Will a precipitate form when 20.0mL of 1.8 x 10^-3M Pb(NO32 is added to 30.0mL of 5.0 x 10^-4M Na2SO4? The Ksp of (PbSO4) is 6.3 x 10^-7.
See your previous post. This problem is the same kind. Only the numbers are different.
To determine if a precipitate will form when Pb(NO3)2 is added to Na2SO4, we need to compare the value of the reaction quotient Qsp with the solubility product constant (Ksp) for PbSO4.
The balanced chemical equation for the reaction is:
Pb(NO3)2 + Na2SO4 -> PbSO4 + 2NaNO3
From the equation, we can see that 1 mole of Pb(NO3)2 reacts with 1 mole of Na2SO4 to form 1 mole of PbSO4.
First, calculate the concentrations of the ions Pb2+ and SO42- in the mixture after mixing the solutions:
For Pb2+:
Concentration of Pb(NO3)2 = 1.8 x 10^-3 M (given)
Volume of Pb(NO3)2 = 20.0 mL = 0.0200 L
Concentration of Pb2+ = (1.8 x 10^-3 M) x (0.0200 L) / (0.0500 L) = 7.2 x 10^-4 M
For SO42-:
Concentration of Na2SO4 = 5.0 x 10^-4 M (given)
Volume of Na2SO4 = 30.0 mL = 0.0300 L
Concentration of SO42- = (5.0 x 10^-4 M) x (0.0300 L) / (0.0500 L) = 3.0 x 10^-4 M
Next, calculate the reaction quotient Qsp for PbSO4:
Qsp = [Pb2+][SO42-] = (7.2 x 10^-4 M) x (3.0 x 10^-4 M) = 2.16 x 10^-7
Finally, compare Qsp with the Ksp value to determine if a precipitate will form:
If Qsp > Ksp, then a precipitate will form.
If Qsp < Ksp, then no precipitate will form.
In this case, Qsp (2.16 x 10^-7) is greater than Ksp (6.3 x 10^-7), meaning that the reaction quotient exceeds the solubility product constant. Therefore, a precipitate of PbSO4 will form.