Question

only for genius ,,,find the value of this question, from ionic eqb

Answer 1

Hey friend, Harish here.

Here is your answer.

Here using the concepts of equivalence will be useful and easy to solve this question.

So , At the end point :

Amount of Salt Produced = Amount of NaOH

                                          = 0.1 × 36.12   m.e

                                          = 3.612  m.e

Amount of HCl added = 0.1 × 18.06 = 1.806  m.e

&

Amount of the Unknown Acid produced = 1.806 m.e

Because addition of 1.806 m.e will reduce the amount of salt by 1.806 m.e and will  produce the exact amount of unknown acid .

Now the Amount of Salt after adding HCl = 3.612 - 1.806 = 1.806 m.e

We know that :

$\mathrm{pH= pK_a + log \frac{m.m\ of\ salt}{acid}} \\ \\\mathrm{ 4.92 = pK_a +log \frac{1.806}{1.806}} \\\\\mathrm{pK_a = 4.92}$

∴ Kₐ = 1.2 × 10⁻⁵

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Hope my answer is helpful to you.