Question

JEE ADVANCED CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

ANSWER.

The minimum molar concentration (m) of H+

required to prevent the precipitation of

ZnS = 0.2 M

EXPLANATION.

$\sf : \implies \: {zn}^{2 + } = 0.05M \: \: and \: \: h_{2}s = 0.1M \: \\ \\ \sf : \implies \: k_{sp} \: (zns) = 1.25 \times 10 {}^{ - 22} \\ \\ \sf : \implies \: h_{2}s \: , k_{net} \: = k_{1} k_{2} \: = 1 \times 10 {}^{ - 21}$

$\sf : \implies \: [zn {}^{2 + } ] = 0.05M \\ \\ \sf : \implies \: [zn {}^{2 + } ] [s {}^{ - 2} ] \: = k_{sp} \\ \\ \sf : \implies \: [s {}^{ - 2} ] \: = \frac{1.25 \times 10 {}^{ - 22} }{0.05} \\ \\\sf : \implies [s {}^{ - 2} ] = 25 \times 10 {}^{ - 22}$

$\sf : \implies \: H _{2}s \: \longrightarrow \: 2 H {}^{ + } + s {}^{ - 2} \\ \\ \sf : \implies \: 1 \times 10 {}^{ - 21} = \frac{ [H {}^{ + } ] {}^{2} [s {}^{ - 2} ] }{[H _{2}s ]} \\ \\ \sf : \implies \: [H {}^{ + } ] {}^{2} \: = \frac{10 {}^{ - 21} \times 10 {}^{ - 1} }{25 \times 10 {}^{ - 22} } \\ \\ \sf : \implies[H {}^{ + } ] {}^{2} = 0.04 \\ \\ \sf : \implies[H {}^{ + } ] {}^{} \: = 0.2M$

Answer 2

Answer:

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