What is the [CrO4] in a 2.5 L saturated solution of lead (II) chromate to which 0.05 mol Pb(NO3)2 is added?
PbCrO4 ==> Pb^2+ + CrO4^2-
...x........x.......x
...........Pb(NO3)2 ==> Pb^2+ + 2NO3^-
initial.....0.05 mol.........0........0
change.....-0.05.......0.05......2*0.05
equil........0..........0.05.....0.10
(Pb^2+) = 0.05/2.5L = 0.02M
Ksp = (Pb^2+)(CrO4^-)
Ksp you know.
(Pb^2+) = x + 0.02 [x from the PbCrO4 and 0.02 from Pb(NO3)2]
(CrO4^2-) = x
Solve for x.
To determine the concentration of [CrO4] in the given saturated solution of lead (II) chromate, we need to consider the stoichiometry of the reaction between lead (II) chromate (PbCrO4) and lead (II) nitrate (Pb(NO3)2).
The balanced equation for the reaction is:
PbCrO4 (s) + Pb(NO3)2 (aq) → 2 Pb2+ (aq) + 2 CrO42- (aq) + 2 NO3- (aq)
From the balanced equation, we can see that each mole of PbCrO4 reacts to produce two moles of CrO42-. Therefore, the molar ratio of [CrO4] to [PbCrO4] is 2:1.
Given that 0.05 mol of Pb(NO3)2 is added, it will react with an equal number of moles of PbCrO4 to produce CrO42-. Thus, we have 0.05 mol of CrO42- present in the solution.
However, the solution is saturated, which means it contains the maximum amount of PbCrO4 that can dissolve at a given temperature. In a saturated solution, some PbCrO4 will remain undissolved, while the extra Pb(NO3)2 added will fully react to produce CrO42-.
Since the question asks for the concentration of [CrO4] rather than the amount, we need to convert the moles of CrO42- to the concentration in terms of molarity.
To do this, divide the number of moles of CrO42- (0.05 mol) by the volume of the solution (2.5 L):
[CrO4] = 0.05 mol / 2.5 L
Therefore, the concentration of [CrO4] in the saturated solution of lead (II) chromate to which 0.05 mol Pb(NO3)2 is added is 0.02 M (mol/L).