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(a) Addition of 0.00 mL of strong base:
The solution contains only 0.100 M weak acid. The concentration of species present will be governed by the equilibrium reaction below. The equilibrium expression for this reaction is: Ka = [H3O+][F-] / [HF]
HF(aq) + H2O(l) <==> H3O+(aq) + F-(aq) where Ka (HF) = 6.8 x 10-4.
[initial] 0.100 ~ 0 0
[final] 0.100 - x ~ x x
Therefore, Ka = [H3O+][F-] / [HF] = [x][x] / [0.100 - x] = 6.8 x 10-4. This equation can be solved for x.
Assuming x is much smaller than 0.100 (i.e. <5%), the equation simplifies to [x][x] / [0.100] = 6.8 x 10-4.
Evaluation gives x = 7.91 x 10-3 M = [H3O+] and therefore thepH = 2.10, ( recall: pH = - log[7.91 x 10-3]).
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(b) Addition of 10.00 mL of strong base: (Buffer Region)
This problem should be broken into two steps. First, react all strong base initially present, (strong base is limiting reagent for this step). The strong base will react with the weak acid and convert it into its conjugate weak base. Assume this reaction goes to completion since K is large for this reaction.
HF(aq) + KOH(aq) -----> H2O(l) + KF(aq) K = (1 / Kb) = large, goes to completion
moles initial 0.00500 0.00200 ~ 0
moles final 0.00300 0 ~ 0.00200
[final] 0.0500 M 0 0.0330 M Total Volume = (50.00 + 10.00) mL = 0.06000 L
Now, the second step involves only the weak acid and its weak conjugate base from the first step. The concentration of these species will be governed by the equilibrium expression below. The net ionic equation for this reaction can be written and a reaction table set up (omitting spectator ions like K+(aq)):
HF(aq) + H2O(l) <==> H3O+(aq) + F-(aq) where Ka (HF) = 6.8 x 10-4.
[initial] 0.0500 ~ 0 0.0330
[final] 0.0500 - x ~ x 0.0330 + x
Therefore, Ka = [H3O+][F-] / [HF] = [0.0330 + x ][x] / [0.0500 - x] = 6.8 x 10-4. This equation can be solved for x.
Assuming x is much smaller than 0.0330 (i.e. <5%), the equation simplifies to [0.0330][x] / [0.0500] = 6.8 x 10-4.
Thus, x = 1.03 x 10-3 M = [H3O+] which is indeed small (<5%) compared to 0.0330. Therefore thepH = 2.99
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(c) Addition of 24.90 mL of strong base:
This problem should be broken into two steps. First, react all strong base initially present, (strong base is limiting reagent for this step). The strong base will react with the weak acid and convert it into its conjugate weak base. Assume this reaction goes to completion since K is large for this reaction.
HF(aq) + KOH(aq) -----> H2O(l) + KF(aq) K = (1 / Kb) = large, goes to completion
moles initial 0.00500 0.00498 ~ 0
moles final 0.00002 0 ~ 0.00498
[final] 3 x 10-4 M 0 0.0665 M Total Volume = 0.07490 L
Now, the second step involves only the weak acid and its weak conjugate base from the first step. The concentration of these species will be governed by the equilibrium expression below. The net ionic equation for this reaction can be written and a reaction table set up (omitting spectator ions like K+(aq)):
HF(aq) + H2O(l) <==> H3O+(aq) + F-(aq) where Ka (HF) = 6.8 x 10-4.
[initial] 3 x 10-4 ~ 0 0.0665
[final] 3 x 10-4 - x ~ x 0.0665 + x
Therefore, Ka = [H3O+][F-] / [HF] = [0.0665 + x ][x] / [3 x 10-4 - x] = 6.8 x 10-4. This equation can be solved for x.
Assuming x is much smaller than 3 x 10-4 (i.e. <5%), the equation simplifies to [0.0665][x] / [3 x 10-4] = 6.8 x 10-4.
