Post Lab Questions: To receive full credit, you must SHOW ALL YOUR WORKI You car
ID: 546686 • Letter: P
Question
Post Lab Questions: To receive full credit, you must SHOW ALL YOUR WORKI You carry out a titration of a pure weak triprotic acid and only see two major pH jumps. One at 10.0 mL and one at 15.0 mL. At what volume is the other equivalence point? 1. 2. You carry out the titration of a weak diprotic solid acid. You take 0.178 g of the solid acid and dissolve it in about 50 mL of distilled water. You titrate with 0.1005 M NaOH and arrive at the second endpoint at 24.20 ml. You know that at the final pH both protons have been titrated. What is the molar mass of the acid?Explanation / Answer
1) The triprotic acid shows a jump in pH when one of the protons is neutralized. The first pH jump occurs at 10.0 mL of the added base when the first proton is lost. The second pH jump occurs at 15.0 mL when the second proton is lost. The triprotic acid will require equal increments of the base between the first and the second equivalence points as well as the second and third equivalence points. Hence, the third equivalence point will occur at 20.0 mL of the added base (ans).
2) Let M be the molar mass of the acid.
Write down the balanced chemical equation for the titration of the diprotic acid (H2A) with NaOH.
H2A(aq) + 2 NaOH(aq) ----------> Na2A(aq) + 2 H2O(l)
As per the stoichiometric equation, for neutralization of 1 mole of H2A, we will require 2 moles of NaOH.
Mole(s) of NaOH required for complete neutralization of H2A = (24.20 mL)*(1 L/1000 mL)*(0.1005 M)*(1 mol/L/1 M) = 2.4321*10-3 mole.
Mole(s) of H2A neutralized = 2.4321*10-3 mole.
As per the problem, moles of H2A taken = (0.178 g)/M. Therefore,
(0.178 g)/M = 2.4321*10-3 mole
====> M = (0.178 g)/(2.4321*10-3 mole) = 73.18778 g/mol 73.19 g/mol (ans).