Question

# The dissociation constants of two acids $$H{ A }_{ 1 }$$ and $$H{ A }_{ 2 }$$ are $$3.0\times { 10 }^{ -4 }$$ and $$1.8\times { 10 }^{ -5 }$$ respectively. The relative strengths of the acids will be:

A
1:4
B
4:1
C
1:16
D
16:1

Solution

## The correct option is B $$4:1$$The dissociation constants of $$HA_1$$ and $$HA_2$$ are $$3\times10^{-4}$$ and $$1.8\times10^{-5}.$$The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are:$$\dfrac { { (Acidic\ Strength })_{ HA_1 } }{ { (Acidic\ Strength })_{ HA_2} } =\dfrac { \sqrt { { (Dissociation\ Constant })_{ HA_1} } }{ \sqrt { { (Dissociation\ Constant })_{ HA_2 } } }$$ Relative Acidic Strength= $$\dfrac { { (Acidic Strength })_{ HA_1 } }{ { (Acidic Strength })_{ HA_2 } } =\dfrac { \sqrt { 3.0\times { 10 }^{ -4 } } }{ \sqrt { 1.8\times { 10 }^{ -5 } } } =4.08=4(approx)$$relative strengths of acids will be $$4:1.$$Chemistry

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