Will a precipitate form upon the addition of 50.00 mL of .010 M Pb(NO3)2 to 75.0

Question

Will a precipitate form upon the addition of 50.00 mL of .010 M Pb(NO3)2 to 75.00 mL of .0075 M NaCl? a. What would the precipitate be? b. Write the reaction representing the solubility equilibrium? c. What are the initial concentrations of the ions that are part of the precipitate? d. Explain which direction the reaction should proceed and what you should observe? e. What is the smallest amount of .0250 M sodium bromide that can be added to 50.00 mL of .0500 M copper(I) nitrate before a precipitate will form?

Explanation / Answer

Qsp = [Pb2+][Cl-]^2

       = (0.01 M x 50 ml/125 ml)(0.0075 M x 75 ml/125 ml)^2

       = 8.1 x 10^-8

Ksp for PbCl2 = 1.7 x 10^-5

Since, Qsp < Ksp, precipitate will not form

a. Precipitate to form = PbCl2

b. PbCl2 <==> Pb2+ + 2Cl-

c. iniitial [Pb2+] = 0.01 M x 50 ml/125 ml = 4 x 10^-3 M

initial [Cl-] = 0.0075 M x 75 ml/125 ml = 4.5 x 10^-3 M

d. With the given concentration, Pb2+ and Cl- would combine to form PbCl2. The direction of reaction would be to the right handside.

e. Ksp for CuBr2 is needed.

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