Question

The pH of mono acidic
weak base is 11.2
calculate its % dissociation in 0.02M
solution.​

Answer 1

Final answer: Per cent dissociation of 0.02 M given weak monoacidic base solution =​ 7.295 %

Given that: We are given $pH$ of a weak monoacidic base is 11.2.

To find: We have to find its per cent dissociation in 0.02 M solution.​

Explanation:

Monoacidic base:

• It is a base that generates one $OH^{-}$ ion when it undergoes complete ionization.

Examples: Sodium hydroxide ($NaOH$), Ammonium hydroxide ($NH_{4}OH$), etc.

• Here, we are given a monoacidic weak base, which means that it does not dissociate $100 \%$ completely.

• $p{H}$ of the given weak monoacidic base = 11.2

$pOH$ of the given solution = $14 -pH$

$pOH= 14 - 11.2 = 2.8$

• We know that $pOH$ expresses the $[OH^{-}]$.

$pOH= -\log_{10}[OH^{-}]$

• From the above expression,

$[OH^{-}] = 10^{-pOH}$

$[OH^{-}] =10^{-2.8} =1.585\times 10^{-3}\ mol\ L^{-1}$    [$pOH$ = 2.8]

• Consider the dissociation of a weak monoacidic base (BOH).

Let 'c' $mol\ L^{-1}$ be the initial concentration of the base and ‘α’ be the degree of dissociation at equilibrium.

                          $c$                  $0$           $0$

The reaction: $BOH (aq)$ ⇄ $B^{+}(aq)+OH^{-}(aq)$

                       $c(1-\alpha )$           $c\alpha$            $c\alpha$

$[OH^{-}] = c\alpha$

$\alpha =\frac{[OH^{-}]}{c}$

• Here, $c= 0.02\ M = 0.03$ $mol\ L^{-1}$ [given]

$[OH^{-}] = 1.585\times 10^{-3} \ mol\ L^{-1}$

• Substitute these values in the equation of $\alpha$.

$\alpha=\frac{1.585\times 10^{-3}}{0.02}=0.07925$

$\%\ \alpha=$ $0.07925\times100$ $= 7.295\ \%$

Hence, per cent dissociation = 7.295 %

To know more about the concept please go through the links

https://brainly.in/question/33862142

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