Question

A 111.0 mL sample of 0.105 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.265 M HNO3. Calculate the pH after the addition of each of the following volumes of acid:

22.0 mL

Express the pH to two decimal places.

44.0 mL

Express the pH to two decimal places.

66.0 mL

Express the pH to two decimal places.

Answer 1

Chemical equation for the acid - base reaction is

CH3NH2(aq) + HNO3(aq) ------ > CH₃NH₃⁺(aq) + NO₃^-(aq)

1 mol -------------- 1 mol ------------- 1 mol ------------- 1 mol

moles of CH3NH2 initially taken = MxV = 0.105 mol/L x 0.111 L = 0.01166 mol

(a): When 22.0 mL 0.265 M HNO3 is added:

Moles of HNO3 added = MxV = 0.265 mol/L x 0.022 L = 0.00583 mol HNO3

--------------CH3NH2(aq) + HNO3(aq) ------ > CH₃NH₃⁺(aq) + NO₃^-(aq)

Init.mol: 0.01166 mol, 0.00583 mol -------- 0 mol

change: - 0.00583 mol , - 0.00583 mol ---- + 0.00583 mol

mol.aft.rxn:0.00583 mol, 0 mol ------------- 0.00583 mol

Applying Hendersen equation

pOH = pKb + log [CH3NH3+] / [CH3NH2] = 3.43 + log( 0.00583 mol / 0.00583 mol) = 3.43

=> pH = 14 - 3.43 = 10.57 (answer)

(b): When 44.0 mL 0.265 M HNO3 is added:

Moles of HNO3 added = MxV = 0.265 mol/L x 0.044 L = 0.01166 mol HNO3

Since moles of HNO3 added is same as the moles of CH3NH2 initially present, all of the CH3NH2 is neutralized and equivalence point is achieved.

Hence moles of CH3NH3+ = 0.01166 mol

Now CH3NH3+ will undergo hydrolysis to form Ch3NH2 and H3O+

-----------------CH3NH3+ ------ > CH3NH2 + H3O+ ; Ka = Kw / Kb = 2.70*10^-11

Initial mol : 0.01166 mol ----- 0 ------------- 0

change: - 0.01166y ----------- +0.01166y, +0.01166y

mol.aft.rxn: 0.01166(1 - y), --- 0.01166y, 0.01166y

Ka = 2.7x10^-11 = 0.01166y * 0.01166y / 0.01166(1 - y)

=> y = 4.81*10^-5

[H3O+] =  0.01166y =  0.01166*4.81*10^-5 = 5.61*10^-7 M

pH = - log[H3O+] = - log 5.61*10^-7 M = 6.25 (answer)

(c): When 66.0 mL 0.265 M HNO3 is added:

Moles of HNO3 added = MxV = 0.265 mol/L x 0.066 L = 0.01749 mol HNO3

Moles of HNO3 remain unreacted = 0.01749 mol - 0.01166 mol = 0.00583 mol HNO3

Total volume = 111.0 mL + 66.0 mL = 177 mL = 0.177 L

[HNO3] = [H3O+] = 0.00583 mol HNO3 / 0.177L = 0.03294 M

=> pH = - log[H3O+] = - log 0.03294 M = 1.48 (answer)