JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 9)
If Ksp of Ag2CO3 is 8 $$ \times $$ 10–12, the molar solubility of Ag2CO3 in 0.1 M AgNO3 is -
8 $$ \times $$ 10–12 M
8 $$ \times $$ 10–10 M
8 $$ \times $$ 10–13 M
8 $$ \times $$ 10–11 M
Explanation
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$$ \therefore $$ 8 $$ \times $$ 10$$-$$12 = (2s + 0.1)2 s
$$ \Rightarrow $$ s $$ \times $$ 10$$-$$2 = 8 $$ \times $$ 10$$-$$12
$$ \Rightarrow $$ s = 8 $$ \times $$ 10$$-$$10
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