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If the molar concentration of Pbl2 is1.5 x10-3 molL-1, the concentration of iodide ionsin g ion L-1 is:(1) 3.0 x 103(3) 0.3 x 10-3(2) 6.0 x 10-3(4) 0.6 x 10-6 |
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Answer» PbI2 (s) <=> Pb2+ (aq) + 2 I- (aq) and the equilibrium condition is given by Ksp = [Pb2+][I-]^2 at equilibrium [Pb2+]= 1.5 x 10^-3 M [I-]= 2 x 1.5 x 10^-3=3.0 x 10^-3 M Ksp = 1.5 x 10^-3 ( 3.0 x 10^-3)^2 =1.4 x 10^-8 |
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