Answer to Question #61310 in General Chemistry for Sedrick

Question

3. The Henry’s law constant for H2S is 0.1 mole/L.atm and
H2S <---> HS- + H+
Where Ka = 10-7. The equilibrium pH of the solution is 4.5 if pure H2S gas is dissolved in water. Find the partial pressure of the H2S gas above the solution.

4. At a wastewater treatment plant, FeCl3(s) is added to remove excess phosphate (PO4-3) from the effluent. Assume the following reactions occur:
FeCl3(s) <---> Fe3+ + 3Cl-
FePO4(s) <---> Fe3+ + PO4-3
The Ksp for the second reaction is . What concentration of Fe3+ (in mg/L) is needed to maintain the phosphate concentration below the limit of 1 mg/L P?

Answer 1

3. Henrys law C = kp, С - concentration in mol/L, k - Henry constant and p - pressure of gas above the solution.
Thus, p = C/k, C can be found using the pH and Ka. [H+] = 10^-pH = 10^-4.5 = 3.2*10^-5 mol/l, Ka = [HS-][H+]/[H2S], Ka = [H+]^2/(C - [H+]), C = 0.01 mol/l
p = 0.01 mol/l / 0.1 mol/(l*atm) = 0.1 atm
4. Ksp = [Fe3+][PO43-], [Fe3+] = Ksp/[PO43-],
[PO43-] in mol/l = 10^-3 g/l / 95 g/mol = 1.1*10^-5 mol/l,
[Fe3+] = 1.3*10^-22 / 1.1*10^-5 = 1.2*10^-17 mol/l or 6.7*10^-13 mg/l