A solution contains 2.2×10−4 M Ag+ and 1.3×10−3 M
If NaI is added, will AgI(Ksp=8.3×10−17) or PbI2(Ksp=7.9×10−9) precipitate first?
I know the answer to this is AgI
Specify the concentration of I− needed to begin precipitation.
thats not the answer i need...
To determine the concentration of I- needed to begin precipitation, we need to compare the value of the reaction quotient (Q) for the two possible precipitation reactions: AgI and PbI2.
First, let's write the balanced equations for the precipitation reactions:
1) Ag+ (aq) + I- (aq) ⇌ AgI (s)
2) Pb2+ (aq) + 2I- (aq) ⇌ PbI2 (s)
To calculate the concentration of I- needed to begin precipitation, we need to find the value of Q for each reaction and compare it to the respective solubility product (Ksp) value.
For reaction 1 (AgI):
Q = [Ag+][I-]
For reaction 2 (PbI2):
Q = [Pb2+][I-]^2
We already know the concentration of Ag+ (2.2×10^-4 M) and Pb2+ (1.3×10^-3 M). Since both reactions involve the same I- ions, we will use the concentration of I- as the variable in both Q calculations.
Since we are looking for the concentration of I- needed to initiate precipitation, we can assume that Q is equal to the corresponding Ksp for the salt that precipitates first.
For AgI (Ksp = 8.3×10^-17):
8.3×10^-17 = (2.2×10^-4) * [I-]
Solving for [I-]:
[I-] = (8.3×10^-17) / (2.2×10^-4) ≈ 3.77×10^-13M
Therefore, the concentration of I- needed to initiate the precipitation of AgI is approximately 3.77×10^-13 M.