Question

A weak monobasic acid is
12% dissociated in 0.05 M solution. What
is percent dissociation in 0.15 M solution.​

Answer 1

Final answer: Per cent dissociation of 0.15 M solution of the given weak monobasic acid = 6.928 %

Given that: We are given a weak monobasic acid is 12 % dissociated in 0.05 M solution.

To find: We have to find the per cent dissociation in 0.15 M solution.​

Explanation:

• Consider a weak monobasic acid ($HA$) that is partially ionised in its aqueous solution. let ‘$c$’ $mol\ L^{-1}$ be the initial concentration of the given acid and $\alpha$ the degree of dissociation at equilibrium.
• The degree of dissociation is the fraction of the total number of moles dissociated at equilibrium.

                   Initial:         $c$ $mol\ L^{-1}$                    0                 0

The chemical reaction: $HA (aq) +H_{2}O (l)$ ⇄ $H_{3}O^{+}(aq)+A^{-}(aq)$

    At equilibrium:    $c(1 -\alpha)\ mol\ L^{-1}$         $c\alpha$ $mol\ L^{-1}$    $c\alpha$ $mol\ L^{-1}$

• The “ionisation constant” $K_{a}$ for acid $HA$ is represented as:

$K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]} = \frac{c\alpha \times c\alpha }{c(1-\alpha )}= \frac{c\alpha^{2}}{1-\alpha}$

• The smaller the value of $K_{a}$ for a given acid weaker the acid is.

• Since, ‘$\alpha$’ << 1 for weak acid, the denominator (1 - $\alpha$) of the expression for  $K_{a}$ can be approximated to 1 and thus,

$K_{a} = c\alpha ^{2}$

Here, $c_{1}$ = 0.05 M

$\alpha _{1}= 12\% = \frac{12}{100} = 0.12$

$K_{a} = c_{1}\alpha_{1}^{2}=0.05\times (0.12)^{2}=7.2\times 10^{-4}$

• Now we have the value of $K_{a}$ and another concentration $c_{2}$ (0.15 M), therefore we can easily calculate the degree of ionization $\alpha_{2}$.

$\alpha _{2} =\sqrt{\frac{K_{a}}{c_{2}}} \\\\\alpha_{2} = \sqrt{\frac{7.2 \times 10^{-4}}{0.15}}=0.06928$

$\% \ \alpha_{2}= \alpha_{2} \times 100 = 0.06928 \times 100=6.928 \%$

• Hence, per cent dissociation = 6.928 %

To know more about the concept please go through the links

https://brainly.in/question/17724232

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