A solution is 0.1M in Cl^-… - JEE Main 2021 Chemistry PYQ

JEE Main 2021 — Chemistry Physical Chemistry

2021mcqhard

A solution is $0.1M$ in ${\mathrm{Cl}}^{-}$ and $0.001M$ in ${\mathrm{CrO}}_{4}^{2-}.$

Solid ${\mathrm{AgNO}}_{3}$ is gradually added to it Assuming that the addition does not change in volume and ${K}_{\mathrm{sp}}(\mathrm{AgCl})=1.7\times {10}^{-10}{M}^{2}$ and

${K}_{\mathrm{sp}}({\mathrm{Ag}}_{2}{\mathrm{CrO}}_{4})=1.9\times {10}^{-12}{M}^{3}.$

Select correct statement from the following:

Official previous-year question

Held on 20 Jul 2021 · Verified 6 Jul 2026.

Options

A
$\mathrm{AgCl}$ precipitates first because its ${K}_{\mathrm{sp}}$ is high.
B
${\mathrm{Ag}}_{2}{\mathrm{CrO}}_{4}$ precipitates first as its ${K}_{\mathrm{sp}}$ is low.
C
${\mathrm{Ag}}_{2}{\mathrm{CrO}}_{4}$ precipitates first because the amount of ${\mathrm{Ag}}^{+}$ needed is low.
D
$\mathrm{AgCl}$ will precipitate first as the amount of ${\mathrm{Ag}}^{+}$ needed to precipitate is low.

Did you get this right?

Solution

$(i)[{\mathrm{Ag}}^{+}]$ required to ppt $\mathrm{AgCl}(s)$

$\mathrm{Ksp}=\mathrm{IP}=[{\mathrm{Ag}}^{+}][{\mathrm{Cl}}^{-}]=1.7\times {10}^{-10}$

$[{\mathrm{Ag}}^{+}]=1.7\times {10}^{-9}$

$(\mathrm{ii})[{\mathrm{Ag}}^{+}]$ required to ppt ${\mathrm{Ag}}_{2}{\mathrm{CrO}}_{4}(s)$

$\mathrm{Ksp}=\mathrm{IP}=[\mathrm{Ag}+{]}^{2}[{\mathrm{CrO}}_{4}^{-2}]=1.9\times {10}^{-12}$

$[{\mathrm{Ag}}^{+}]=4.3\times {10}^{-5}$

$[{\mathrm{Ag}}^{+}]$ required to ppt $\mathrm{AgCl}$ is low so $\mathrm{AgCl}$ will ppt ${1}^{\mathrm{st}}.$

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Options

$\mathrm{AgCl}$ precipitates first because its ${K}_{\mathrm{sp}}$ is high.

${\mathrm{Ag}}_{2}{\mathrm{CrO}}_{4}$ precipitates first as its ${K}_{\mathrm{sp}}$ is low.

${\mathrm{Ag}}_{2}{\mathrm{CrO}}_{4}$ precipitates first because the amount of ${\mathrm{Ag}}^{+}$ needed is low.

$\mathrm{AgCl}$ will precipitate first as the amount of ${\mathrm{Ag}}^{+}$ needed to precipitate is low.

Solution

$(i)[{\mathrm{Ag}}^{+}]$ required to ppt $\mathrm{AgCl}(s)$

$\mathrm{Ksp}=\mathrm{IP}=[{\mathrm{Ag}}^{+}][{\mathrm{Cl}}^{-}]=1.7\times {10}^{-10}$

$[{\mathrm{Ag}}^{+}]=1.7\times {10}^{-9}$

$(\mathrm{ii})[{\mathrm{Ag}}^{+}]$ required to ppt ${\mathrm{Ag}}_{2}{\mathrm{CrO}}_{4}(s)$

$\mathrm{Ksp}=\mathrm{IP}=[\mathrm{Ag}+{]}^{2}[{\mathrm{CrO}}_{4}^{-2}]=1.9\times {10}^{-12}$

$[{\mathrm{Ag}}^{+}]=4.3\times {10}^{-5}$

$[{\mathrm{Ag}}^{+}]$ required to ppt $\mathrm{AgCl}$ is low so $\mathrm{AgCl}$ will ppt ${1}^{\mathrm{st}}.$

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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

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