Titration of a Weak Acid Pre-Lab Assignment The titration curve shown was obtain
ID: 1027912 • Letter: T
Question
Titration of a Weak Acid Pre-Lab Assignment The titration curve shown was obtained for the titration of a 0.297 g sample of a solid acid with 0.150 M KOH 14 12 10 25 30 15 Volume of KOH added (mL 10 1. What volume of KOH was added to reach the equivalence point? 2. How many moles of acid are present? Assume a 1:1 stoichiometry between acid and base. 3. What is the molar mass of this acid? 4. What is the pKa? 5. In Part A of the experiment, what volume of unknown acid will you use? 6. In Part B of the experiment, what mass of unknown acid will you use?Explanation / Answer
1) The pH at the equivalence point is the mid-point of the steeply rising portion of the titration curve. The pH at the equivalence point is approximately 8.0 and the corresponding volume of KOH dispensed is approximately 17.0 mL (ans).
2) We require 17.0 mL of 0.150 M KOH to reach the equivalence point. The mole(s) of KOH required to reach the equivalence point is (volume of KOH in L)*(molarity of KOH) = (17.0 mL)*(1 L/1000 mL)*(0.150 M) = 0.00255 mole.
Let us denote the weak acid as HA. HA reacts with KOH on a 1:1 molar ratio as
HA (aq) + KOH (aq) ------> KA (aq) + H2O (l)
As per the stoichiometry of the reaction,
1 mole HA = 1 mole KOH.
Therefore, mole(s) of the weak acid neutralized by 0.00255 mole KOH = mole(s) of KOH present = 0.00255 mole (ans).
3) We took 0.297 g of the acid. Again, we have 0.00255 mole of the acid; therefore, we must have,
0.00255 mole = 0.297 g
=====> 1 mole = (0.297 g)/(0.00255 mole) = 116.47 g
The molar mass of the acid is 116.47 g/mol (ans).
4) We took solid acid; hence, the volume change at the end point is negligible and the volume of the acid solution can be taken as 17.0 mL.
The pH at the equivalence point is 8.0; therefore, [H+] = antilog (-pH) = 1.0*10-8 M.
Consequently, [OH-] = Kw/[H+] = (1.0*10-14)/(1.0*10-8) = 1.0*10-6 M.
At the equivalence point, we have KA (A- is the conjugate base of the weak acid HA) which establishes equilibrium as
A- (aq) + H2O (l) ---------> HA (aq) + OH- (aq)
Hence, we obtained [OH-] = 1.0*10-6 M = [HA] (due to the 1:1 nature of the ionization).
Mole(s) of A- formed = moles of HA taken = 0.00255 mole.
Molarity of A- = (moles of A- formed)/(volume of the solution in L) = (0.0255 mole)/[(17.0 mL)*(1 L/1000 mL)] = 0.15 mol/L.
Since OH- is formed, we will work with Kb (base ionization constant of A-) with the assumption that the extent of ionization is much smaller than the initial concentration of HA.
Kb = [HA][OH-]/[A-] = (1.0*10-6)*(1.0*10-6)/(0.15) = (1.0*10-6)2/(0.15) = 6.667*10-12
pKb = -log Kb = -log (6.667*10-12) = 11.176.
Consequently, pKa = 14 – pKb = 14 – 11.176 = 2.824 2.82
The pKa of the weak acid is 2.82 (ans).
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