Thus, x = 3 x 10-6 M = [H3O+] which is indeed small (<5%) compared to 3 x 10-4. Therefore thepH = 5.5
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(d) Addition of 25.00 mL of strong base: (The Equivalence Point)
Again, this problem should be broken into two steps. First, react all strong base initially present. There is exactly enough strong base to react with the weak acid available. The strong base will react with the weak acid and convert it into its conjugate weak base(F-). Again, assume this reaction goes to completion since K is large for this reaction.
HF(aq) + KOH(aq) -----> H2O(l) + KF(aq) K = (1 / Kb) = large, goes to completion
moles initial 0.00500 0.00500 ~ 0
moles final 0 0 ~ 0.00500
[final] 0 0 0.0667 M Total Volume = 0.07500 L
Notice that the second step involves only the weak conjugate base from the first step. The concentration of species present will be governed by the equilibrium expression below. The net ionic equation for this reaction can be written and a reaction table set up (omitting spectator ions like K+(aq)):
F-(aq) + H2O(l) <==> OH-(aq) + HF(aq) where Kb (F-) = 1.47 x 10-11.
[initial] 0.0667 ~ 0 0
[final] 0.0667 - x ~ x x
Therefore, Kb = [OH-][HF] / [F-] = [x][x] / [0.0667 - x] = 1.47 x 10-11. This equation can be solved for x.
Assuming x is much smaller than 0.0667 (i.e. <5%), the equation simplifies to [x][x] / [0.0667] = 1.47 x 10-11.
Thus, x = 9.90 x 10-7 = [OH-] which is small (<5%) compared to 0.0667.
Finally, pH = 14.00 - pOH = 14.00 - (-log[9.90 x 10-7]) = 14.00 - 6.004. Therefore thepH = 8.00
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(e) Addition of 25.10 mL of strong base: (Excess strong base added)
Again, this problem should be broken into two steps. First, react all strong base initially present. There is exactly enough strong base to react with the weak acid available. The strong base will react with the weak acid and convert it into its conjugate weak base(F-). Again, assume this reaction goes to completion since K is large for this reaction.
HF(aq) + KOH(aq) -----> H2O(l) + KF(aq) K = (1 / Kb) = large, goes to completion
moles initial 0.00500 0.00502 ~ 0
moles final 0 0.00002 ~ 0.00500
[final] 2.66 x 10-4 M *Note below Total Volume = 0.07510 L
Notice that the first step leaves 2.66 x 10-4 M hydroxide ion (from KOH) unreacted. In step (d) above the hydroxide ion produced by the 0.00500 moles of flouride ion (from KF) was only 9.90 x 10-7 M. This value is small compared to the remaining KOH now in solution so the pH is really only determined by the excess KOH.
Therefore, [OH-] = 2.66 x 10-4 M.
The pH = 14.00 - pOH = 14.00 - (-log[2.66 x 10-4]) = 14.00 - 3.575. Therefore thepH = 10.42
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(f) Addition of 30.00 mL of strong base:
The same logic follows from part (e) above. Notice that the first step now leaves 0.0125 M hydroxide ion (from KOH) unreacted (0.00600 moles KOH - 0.00500 moles HF in a total volume of 80.00 mL). In step (d) above the hydroxide ion produced by the 0.00500 moles of flouride ion (from KF) was only 9.90 x 10-7 M. This value is small compared to the remaining KOH now in solution so the pH is really only determined by the excess KOH.
Therefore, [OH-] = 0.0125 M.
The pH = 14.00 - pOH = 14.00 - (-log[0.0125]) = 14.00 - 1.903. Therefore thepH = 12.10
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(g) Addition of 40.00 mL of strong base:
The same logic follows from part (e) above. Notice that the first step now leaves 0.0333 M hydroxide ion (from KOH) unreacted (0.00800 moles KOH - 0.00500 moles HF in a total volume of 90.00 mL). In step (d) above the hydroxide ion produced by the 0.00500 moles of flouride ion (from KF) was only 9.90 x 10-7 M. This value is small compared to the remaining KOH now in solution so the pH is really only determined by the excess KOH.
Therefore, [OH-] = 0.0333 M.
The pH = 14.00 - pOH = 14.00 - (-log[0.0333]) = 14.00 - 1.477. Therefore thepH = 12.52
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