JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

$25.0\mathrm{mL}$ of $0.050M\mathrm{Ba}{({\mathrm{NO}}_{3})}_{2}$ is mixed with $25.0\mathrm{mL}$ of $0.020M\mathrm{NaF}.{K}_{\mathrm{sp}}$ of ${\mathrm{BaF}}_{2}$ is $0.5\times 10–6$ at $298K.$ The ratio of $[{\mathrm{Ba}}^{2+}]{[{F}^{-}]}^{2}$ and ${K}_{\mathrm{sp}}$ is $______.$

Official previous-year question

Held on 13 Apr 2023 · Verified 6 Jul 2026.

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Solution

Given:

Volume of $\mathrm{Ba}({\mathrm{NO}}_{3}{)}_{2}$ solution = $25.0$ mL Concentration of $\mathrm{Ba}({\mathrm{NO}}_{3}{)}_{2}$ solution = $0.050$ M Volume of $\mathrm{NaF}$ solution = $25.0$ mL Concentration of $\mathrm{NaF}$ solution = $0.020$ M

$[{\mathrm{Ba}}^{+2}]=(\frac{25\times 0.05}{50})=0.025M$

$[{F}^{-}]=(\frac{25\times 0.02}{50})=0.01M$

$[{\mathrm{Ba}}^{+2}]$ $[{F}^{-}]=25\times {10}^{-7}$

${K}_{\mathrm{sp}}=0.5\times {10}^{-6}=5\times {10}^{-7}$

Ratio= $\frac{[{\mathrm{Ba}}^{+2}][{F}^{-}]}{{K}_{sp}}=\frac{25\times {10}^{-7}}{5\times {10}^{-7}}=5$

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JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

Official previous-year question

Held on 13 Apr 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

Given:

$[{\mathrm{Ba}}^{+2}]=(\frac{25\times 0.05}{50})=0.025M$

$[{F}^{-}]=(\frac{25\times 0.02}{50})=0.01M$

$[{\mathrm{Ba}}^{+2}]$ $[{F}^{-}]=25\times {10}^{-7}$

${K}_{\mathrm{sp}}=0.5\times {10}^{-6}=5\times {10}^{-7}$

Ratio= $\frac{[{\mathrm{Ba}}^{+2}][{F}^{-}]}{{K}_{sp}}=\frac{25\times {10}^{-7}}{5\times {10}^{-7}}=5$

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

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