## Answers

**Acid A**

pKa = 8.515

Ka = 10^{-pKa} = 10^{-(8.515)} = 3.05 x 10^{-9}

**Acid B**

Ka = 2 x 10^{-11}

pKa = -log[Ka] = -log(2 x 10^{-11}) = 10.699

**Acid C**

pKa = 7.04

Ka = 10^{-pKa} = 10^{-(7.04)} = 9.120 x 10^{-8}

It is well known that the relative strength of acid depends upon the dissociation constant of the acid. Higher the Ka value, higher the acidic strength.

since the decreasing order of acids, A, B and C is as

C (9.120 x 10^{-8}) > A ( 3.05 x 10^{-9}) > B (2 x 10^{-11})

Thus, the relative strength of acids are as

**C > A > B**

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