A 1.0L solution of lead chromate is prepared by mixing 0.50 mg of lead nitrate(Pb(NO3)2) with 0.020 mg of potassium chromate (K2CrO4). Will a precipitate form? Ksp for PbCrO4 is 2.0 * 10-14.
Question
A 1.0L solution of lead chromate is prepared by mixing 0.50 mg of lead nitrate(Pb(NO3)2) with 0.020 mg of potassium chromate (K2CrO4). Will a precipitate form? Ksp for PbCrO4 is 2.0 * 10-14.
Solution
Sure, let's solve this step by step.
Step 1: Identify the possible reaction The reaction between lead nitrate and potassium chromate can produce lead chromate and potassium nitrate. The reaction is as follows:
Pb(NO3)2 + K2CrO4 -> PbCrO4 + 2KNO3
Step 2: Determine the initial concentrations The initial concentration of Pb2+ from Pb(NO3)2 is 0.50 mg/L = 0.50 g/m^3 = 0.50 * 10^-6 kg/m^3. The molar mass of Pb(NO3)2 is 331.2 g/mol, so the initial molar concentration is (0.50 * 10^-6 kg/m^3) / (331.2 * 10^-3 kg/mol) = 1.51 * 10^-6 M.
The initial concentration of CrO4^2- from K2CrO4 is 0.020 mg/L = 0.020 g/m^3 = 0.020 * 10^-6 kg/m^3. The molar mass of K2CrO4 is 194.2 g/mol, so the initial molar concentration is (0.020 * 10^-6 kg/m^3) / (194.2 * 10^-3 kg/mol) = 1.03 * 10^-7 M.
Step 3: Determine if a precipitate will form A precipitate will form if the reaction quotient Q is greater than the solubility product constant Ksp. The reaction quotient Q is given by the product of the concentrations of the ions in the solution. For the reaction above, Q = [Pb2+][CrO4^2-].
Substituting the initial concentrations, Q = (1.51 * 10^-6 M) * (1.03 * 10^-7 M) = 1.56 * 10^-13.
Since Q > Ksp (1.56 * 10^-13 > 2.0 * 10^-14), a precipitate of PbCrO4 will form.
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