Calculate the pH for each of the following cases in the titration of 25.0 mL of
ID: 495858 • Letter: C
Question
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.210 M pyridine, C_5H_5N(aq) with 0.210 M HBr(aq): (a) before addition of any HBr (b) after addition of 12.5 mL of HBr (c) after addition of 14.0 mL of HBr (d) after addition of 25.0 mL of HBr (e) after addition of 33.0 mL of HBr At point (a), this is a typical weak base question. B(aq) + H_2O(I) BH^+(aq) + OH^-(aq) k_b = x^2/0.210 M - x almostequalto x^2/0.120 M where k_b = 1.7 times 106-9 and B(aq) is pyridine, C_5H_5N(aq). Solve for x, which is equal to [OH^-], then take the negative logarithm to find pOH which can be converted into pH for the answer to (a). pH = 14.00 - pOHExplanation / Answer
a ) before addition of HBr
It is only a solution of weak base and its pH is given by
pH = 14 - 1/2[pKb -log c]
=14 - 1/2 [9-log 1.7 - log 0.21]
= 9.2763
b) B + HBr ------------------> BH+ + Br-
25x0.210 0 0 0 initial mmoles
12.5x 0.210 change
2.625 0 2.625 - after reaaction
thus it forms a buffer solution , whose pH is given by Hendersen equation as
pH = 14 - {pkb + log [conjugate acid] /[base]}
= 14 - { 8.7696 + log 2.625/2.625}
= 5.2304
c) affter adding 14mL of Hbr
B + HBr ------------------> BH+ + Br-
25x0.210 0 0 0 initial mmoles
14x 0.210 change
2.31 0 2.94 - after reaaction
thus it forms a buffer solution , whose pH is given by Hendersen equation as
pH = 14 - {pkb + log [conjugate acid] /[base]}
= 14 - { 8.7696 + log 2.94/2.31}
=5.1256
d) after addition of 25 mL of HBr
B + HBr ------------------> BH+ + Br-
25x0.210 0 0 0 initial mmoles
25x 0.210 change
0 0 5.25 - after reaaction
The solution now contains only salt of weak base and strong acid , whose pH is given by
pH = 1/2 { pKw -pkb - log C]
the concentration of salt = 5.25/50 M
thus pH = 1/2 [ 14 - 8.7696 - log 5.25/50]
= 4.2515
e) after addition of 33 ml of HBr
B + HBr ------------------> BH+ + Br-
25x0.210 0 0 0 initial mmoles
33x 0.210 change
0 1.68 5.25 - after reaaction
Now the solution contains a strong acid and a weak acid .thus thepH depends only on [strong acid]
The [strong acid] = 1.68/58 =0.0247 M
thus pH of solution = - log [H+]
= - log (0.0247)
= 1.607