Question

A chemist titrates 250.0 mL of a 0.5659 M lidocaine (C14H21NONH) solution with 0.4276 M HBr solution at 25 °C. Calculate the pH at equivalence. The pK, of lidocaine is 7.94. Round your answer to 2 decimal places Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added. PH -

Answer 1

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