What is the $p H$ of a solution of $0.28 M$ acid and $0.84 M$ of its conjugate base if the ionization constant of acid is $4 \times 10^{- 4}$ ?

3.88

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3.34

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10.12

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Solution

The correct option is A $3.88$
As we know,
$p H = p K_{a} + l o g (s a l t / a c i d)$
and by putting given value
$p H = - l o g 4 * 10^{- 4} + l o g 0.84 / 0.28 = 3.88$

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What is the pH of a solution of 0.28 M acid and 0.84 M of its conjugate base if the ionization constant of acid is 4×10−4?

The correct option is A 3.88As we know,pH=pKa+log(salt/acid)and by putting given valuepH=−log4∗10−4+log0.84/0.28=3.88

What is the $p H$ of a solution of $0.28 M$ acid and $0.84 M$ of its conjugate base if the ionization constant of acid is $4 \times 10^{- 4}$ ?

The correct option is A $3.88$
As we know,
$p H = p K_{a} + l o g (s a l t / a c i d)$
and by putting given value
$p H = - l o g 4 * 10^{- 4} + l o g 0.84 / 0.28 = 3.88$