The solubility product of PbI2 is 7.2×10−9 Calculate the maximum mass of NaI which may be added in 500 mL of 0.005 M−Pb(NO3)2solution without any precipitation of PbI2Atomic weight of Na=23, I=127
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a
0.09 g
b
0.9 g
c
9 g
d
90 g
answer is A.
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Detailed Solution
Given,The solubility product of PbI2 ,Ksp=7.2×10−9Volume,v=500 mLmolarity, M = 0.005 M maximum mass of , m = ? prevent any precipitation of PbI2,Q