A vessel contains 500. milliliters of a 0.100-molar H2S solution. For H2S, Ka1 = 1.0 X 10-7 and
Ka2 = 1.3 X 10-13
(a) What is the pH of the solution?
(b) How many milliliters of 0.100-molar NaOH solution must be added to the solution to create a solution with a pH of 7?
(c) What will be the pH when 800. milliliters of 0.100-molar NaOH has been added?
(Hint: What is the only species left in solution after 500. mL of the 0.100M NaOH solution is added? It is now a new titration problem using Ka2.)
(d) What is the value of Keq for the following reaction?
H2S ↔ 2H+ + S-2
(Hint; Think about Hess’s Law and what two reactions when added together equal the reaction above. What do you do with equilibrium constants of reactions you add together.)
a. ...................H2S --> H^+ + HS^-
I........................0.1.......0...........0
C.......................-x.........x............x
E......................0.1-x......x............x
Plug the E line into Ka1 expression and solve for x = (H^+) and convert to pH.
b. ...........H2S + NaOH ==> NaHS + H2O
Ka1 for H2S = 1E-7 and pK1 = 7. The pH of the solution is guided by
pH = pK1 + log (base)/(acid)
7 = 7 + log (b/a) so you want b/a = 1 so log b/a = 0.
You have 500 mL of 0.100 M H2S so you want to add 250 mL of 0.100 M NaOH. You had 500 x 0.1 = 50 millimoles H2S to start, you add 25 mmols NaOH which produces 25 mmols NaHS and leaves 25 mmols H2S
c. We add 800 mL of 0.1 M = 80 mmolsThe first titration is with 50 and I'm repeating this.
.....................H2S + OH^- ==> HS^- + H2O
I.....................50...........0...........0
add..............................50.......................
C....................-50........-50.........+50
E......................0............0..........50 so at this point we have ONLY HS^-
Second part of part c. The first 500 mL was used for the first part so the next 300 of 0.1 M (30 mmols) used for the second part.
.............HS^- + OH^- ==> S^= + H2O
I.............50.........................0
add.....................30............................
C............-30.......-30...........+30
E..............20.........0.............30
So plug the E line into the HH equation and solve for pH.
d. Keq = ka1 x ka2 = ?
Post your work if you get stuck.