NEET UG 2016 — Chemistry Physical Chemistry

2016mcqmedium

MY and $N{Y}_{3}$ , two nearly insoluble salts, have the same ${K}_{sp}$ values of $6.2\times {10}^{-13}$ at room temperature. Which statement would be true in regard to MY and $N{Y}_{3}$ ?

Official previous-year question

Held on 30 Apr 2016 · Verified 9 Jul 2026.

Options

A
The molar solubilities of MY and $N{Y}_{3}$ in water are identical.
B
The molar solubility of MY in water is less than that of $N{Y}_{3}$ .
C
The salts MY and $N{Y}_{3}$ are more soluble in 0.5 M KY than in pure water.
D
The addition of the salt of KY to solution of MY and $N{Y}_{3}$ will have no effect on their solubilities.

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Solution

NEET 2016 Chemistry, Ionic Equilibrium — solution figure 1

MY insoluble salt ${K}_{ sp} ={s}^{2}$

NEET 2016 Chemistry, Ionic Equilibrium — solution figure 2

$N{Y}_{3}$ salt ${K}_{ sp} =s{( 3s )}^{3}$

$=27{s}^{4}$

$S={ ( \frac{ {k}_{ sp} }{ 27} )}^{ \frac{1}{n} }$

${S}_{ N{Y}_{3} }$ $={( \frac{ {K}_{ sp} }{ 27} )}^{ \frac{1}{4} }$ $={ ( \frac{ 6.2\times { 10}^{ -13} }{ 27} )}^{ \frac{1}{4} }$

$\therefore {S}_{(MY)}=\sqrt{6.2\times {10}^{-13}}=7.8\times {10}^{-7}$

Solubility values

$\therefore {S}_{(MY)}<{S}_{(N{Y}_{3})}$

Your note

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NEET UG 2016 — Chemistry Physical Chemistry

2016mcqmedium

MY and $N{Y}_{3}$ , two nearly insoluble salts, have the same ${K}_{sp}$ values of $6.2\times {10}^{-13}$ at room temperature. Which statement would be true in regard to MY and $N{Y}_{3}$ ?

Official previous-year question

Held on 30 Apr 2016 · Verified 9 Jul 2026.

Options

A
The molar solubilities of MY and $N{Y}_{3}$ in water are identical.
B
The molar solubility of MY in water is less than that of $N{Y}_{3}$ .
C
The salts MY and $N{Y}_{3}$ are more soluble in 0.5 M KY than in pure water.
D
The addition of the salt of KY to solution of MY and $N{Y}_{3}$ will have no effect on their solubilities.

Did you get this right?

Solution

NEET 2016 Chemistry, Ionic Equilibrium — solution figure 1

MY insoluble salt ${K}_{ sp} ={s}^{2}$

NEET 2016 Chemistry, Ionic Equilibrium — solution figure 2

$N{Y}_{3}$ salt ${K}_{ sp} =s{( 3s )}^{3}$

$=27{s}^{4}$

$S={ ( \frac{ {k}_{ sp} }{ 27} )}^{ \frac{1}{n} }$

${S}_{ N{Y}_{3} }$ $={( \frac{ {K}_{ sp} }{ 27} )}^{ \frac{1}{4} }$ $={ ( \frac{ 6.2\times { 10}^{ -13} }{ 27} )}^{ \frac{1}{4} }$

$\therefore {S}_{(MY)}=\sqrt{6.2\times {10}^{-13}}=7.8\times {10}^{-7}$

Solubility values

$\therefore {S}_{(MY)}<{S}_{(N{Y}_{3})}$

Your note

NEET UG Physical Chemistry in other years

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NEET UG 2016 — Chemistry Physical Chemistry

Options

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Solution

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NEET UG Physical Chemistry in other years

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NEET UG 2016 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

NEET UG Physical Chemistry in other years

More NEET UG Physical Chemistry questions

Other Chemistry topics for NEET UG

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry