Acidic\ Strength })_{ CH_{ 2 }COOH } } =\frac { \sqrt { 2.1\times { 10 }^{ -4 } } }{ \sqrt { 1.8\times { 10 }^{ -5 } } } =3.415$$Chemistry

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