JEE MAIN - Chemistry (2004 - No. 7)
The molar solubility (in ol L-1) of a sparingly soluble salt MX4 is "s". The corresponding solubility product is Ksp. 's' is given in term of Ksp by the relation :
s = (256 Ksp)1/5
s = (128 Ksp)1/4
s = ( Ksp / 128)1/4
s = (Ksp / 256)1/5
Explanation
$$M{X_4}\rightleftharpoons\,\mathop {{M^{4 + }}}\limits_{S\,\,\,\,\,\,\,} + \mathop {4{X^ - }}\limits_{4S} $$
$${K_{sp}} = \left[ s \right]{\left[ {4s} \right]^4} = 256\,{s^5}$$
$$\therefore$$ $$\,\,\,\,\,s = {\left( {{{{K_{sp}}} \over {256}}} \right)^{1/5}}$$
$${K_{sp}} = \left[ s \right]{\left[ {4s} \right]^4} = 256\,{s^5}$$
$$\therefore$$ $$\,\,\,\,\,s = {\left( {{{{K_{sp}}} \over {256}}} \right)^{1/5}}$$
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