JEE MAIN - Chemistry (2005 - No. 5)
The solubility product of a salt having general formula MX2, in water is: 4 $$\times$$ 10-12 . The
concentration of M2+ ions in the aqueous solution of the salt is :
2.0 $$\times$$ 10-6 M
4.0 $$\times$$ 10-10 M
1.0 $$\times$$ 10-4 M
1.6 $$\times$$ 10-4 M
Explanation
$$M{X_2}\,\rightleftharpoons\,\,\mathop {{M^{ 2 + }}}\limits_{s\,\,\,\,\,\,\,\,} \,\, + \,\,\mathop {2{X^ - }}\limits_{2s} $$
Where $$s$$ is the solubility of $$M{X_2}$$
then $${K_{sp}} = $$ $$s \times {\left( {2s} \right)^2}$$ = $$4{s^3}$$;
$$ \Rightarrow $$$$ 4 \times {10^{ - 12}} = 4{s^3};$$
$$ \Rightarrow $$ $$s = 1 \times {10^{ - 4}}$$
$$\left[ {{M^{ 2+ }}} \right] = s $$ $$ = 1 \times 10^ {- 4}$$
Where $$s$$ is the solubility of $$M{X_2}$$
then $${K_{sp}} = $$ $$s \times {\left( {2s} \right)^2}$$ = $$4{s^3}$$;
$$ \Rightarrow $$$$ 4 \times {10^{ - 12}} = 4{s^3};$$
$$ \Rightarrow $$ $$s = 1 \times {10^{ - 4}}$$
$$\left[ {{M^{ 2+ }}} \right] = s $$ $$ = 1 \times 10^ {- 4}$$
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