Consider the titration of 50.0 mL of 0.100 M HC3H502 by 0.100 M KOH for the next
ID: 564122 • Letter: C
Question
Consider the titration of 50.0 mL of 0.100 M HC3H502 by 0.100 M KOH for the next five questions (Ka for HC3H502 1.3 x 10-5). Calculate all pH values to two decimal places. 1) Calculate the pH after 0.0 mL of KOH has been added. 2.94 As always, determine the major species present, then determine the pH. See the text for discussion of calculating the pH of weak acid solutions ou are correct. Previous Tries r receipt no. is 168-4920 2) Calculate the pH after 20.0 mL of KOH has been added. 3.13 Assume the added strong base reacts completely with the weak acid. Determine the major species present after the added KOH reacts completely with the weak acid, then determine how to solve for the plH At some point you will have to switch from using the Ka reaction to using the Kb reaction Subrmit Answer Incorrect. Tries 4/45 Previous Tries 3) Calculate the pH after 25.0 mL of KOH has been added. Submit Answer Tries 0/45 Submit Answer Tries 0/45 Submit Answer Tries 0/45 4) Calate the pH at the equivalence point. 5) Calculate the pH after 100.0 mL of KOH has been addedExplanation / Answer
2) Ka of HC3H5O2 = 1.3*10-5; pKa = -log Ka = -log (1.3*10-5) = 4.89.
Moles of HC3H5O2 = (50.0 mL)*(1 L/1000 mL)*(0.100 M) = 0.005 mole.
Write the balanced chemical equation as below.
HC3H5O2 (aq) + KOH (aq) --------> KC3H5O2 (aq) + H2O (l) …..(1)
As per the stoichiometric equation (1),
1 mole HC3H5O2 = 1 mole KOH = 1 mole KC3H5O2.
Moles of KOH added = (20.0 mL)*(1 L/1000 mL)*(0.100 M) = 0.002 mole.
Moles HC3H5O2 neutralized = moles KOH added = moles KC3H5O2 formed = 0.002 mole; moles HC3H5O2 retained at equilibrium = (0.005 – 0.002) mole = 0.003 mole.
Since the volume of the solution is constant at (50.0 + 20.0) mL = 70.0 mL, we must have
[KC3H5O2]/[HC3H5O2] = (moles KC3H5O2)/(moles HC3H5O2) = (0.002 mole)/(0.003 mole) = 0.667.
Use the Henderson-Hasslebach equation as below.
pH = pKa + log [KC3H5O2]/[HC3H5O2] = 4.89 + log (0.667) = 4.89 + (-0.1759) = 4.7141 4.71 (ans).
3) Moles of KOH added = (25.0 mL)*(1 L/1000 mL)*(0.100 M) = 0.0025 mole.
Moles HC3H5O2 neutralized = moles KOH added = moles KC3H5O2 formed = 0.0025 mole; moles HC3H5O2 retained at equilibrium = (0.005 – 0.0025) mole = 0.0025 mole.
Since the volume of the solution is constant at (50.0 + 25.0) mL = 75.0 mL, we must have
[KC3H5O2]/[HC3H5O2] = (moles KC3H5O2)/(moles HC3H5O2) = (0.0025 mole)/(0.0025 mole) = 1.00.
Use the Henderson-Hasslebach equation as below.
pH = pKa + log [KC3H5O2]/[HC3H5O2] = 4.89 + log (1.00) = 4.89 + (0.00) = 4.89 (ans).
4) At the equivalence point, moles HC3H5O2 = moles KOH = moles KC3H5O2 = 0.005 and no HC3H5O2 is left unreacted.
Volume of 0.100 M KOH required = (0.005 mole)/(0.100 M) = 0.05 L = (0.5 L)*(1000 mL/1 L) = 50.0 mL and the total volume of the solution is (50.0 + 50.0) mL = 100.0 mL.
[KC3H5O2] = (moles KC3H5O2)/(volume of solution in L) = (0.005 mole)/[(100.0 mL)*(1 L/1000 mL)] = 0.05 M.
KC3H5O2 (C3H5O2- is the active species) and re-establishes equilibrium as
C3H5O2- (aq) + H2O (l) -------> HC3H5O2 (aq) + OH- (aq)
Since OH- is formed, we must work with Kb. Given Ka, we can obtain Kb as Kb = Kw/Ka = (1.0*10-14)/(1.3*10-5) = 7.692*10-10
The base ionization constant is given as
Kb = [HC3H5O2][OH-]/[C3H5O2-] = (x).(x)/(0.05 – x)
Since Kb is small, we can assume (0.05 – x) 0.05; therefore,
7.692*10-10 = x2/(0.05)
=====> x2 = 3.846*10-11
=====> x = 6.202*10-6
Thus, [OH-] = 6.202*10-6 M and pOH = -log [OH-] = -log (6.202*10-6) = 5.2075; therefore, pH = 14 – pOH = 14 – 5.2075 = 8.7925 8.79 (ans).
5) 100.00 mL of 0.100 M KOH is added; 50.00 mL KOH neutralizes HC3H5O2. The volume of excess HC3H5O2 = (100.00 – 50.00) mL = 50.00 mL.
Since KOH is a strong base (as compared to C3H5O2-, hence, the pH of the solution is decided by KOH.
Molarity of excess KOH = (50.00 mL)*(0.100 M)/(150.00 mL) = 0.033 M.
pOH = -log [OH-] = -log (0.033) = 1.4815 and pH = 14 – pOH = 14 – 1.4815 = 12.5185 12.52 (ans).