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