a sample is prepared by combining 1 ml of .0050 M Pb(NO3)2, 1 ml of .0075 M KI a

Question

a sample is prepared by combining 1 ml of .0050 M Pb(NO3)2, 1 ml of .0075 M KI and 5 ml of water. A. determine the concentration of Pb2+ and I- in this solution B. The Ksp value of PbI2 is 9.8x10^-9. Based on the concentrations you found above, calculate the Q value. Would the precipitate form in this case? Explain a sample is prepared by combining 1 ml of .0050 M Pb(NO3)2, 1 ml of .0075 M KI and 5 ml of water. A. determine the concentration of Pb2+ and I- in this solution B. The Ksp value of PbI2 is 9.8x10^-9. Based on the concentrations you found above, calculate the Q value. Would the precipitate form in this case? Explain A. determine the concentration of Pb2+ and I- in this solution B. The Ksp value of PbI2 is 9.8x10^-9. Based on the concentrations you found above, calculate the Q value. Would the precipitate form in this case? Explain B. The Ksp value of PbI2 is 9.8x10^-9. Based on the concentrations you found above, calculate the Q value. Would the precipitate form in this case? Explain

Explanation / Answer

A) Pb(NO3)   +2KI-----------à PbI2 + 2K(NO3)

In the solution of given samples lead ions, iodide ions,potassium ions and nitrate ions are present.

[ Pb2+] =0.005M,[I-]=2*0.0075=0.015M

B) Given that Ksp=9.8x10^-9

Q=[Pb2+][I-]2 =0.005X(0.015)2 =0.005X0.000225=11.25*10^-7

So the ionic product,Q =11.25*10^-7

The precipitate will form in this case because ionic product exceeded the Ksp value.

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