Question

8:21 PM Wed Oct 30 85%O + In a titration in which 0.1037 FHCI is added to 25.00 mL of a weak base with pK 4.50, the end point is reached after the addition of 32.65 mL of the acid. wB S A [2 pts] What is the concentration of the weak base? (a) C v C2 V U lo37xo. 63265 V 2 E01354 o25 [4 pts] Calculate the pH in the titration flask .00 mL before the equivalence point OR 1.00 mL (b) after the equivalence point? (Choose either) 2. 75 (c) [3 pts] What indicator should be used? Why? (Show calculations. A

Answer 1

al - Concentration added 6 = 0.1037F ci = 0.1037M Volume of added HCl = V = esoooh 32.65ml Given volume of weak bare = ₂ = 25.00ml Suppose Concentration of weak ban be 'G' Molar At End point ciri = &v2 0.103+M X 32.65 ml = SX 25-comL 18 = 0.13sum | Answer - ImL before equivalence point Volume of added Hel = (32.65-1) mL = 31.65mL No of m.moles of added HCl = 0.1034 MX 31-65 mL = 3.282 m.mol weak baue =0,1354mX 25.00mL - 3-osmomok No of m.mole of weak bace left = ( 3.395 – 3-222)m.cool = 0.103m.mob. No of mmole of salt formed = No of mmole of acid added This is basic buffer solution = 3-282 mmole [w.Bt S.A [Bar] J pH = 14 - (pkb +log Exalt) - 14 = [9:50 tlong le enam PH = 7.gg 1 Anoon

ImL aften Equivalence Point Volume of Hel added = (32-65 + 1) ML = 33.65 mL No of m.moles of He added = 0.1037M X 33-65ML = 3.48 m-mole No of momoles of weak base = 3.385m. mole Ich left = 13.48 -3.385) momol = 0.095m.mok = o-ogsx 103 mole Imole = 10 m-mol Total volume of solution (M = (33-65+25) mL - 58-65mL -0.05865L. Concentration of Hol = o.ogsxlo mole 1.62x10 0.05865L [Hol] = 1.628103 [14+] = 1.628103 pH = -log[ht ] = -log(1-62 X10°) pH = 2.79 1 Answer o The selection of indicator is done in such a way that it should encompars the PH interval change at Equivalence point Red Methyl sange can be used for this titration. As its PH range is AA 4-2- 6.0 At equivalence point, to of mindles of salt formed - 3.38 sm.mole Total volume of solution = (25+ 32-65) mL = 57-65ML Concentration of salt formed = 3.385m-mole - 57.65mL OSIM

- PH - pko - logo = = [14 - 4.50 - logo-os et] PH = 5.36 Ansunes