Question

Can somebody please help me with this? I already asked this and nobody answered.

2. Pyridine is a weak base. Its conjugate acid (call it BH*) has pk, 5.23. Consider the titration of 20.0 mL of 0.0500 M B with 0.100 M HCI (a) Write the net ionic equation for the titration reaction. What is the equilibrium constant for this reaction, and why can we assume that it "goes to completion"? Why should you expect pH < 7 at the equivalence point? (b) Calculate the volume of titrant required to reach the equivalence point. Include units throughout with an explicit mole ratio. (c) Calculate the initial pH, using our "hopeful approximation." Check this approximation according to the 5 % rule, but do not attempt a more accurate calculation if it does not hold up. (d) Calculate the total concentration (after dilution) of [B]+ (HB ] at the equivalence point. Then calculate the pH at the equivalence point, again using our hopeful approximation. (e) Why will pH 5.23 when 5.00 mL of titrant has been added? Calculate the pH after 4.00 mL and 7.50 mL of titrant have been added. Why do we not have to correct for dilution here? (f) Calculate the pH after 20.0 mL of titrant have been added. What will the "asymptotic" pH be when a large excess of titrant has been added? Make a reasonably accurate sketch of the titration curve on a separate sheet of paper.

Answer 1

s as CGH5 Neast at Caths NH Can 1614 Ko se o 16."= 1.4 %169 -7X107 105:23 this is weak base & strong acid titration. conjugate acid is acidic.so &H <7) at Equivales point 20.0 0.0 500 = 0.100X V V 10 volume of titraat at equivalena point Tv=lome pont-'s (ples-log c) = z(8-79 -logo.as) pot = 5.04 pH = 8.96 -0-033 M [BH+] = 2040-0500/20+10 [BH+] = 0.033 M pH = 7-% (pkb +log c) --7-% (8.77 +log 0.033) pH = 3.35]

e)

mmoles of HCl = 5 x 0.100 = 0.5

mmoles of B = 20 x 0.05 = 1

B   +   H+     ------------->   BH+

1.0     0.5                             0

0.5       0                             0.5

this is half - equivalence point. here pH = pKa

pH = 5.23

after 4 ml of HCl added :

mmoles of B = 20 x 0.05 = 1

B   +   H+     ------------->   BH+

1.0     0.4                             0

0.6      0                             0.4

pH = pKa + log [0.6 / 0.4]

pH = 5.41

after 7.50 mL added :

pH = 4.75

f)

pH = 1.60