use:
pKb = -log Kb
7.94= -log Kb
Kb = 1.148*10^-8
find the volume of HBr used to reach equivalence point
M(C14H21NONH)*V(C14H21NONH) =M(HBr)*V(HBr)
0.5659 M *250.0 mL = 0.4276M *V(HBr)
V(HBr) = 330.86 mL
Given:
M(HBr) = 0.4276 M
V(HBr) = 330.86 mL
M(C14H21NONH) = 0.5659 M
V(C14H21NONH) = 250 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.4276 M * 330.86 mL = 141.475 mmol
mol(C14H21NONH) = M(C14H21NONH) * V(C14H21NONH)
mol(C14H21NONH) = 0.5659 M * 250 mL = 141.475 mmol
We have:
mol(HBr) = 141.475 mmol
mol(C14H21NONH) = 141.475 mmol
141.475 mmol of both will react to form C14H21NONH2+ and H2O
C14H21NONH2+ here is strong acid
C14H21NONH2+ formed = 141.475 mmol
Volume of Solution = 330.8583 + 250 = 580.8583 mL
Ka of C14H21NONH2+ = Kw/Kb = 1.0E-14/1.1481536214968817E-8 = 8.71*10^-7
concentration ofC14H21NONH2+,c = 141.475 mmol/580.8583 mL = 0.2436 M
C14H21NONH2+ + H2O -----> C14H21NONH + H+
0.2436 0 0
0.2436-x x x
Ka = [H+][C14H21NONH]/[C14H21NONH2+]
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 ((8.71*10^-7)*0.2436) = 4.606*10^-4
since c is much greater than x, our assumption is correct
so, x = 4.606*10^-4 M
[H+] = x = 4.606*10^-4 M
use:
pH = -log [H+]
= -log (4.606*10^-4)
= 3.3367
Answer: 3.34
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