what is the molar solubility of lead(II) chromate (ksp=1.8x10^-14) in 0.13 M potassium chromate?
See your other post below.
To determine the molar solubility of lead(II) chromate (PbCrO4) in 0.13 M potassium chromate (K2CrO4), we need to consider the solubility product constant (Ksp) and the reaction equation:
PbCrO4(s) ⇌ Pb2+(aq) + CrO42-(aq)
The Ksp expression for the reaction is:
Ksp = [Pb2+][CrO42-]
Given that the Ksp value for PbCrO4 is 1.8x10^-14, we can write:
1.8x10^-14 = [Pb2+][CrO42-]
However, we are also provided with the concentration of potassium chromate (K2CrO4), which is 0.13 M. This means that [CrO42-] = 0.13 M.
Now, we need to solve for [Pb2+].
By substituting these values into the Ksp expression, we get:
1.8x10^-14 = [Pb2+](0.13)
Rearranging the equation to solve for [Pb2+], we have:
[Pb2+] = (1.8x10^-14) / (0.13)
Calculating this, we find the molar solubility of PbCrO4 in 0.13 M K2CrO4 to be approximately 1.38x10^-13 M.