Question

In the titration of a 25.00 mL of 0.245 M weak base (Kb= 1.0 *10^-4) being titrated by 0.365 M HCl determinethe initial pH, the pH after 12.3 mL of HCl has been added, the pH at equivalence point, and the pH after 22.4 mL of HCl has been reached.

Please explain! I'm having a hard time with these type of questions.

Answer 1

a) when no HCL is added
BOH ----> B+ + OH-
0.245 M 0 0
0.245-x x x

Kb= [B+][OH-]/[BOH]
1*10^-4 =x*x / (0.245-x)

since Kb is small, x will be very small and it can be ignored as compared to 0.245
above expression becomes,
1*10^-4 =x*x / (0.245)
x = 4.95*10^-3 M
[OH-] = 4.95*10^-3 M
pOH = -log [OH-]
=- -log (4.95*10^-3)
=2.3
pH = 14 -pOH
=14 - 2.3
= 11.7
Answer: 11.7

b)
initial mmol of base = 0.245*25 = 6.125 mmol
moles of acid added = 0.365*12.3 = 4.49 mmol

4.49 mmol of each will raect
remaining base = 6.125 - 4.49 = 1.635 mmol

total volume = 12.3 + 25 = 37.3 mL

[BOH] = 1.635/37.3= 0.044 M

BOH ----> B+ + OH-
0.044 0 0
0.044-x x x

Kb= [B+][OH-]/[BOH]
1*10^-4 =x*x / (0.044-x)

since Kb is small, x will be very small and it can be ignored as compared to 0.245
above expression becomes,
1*10^-4 =x*x / (0.044)
x = 2.1*10^-3 M
[OH-] = 2.1*10^-3 M
pOH = -log [OH-]
=- -log (2.1*10^-3)
=2.68
pH = 14 -pOH
=14 - 2.68
= 11.32

you can solve remaining by following same steps

Question is becoming very lengthy

I hope it helps