The ionization constant of a monobasic acid is 2-24- 10-2- The pH of 0-01 M acid solution is-

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The ionizatio...

Question

The ionization constant of a monobasic acid is $2.24 \times 10^{- 2}$ . The $p H$ of $0.01 M$ acid solution is:

1.3

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3.3

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1.82

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Solution

The correct option is A 1.82
For monobasic acid, by applying Ostwald's dilution law equation, $[H^{+}] = \sqrt{c \times K_{a}}$ .
Substituting the given values in the above equation,
$[H^{+}] = \sqrt{(0.01 \times 2.24 \times 10^{- 2})} = 0.014966 M$ .
The $p H$ is a negative logarithm of hydrogen ion concentration.
$p H = - l o g [H^{+}] = - l o g (0.014966) = 1.82$ .

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The ionization constant of a monobasic acid is 2.24×10−2. The pH of 0.01 M acid solution is:

The correct option is A 1.82For monobasic acid, by applying Ostwald's dilution law equation, [H+]=√c×Ka.Substituting the given values in the above equation,[H+]=√(0.01×2.24×10−2)=0.014966M.The pH is a negative logarithm of hydrogen ion concentration.pH=−log[H+]=−log(0.014966)=1.82.

The ionization constant of a monobasic acid is $2.24 \times 10^{- 2}$ . The $p H$ of $0.01 M$ acid solution is: