Question

Consider the titration of a 28.0 -ml sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following

-the pH at the equivalence point

please explain. thank you!

Answer 1

find the volume of HBr used to reach equivalence point

M(CH3NH2)*V(CH3NH2) =M(HBr)*V(HBr)

0.175 M *28.0 mL = 0.155M *V(HBr)

V(HBr) = 31.6129 mL

Given:

M(HBr) = 0.155 M

V(HBr) = 31.6129 mL

M(CH3NH2) = 0.175 M

V(CH3NH2) = 28 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.155 M * 31.6129 mL = 4.9 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.175 M * 28 mL = 4.9 mmol

We have:

mol(HBr) = 4.9 mmol

mol(CH3NH2) = 4.9 mmol

4.9 mmol of both will react to form CH3NH3+ and H2O

CH3NH3+ here is strong acid

CH3NH3+ formed = 4.9 mmol

Volume of Solution = 31.6129 + 28 = 59.6129 mL

Ka of CH3NH3+ = Kw/Kb = 1.0E-14/4.4E-4 = 2.273*10^-11

concentration ofCH3NH3+,c = 4.9 mmol/59.6129 mL = 0.0822 M

CH3NH3+ + H2O -----> CH3NH2 + H+

8.22*10^-2 0 0

8.22*10^-2-x x x

Ka = [H+][CH3NH2]/[CH3NH3+]

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 ((2.273*10^-11)*8.22*10^-2) = 1.367*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.367*10^-6 M

[H+] = x = 1.367*10^-6 M

use:

pH = -log [H+]

= -log (1.367*10^-6)

= 5.8643

Answer: 5.86