JEE MAIN - Chemistry (2020 - 9th January Morning Slot - No. 5)
The Ksp for the following dissociation is
1.6 × 10–5
$$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $$
Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl ?
$$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $$
Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl ?
Q > Ksp
Not enough data provided
Q < Ksp
Q = Ksp
Explanation
[Pb2+] = $${{300 \times 0.134} \over {400}}$$
= 1.005 × 10–1 M
[Cl-] = $${{100 \times 0.4} \over {400}}$$
= 10–1 M
$$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $$
Q = [Pb2+] × [Cl–]2
= 0.1005 × (0.1)2
= 1.005 × 10–3
Given Ksp = 1.6 × 10–5
$$ \therefore $$ Q > Ksp
= 1.005 × 10–1 M
[Cl-] = $${{100 \times 0.4} \over {400}}$$
= 10–1 M
$$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $$
Q = [Pb2+] × [Cl–]2
= 0.1005 × (0.1)2
= 1.005 × 10–3
Given Ksp = 1.6 × 10–5
$$ \therefore $$ Q > Ksp
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