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DM Sherman Practical 3: Solubility and Acid base Equilibria (Solutions) Suppose we have an acid-dissociation reaction HA = A-^ + H+ With โฌ
K =
[ A โ][ H +]
[ HA ]
we can take the โ log (=p) of both sides to get pK = pA + pH - pHA If we know the total concentration (Stot) of our dissolved acid (i.e., Stot = [A] + [HA]), then we can make a plot of pA vs pH and pHA vs pH. The reason we can do this in log space is because when pH < pK, we have that (to a really good approximation) pHA = pStot but with pHA = pStot, if follows that pA = pK โ pH + pStot Letโs plot this up using, for illustrative purposes, an acid with pK = 5 and pStot = 1. We set up a diagram with pC on the y-axis (here, C stands for concentration of either HA, A or H or OH) and pH on the x-axis.
DM Sherman On this diagram, I have plotted three lines: pH = pK, pOH =14-pH and pH=pH. I can plot the equations for pA and pHA in the region where pH < pK: Now we can plot the speciation under the conditions when pH > pK. Under those conditions, we have [A-] >> [HA] so that, to a very good approximation, pA = pStot. If pA = pStot, then it follows that pHA = pStot + pH โ pK We can now plot pA and pHA in the region pH > pK: What about the region where pH = pK? When pH = pK, we have
DM Sherman
- (a) Construct a pC-pH diagram for H 2 S system, with [S]tot = [H 2 S] + [HS] + [S-^2 ] 0.001 m, given the following equilibria: โฌ
K 2 =
[ H +] [ S โ^2 ]
[^ HS โ]
= 10 โ^18
(b). Now plot the concentration of dissolved Zn (as pZn) on the same diagram, given the equilbrium constant for the reaction ZnS + H+ = Zn2+ + HS-: โฌ
K 2 =
[ Zn +^2 ] [ HS โ]
[^ H +]
= 10 โ11.^4
and the condition that [S]tot = 0.001 m. c. Finally plot pZn on a new diagram for the case where there is no other source of sulfur (i.e. [Zn] = [H2S]+[HS]+[S]). Solution: pK 1 = pH + pHS โ pH 2 S = 7. pK 2 = pH + pS โ pHS = 18 Under acidic conditions (pH < 7.03), pH 2 S = pStot = 3. When pH 2 S = pHS = p(Stot/2), pH = 7.03. When H2S is dominant, pHS = 7.03+pStot-pH. When HS is dominant, pH 2 S = pH + pHS โ 7.03. When pHS = pS, pH wil be 18 (i.e., never..). Hence between pH = 7-14, pHS = pStot. Over this range, pS = 18+3 โ pH = 21-pH.
K 1 =
[ H +] [ HS โ]
[^ H 2 S ]
DM Sherman b) For the ZnS dissolution when Stot = 0.001, pZn + pHS โ pH = 11.4 or pZn = 11.4 โ pHS + pH. At pH = 0, pHS = 10.03 and pZn = 1.43. At pH = 4, pHS = 5.97 so pZn = 11.4 + 4 โ 5.97 = 9.43. When HS is dominant, pHS = 3, so pZn = 8.4 + pH c) For the ZnS dissolution when [Zn] = [H2S] + [HS] + [S] pZn + pHS โ pH = 11.4. When HS is dominant, pZn = pHS so that 2pZn โ pH = 11.4 or pZn = (11.4 + pH)/ When H 2 S is dominant, pZn = pH2S. Since pHS + pH โ pH2S = 7.03 we get pHS = 7.03 โ pH + pH2S. If we combine this with pZn + pHS โ pH = 11.4, we get pZn + 7.03 โ pH + pH 2 S โ pH =11.4 or (since, pZn = pH 2 S), 2pZn โ 2pH = 4.37 or pZn = (4.37 + 2pH)/
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