MY and N Y 3 , two nearly insoluble salts, have the same K s p values of 6.2 × 10 - 13 at room temperature.…

1. The molar solubilities of MY and $\mathrm{N}{\mathrm{Y}}_3$ in water are identical.
2. The molar solubility of MY in water is less than that of $\mathrm{N}{\mathrm{Y}}_3$ .
3. The salts MY and $\mathrm{N}{\mathrm{Y}}_3$ are more soluble in 0.5 M KY than in pure water.
4. The addition of the salt of KY to solution of MY and $\mathrm{N}{\mathrm{Y}}_3$ will have no effect on their solubilities.

Solution

MY insoluble salt $K_{sp}=s^2$

$N{Y}_3$ salt $K_{sp}=s{\left(3s\right)}^3$
$=27s^4$
$S={\left(\frac{k_{sp}}{27}\right)}^{\frac{1}{n}}$
$S_{N{Y}_3}$ $={\left(\frac{K_{sp}}{27}\right)}^{\frac{1}{4}}$ $={\left(\frac{6.2\times {10}^{-13}}{27}\right)}^{\frac{1}{4}}$
$\therefore {\mathrm{S}}_{\left(\mathrm{M}\mathrm{Y}\right)}=\sqrt{6.2\times {10}^{-13}}=7.8\times {10}^{-7}$
Solubility values
$\therefore {\mathrm{S}}_{\left(\mathrm{M}\mathrm{Y}\right)}<{\mathrm{S}}_{\left(\mathrm{N}{\mathrm{Y}}_3\right)}$

Asked in: NEET 2016 (Phase 1)

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