Question

Consider the titration of 100.0 mL of 0.500 M CH3NH2 by 0.250 M HCl. (Kb for CH3NH2 = 4.4×10-4) Part 1 Calculate the pH after 0.0 mL of HCl added. pH = Part 2 Calculate the pH after 25.0 mL of HCl added. pH = Part 3 Calculate the pH after 80.0 mL of HCl added. pH = Part 4 Calculate the pH at the equivalence point. pH = Part 5 Calculate the pH after 300.0 mL of HCl added. pH = Part 6 At what volume of HCl added, does the pH = 10.64? mL

Answer 1

solution:-

Part (1) when no HCl is added,

CH3NH2 + H2O <------> CH3NH3+ + OH-

I 0.500 0 0

C -X +X +X

0.500 - X X X

4.4 * 10^-4 = (X)^2/(0.500 - X)

X^2 + 0.00044X - 0.00022 = 0

on solving this quadratic equation,

X = 0.0146

[OH-] = 0.0146

pOH = -log(0.0146)

pOH = 1.84

pH = 14 - 1.84 = 12.16 (answer)

Part (2)

moles of CH3NH2 = 0.500 * 0.1ml = 0.0500

moles of HCl = 0.250 * 0.025 = 0.00625

CH3NH2 + H+ <-------> CH3NH3+

I 0.0500 0.00625 0

C -0.00625 -0.00625 +0.00625

E 0.04375 0 0.00625

here we have a weak base and it's conjugate acid, so it's a buffer solution. Handerson equation could be applied for this.

pOH = Pkb + log(salt/base)

from Kb, Pkb = - log(4.4 * 10^-4) = 3.36

so, pOH = 3.36 + log(0.00625/0.04375)

pOH = 3.36 - 0.845

pOH = 2.515

pH = 14 - 2.515 = 11.485 (answer)

Part(3)

moles of CH3NH2 = 0.050

moles of HCl added = 0.250 * 0.08 = 0.020

moles of salt (CH3NH3+) formed = 0.020

excess moles of CH3NH2 = 0.050 - 0.020 = 0.030

Here we again have a buffer solution so Handerson equation could be used.

pOH = 3.36 + log(0.02/0.03)

pOH = 3.36 - 0.176 = 3.184

pH = 14 - 3.184 = 10.816(answer)

Part (4)

at equivalence point moles of acid and base would be in their stoichiometry ratio. (1:1 ratio from balanced equation)

so, 0.050 moles of HCl would be used since we have 0.050 moles pf CH3NH2.

voluem of HCl used = moles/molarity = 0.050/0.250 = 0.2 L = 200 ml

total volume = 100 ml + 200 ml = 300 ml = 0.3L

moles of CH3NH3+ formed = 0.050

molarity of salt = 0.050/0.3 = 0.167M

Hydrolysis of this salt would take place now.

CH3NH3+ + H2O ----> CH3NH2 + H3O+

I 0.167 0 0

C -X +X +X

E 0.167 - X X X

Ka = (X)^2/(0.167 - X)

since ka = kw/kb = 1.0 x 10^-4/(4.4 x 10^-4) = 2.27 x 10^-11

2.27 x 10^-11 = X^2/(0.167-X)

since value of ka is very low so, 0.167 - X could be taken as 0.167

so, 2.27 x 10^-11 = X^2/(0.167)

on solving it, X = 1.95 x 10^-6

[H3O+] = 1.95 x 10^-6

pH = - log(1.95 x 10^-6) = 5.71 (answer)

Part(5)

moles of CH3NH2 = 0.050

moles of HCl added = 0.250 * 0.3 = 0.075

excess moles of HCl = 0.075 - 0.050 = 0.025

total volume = 100 ml + 300 ml = 400 ml = 0.4L

HCl concentration in solution = 0.025/0.4 = 0.0625

pH = - log(0.0625) = 1.20(answer)

Part (6)

we must have a buffer solution when the pH is 10.64

let's say X moles of HCl are added.

then moles of CH3NH3+ formed would = X

and moles of excess CH3NH2 = 0.050 - X

pOH = 14 - 10.64 = 3.36

3.36 = 3.36 + log[X/(0.050-X)

3.36 - 3.36 = log[X/(0.050-X)

0 = log[X/(0.050-X)

10^0 = [X/(0.050-X)

1 = [X/(0.050-X)

0.050 - X = X

0.050 = 2X

X = 0.050/2 = 0.025

so, Moles of HCl added = 0.025

volume of HCl added = moles/molarity = 0.025/0.25 = 0.1L = 100 ml (answer)