Question

Answer this question

Answer 1

The correct answer is 2) 4.2

Explanation:

We are given:

Molarity of weak acid = 0.1 N

$K_a=4.9\times 10^{-8}$

The chemical equation for the dissociation of weak acid, HA follows:

                     $HA\rightleftharpoons H^++A^-$

I:                    0.1

C:                   -x       +x      +x

E:                 0.1-x       x        x

The expression for equilibrium constant follows:

$K_a=\frac{[H^+][A^-]}{[HA]}$

Putting values in above expression, we get:

$4.9\times 10^{-8}=\frac{x^2}{0.1-x}\\\\4.9\times 10^{-8}(0.1-x)=x^2\\\\x=6.99\times 10^{-5}, -7.00\times 10^{-5}$

Neglecting the negative value of x as concentration cannot be negative.

So, $[H^+]=x=6.99\times 10^{-5}M$

pH is defined as the negative logarithm of hydrogen ion concentration present in the solution.

$pH=-\log [OH^+]$

Putting values in above equation, we get:

$pH=-\log (6.99\times 10^{-5})\\\\pH=4.2$

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