JEE Main 2018 — Chemistry Physical Chemistry

2018mcqmedium

An aqueous solution contains an unknown concentration of ${\mathrm{Ba}}^{2+}$ . When $50\mathrm{mL}\mathrm{of}a1M$ solution of ${\mathrm{Na}}_{2}{\mathrm{SO}}_{4}$ is added, ${\mathrm{BaSO}}_{4}$ just begins to precipitate. The final volume is $500\mathrm{mL}$ . The solubility product of ${\mathrm{BaSO}}_{4}$ is $1\times {10}^{-10}$ . What is the original concentration of ${\mathrm{Ba}}^{2+}$ ?

Official previous-year question

Held on 8 Apr 2018 · Verified 6 Jul 2026.

Options

A
$1.0\times {10}^{-10}M$
B
$5\times {10}^{-9}M$
C
$2\times {10}^{-9}M$
D
$1.1\times {10}^{-9}M$

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Solution

${\mathrm{Ba}}^{+2}+{\mathrm{SO}}_{4}^{2-}\rightleftharpoons \underset{\mathrm{ppt}}{{\mathrm{BaSO}}_{4(s)}}$

Final conc. of $[{\mathrm{SO}}_{4}^{2-}]=\frac{{\mathrm{MV}}_{1}}{{V}_{1}+{V}_{2}}=\frac{1\times 50}{500}=0.1M$

Final conc.of $[{\mathrm{Ba}}^{+2}]$ When ${\mathrm{BaSO}}_{4}$ start precipitating

${K}_{\mathrm{SP}}={Q}_{\mathrm{SP}}=[{\mathrm{Ba}}^{+2}][{\mathrm{SO}}_{4}^{2-}]$

${10}^{-10}=[{\mathrm{Ba}}^{+2}](0.1M)$

Initial conc. $[{\mathrm{Ba}}^{+2}]$ ; initial volume was $=500-50=450\mathrm{ml}$

${M}_{1}{V}_{1}={M}_{2}{V}_{2}$

${M}_{1}=\frac{{M}_{2}{V}_{2}}{{V}_{1}}=\frac{{10}^{-9}\times 500}{450}$

${M}_{1}=1.1\times {10}^{-9}M$

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JEE Main 2018 — Chemistry Physical Chemistry

2018mcqmedium

Official previous-year question

Held on 8 Apr 2018 · Verified 6 Jul 2026.

Options

A
$1.0\times {10}^{-10}M$
B
$5\times {10}^{-9}M$
C
$2\times {10}^{-9}M$
D
$1.1\times {10}^{-9}M$

Did you get this right?

Solution

${\mathrm{Ba}}^{+2}+{\mathrm{SO}}_{4}^{2-}\rightleftharpoons \underset{\mathrm{ppt}}{{\mathrm{BaSO}}_{4(s)}}$

Final conc. of $[{\mathrm{SO}}_{4}^{2-}]=\frac{{\mathrm{MV}}_{1}}{{V}_{1}+{V}_{2}}=\frac{1\times 50}{500}=0.1M$

Final conc.of $[{\mathrm{Ba}}^{+2}]$ When ${\mathrm{BaSO}}_{4}$ start precipitating

${K}_{\mathrm{SP}}={Q}_{\mathrm{SP}}=[{\mathrm{Ba}}^{+2}][{\mathrm{SO}}_{4}^{2-}]$

${10}^{-10}=[{\mathrm{Ba}}^{+2}](0.1M)$

Initial conc. $[{\mathrm{Ba}}^{+2}]$ ; initial volume was $=500-50=450\mathrm{ml}$

${M}_{1}{V}_{1}={M}_{2}{V}_{2}$

${M}_{1}=\frac{{M}_{2}{V}_{2}}{{V}_{1}}=\frac{{10}^{-9}\times 500}{450}$

${M}_{1}=1.1\times {10}^{-9}M$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2018 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2018 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry