Question

A 1.25 m solution of the weak acid ha is 9.2% dissociated. What is the ph of the solution?

Answer 1

Answer:

0.94 is the pH of the solution.

Explanation:

The equilibrium reaction for dissociation of $HA$ (weak acid) is,

                           $HA+H_2O\rightleftharpoons A^-+H_3O^+$

initially conc.         c                       0         0

At eqm.              $c(1-\alpha)$                $c\alpha$        $c\alpha$

Concentration (c) = 1.25 M

Degree of dissociation = $\alpha$ = 9.2% = 0.092

$[H_3O^+]=c\alpha$

$= 1.25 M\times 0.092=0.115 M$

The pH of the solution is :

$pH=-\log[H_3O^+]$

$=-\log[0.115 M}=0.94$

Answer 2

pH of 1.25 m solution of the weak acid HA is 9.2% dissociated is 0.94.

What is pH?

pH of a solution tells about the acidity or basicity of the given solution.

Given reaction can be expressed in the initial and equilibrium state as follow:

$\[\mathop {{\rm{HA}}}\limits_{\scriptstyle{\rm{C}}\hfill\atop\scriptstyle{\rm{C(1 - \alpha )}}\hfill} {\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to \mathop {{{\rm{A}}^{\rm{ - }}}}\limits_{\scriptstyle{\rm{0}}\hfill\atop\scriptstyle{\rm{C\alpha }}\hfill} {\rm{ + }}\mathop {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}}\limits_{\scriptstyle{\rm{0}}\hfill\atop\scriptstyle{\rm{C\alpha }}\hfill} \]$

pH of the solution can be calculated as pH = -log[H₃O⁺]

For this first of all we have to determine the concentration of H₃O⁺, which is equal to C∝. Where,

C = concentration of weak acid = 1.25M (given)

∝ = degree of dissociation = 9.2% = 0.092

So, concentration of H₃O⁺ = 1.25 × 0.092 = 0.115M

Now put this value in the pH equation to calculate;

pH = -log[H₃O⁺]

pH = -log(0.115)

pH = 0.94

Hence, pH of the solution is 0.94.

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