I know pH=pKa+log(base/acid) but for base/acid I keep getting 0/0.1 and when cal
ID: 571182 • Letter: I
Question
I know pH=pKa+log(base/acid) but for base/acid I keep getting 0/0.1 and when calculated i get an undefined error.
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.200 M pyridine, C5H5N(aq) with 0.200 M HBr(aq) At point (a), this is a typical weak base question Number Blaq) +H2O(l) M'(aq) + OH-(aq) (a) before addition of any HBr 4.7 0.200 M 0 Number + X (b) after addition of 12.5 mL of HBr|5.23 0.200 M-x Number K, (c) after addition of 24.0 mL of HBr 3.85 "0.200 M-r 0.200 M Number where Kb = 1.7 x 10-9 M and B(aq) is pyridine, (d) after addition of 25.0 mL of HBr 5.23 C5H5N(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) Number (e) after addition of 32.0 mL of HBr pH= 14.00-p0HExplanation / Answer
a)when 0.0 mL of HBr is added
C5H5N dissociates as:
C5H5N +H2O -----> C5H5NH+ + OH-
0.2 0 0
0.2-x x x
Kb = [C5H5NH+][OH-]/[C5H5N]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.7*10^-9)*0.2) = 1.844*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.844*10^-5 M
So, [OH-] = x = 1.844*10^-5 M
use:
pOH = -log [OH-]
= -log (1.844*10^-5)
= 4.7343
use:
PH = 14 - pOH
= 14 - 4.7343
= 9.27
Answer: 9.27
b)when 12.5 mL of HBr is added
Given:
M(HBr) = 0.2 M
V(HBr) = 12.5 mL
M(C5H5N) = 0.2 M
V(C5H5N) = 25 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.2 M * 12.5 mL = 2.5 mmol
mol(C5H5N) = M(C5H5N) * V(C5H5N)
mol(C5H5N) = 0.2 M * 25 mL = 5 mmol
We have:
mol(HBr) = 2.5 mmol
mol(C5H5N) = 5 mmol
2.5 mmol of both will react
excess C5H5N remaining = 2.5 mmol
Volume of Solution = 12.5 + 25 = 37.5 mL
[C5H5N] = 2.5 mmol/37.5 mL = 0.0667 M
[C5H5NH+] = 2.5 mmol/37.5 mL = 0.0667 M
They form basic buffer
base is C5H5N
conjugate acid is C5H5NH+
Kb = 1.7*10^-9
pKb = - log (Kb)
= - log(1.7*10^-9)
= 8.77
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 8.77+ log {6.667*10^-2/6.667*10^-2}
= 8.77
use:
PH = 14 - pOH
= 14 - 8.7696
= 5.2304
Answer: 5.23
c)when 24.0 mL of HBr is added
Given:
M(HBr) = 0.2 M
V(HBr) = 24 mL
M(C5H5N) = 0.2 M
V(C5H5N) = 25 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.2 M * 24 mL = 4.8 mmol
mol(C5H5N) = M(C5H5N) * V(C5H5N)
mol(C5H5N) = 0.2 M * 25 mL = 5 mmol
We have:
mol(HBr) = 4.8 mmol
mol(C5H5N) = 5 mmol
4.8 mmol of both will react
excess C5H5N remaining = 0.2 mmol
Volume of Solution = 24 + 25 = 49 mL
[C5H5N] = 0.2 mmol/49 mL = 0.0041 M
[C5H5NH+] = 4.8 mmol/49 mL = 0.098 M
They form basic buffer
base is C5H5N
conjugate acid is C5H5NH+
Kb = 1.7*10^-9
pKb = - log (Kb)
= - log(1.7*10^-9)
= 8.77
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 8.77+ log {9.796*10^-2/4.082*10^-3}
= 10.15
use:
PH = 14 - pOH
= 14 - 10.1498
= 3.8502
Answer: 3.85
d)when 25.0 mL of HBr is added
Given:
M(HBr) = 0.2 M
V(HBr) = 25 mL
M(C5H5N) = 0.2 M
V(C5H5N) = 25 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.2 M * 25 mL = 5 mmol
mol(C5H5N) = M(C5H5N) * V(C5H5N)
mol(C5H5N) = 0.2 M * 25 mL = 5 mmol
We have:
mol(HBr) = 5 mmol
mol(C5H5N) = 5 mmol
5 mmol of both will react to form C5H5NH+ and H2O
C5H5NH+ here is strong acid
C5H5NH+ formed = 5 mmol
Volume of Solution = 25 + 25 = 50 mL
Ka of C5H5NH+ = Kw/Kb = 1.0E-14/1.7E-9 = 5.882*10^-6
concentration ofC5H5NH+,c = 5 mmol/50 mL = 0.1 M
C5H5NH+ + H2O -----> C5H5N + H+
0.1 0 0
0.1-x x x
Ka = [H+][C5H5N]/[C5H5NH+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.882*10^-6)*0.1) = 7.67*10^-4
since c is much greater than x, our assumption is correct
so, x = 7.67*10^-4 M
[H+] = x = 7.67*10^-4 M
use:
pH = -log [H+]
= -log (7.67*10^-4)
= 3.1152
Answer: 3.12
5)when 32.0 mL of HBr is added
Given:
M(HBr) = 0.2 M
V(HBr) = 32 mL
M(C5H5N) = 0.2 M
V(C5H5N) = 25 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.2 M * 32 mL = 6.4 mmol
mol(C5H5N) = M(C5H5N) * V(C5H5N)
mol(C5H5N) = 0.2 M * 25 mL = 5 mmol
We have:
mol(HBr) = 6.4 mmol
mol(C5H5N) = 5 mmol
5 mmol of both will react
excess HBr remaining = 1.4 mmol
Volume of Solution = 32 + 25 = 57 mL
[H+] = 1.4 mmol/57 mL = 0.0246 M
use:
pH = -log [H+]
= -log (2.456*10^-2)
= 1.6097
Answer: 1.61