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Question

A monoprotic acid in $$1.00\ M$$ solution is $$0.01\%$$ ionised. The dissociation of this acid is:


A
1×108
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B
1×104
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C
1×106
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D
1×105
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Solution

The correct option is A $$1\times 10^{-8}$$
$$K=\frac { { \alpha  }^{ 2 }C }{ 1-\alpha  } \\ \\ \alpha =\frac { 0.01 }{ 100 } \approx 1\\ \\ \therefore K={ \alpha  }^{ 2 }C\quad =\left[ \frac { 0.01 }{ 100 }  \right] ^{ 2 }\times 1\\ \quad \quad \quad \quad \quad \quad \quad \quad =1\times 1{ 0 }^{ -8 }$$

Chemistry

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