Question

Consider the titration of a 28.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Initial pH is 11.95.

Find:

a) the volume of added acid required to reach the equivalence point

b) the pH at 6.0 mL of added acid

c) the pH at one-half of the equivalence point

d) the pH at the equivalence point

e) the pH after adding 5.0 mL of acid beyond the equivalence point

Answer 1

(a) : The balanced chemical equation for the neutralization reaction is

CH3NH2(aq) + HBr(aq) ------ > CH₃NH₃⁺(aq) + Br-(aq)

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

moles of CH3NH2 initially taken = MxV = 0.180 mol/L x 0.0280 L = 0.00504 mol

From the above neutralization reaction it is clear that equal mole of CH3NH2 and HBr react with eachother. Hence

Moles of CH3NH2 = moles of HBr

=> 0.00504 mol = MxV for HBr

=> 0.00504 mol = 0.155 ml/L x V

=> V = 0.0325 L = 32.5 mL

Hence volume of acid required to reach equivalence point is 32.5 mL (answer)

(b): pKb of CH3NH2 = 3.36

Moles of acid added = MxV = 0.155 mol/L x 0.006 L = 0.00093 mol

--------------CH3NH2(aq) + HBr(aq) ------ > CH₃NH₃⁺(aq) + Br-(aq)

Init.mol: 0.00504 mol, 0.00093 mol ------ 0 mol

change: - 0.00093 mol, - 0.00093 mol---- + 0.00093 mol

mol.aft.rxn:0.00411 mol, 0 mol ------------- 0.00093 mol

pOH = pKb + log [CH3NH3+] / [CH3NH2] = 3.36 + log(0.00093 / 0.00411) = 2.715

=> pH = 14 - 2.715 = 11.285 (answer)

(c): At one - half equivalence point, 0.00504 mol / 2 = 0.00252 mol of CH3NH2 is neutralized.

--------------CH3NH2(aq) + HBr(aq) ------ > CH₃NH₃⁺(aq) + Br-(aq)

Init.mol: 0.00504 mol, 0.00252 mol ------ 0 mol

change: - 0.00252 mol, - 0.00252 mol---- + 0.00252 mol

mol.aft.rxn:0.00252 mol, 0 mol ------------- 0.00252 mol

pOH = pKb + log [CH3NH3+] / [CH3NH2] = 3.36 + log(0.00252 / 0.00252) = 3.36

=> pH = 14 - 3.36 = 10.64 (answer)

(d): At equivalence point all of the CH3NH2 is neutralized.

Hence moles of CH3NH3+ = 0.00504 mol

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

-----------------CH3NH3+ ------ > CH3NH2 + H3O+ ; Ka = Kw / Kb = 2.29x10^-11

Initial mol : 0.00504 mol ----- 0 ------------- 0

change: - 0.00504x ----------- +0.00504x, +0.00504x

mol.aft.rxn: 0.00504(1 - x), 0.00504x, 0.00504x

Ka = 2.29x10^-11 = 0.00504x*0.00504x / 0.00504(1 - x)

=> x = 6.74*10^-5

[H3O+] =  0.00504x =  0.00504*6.74*10^-5 = 3.40*10^-7 M

pH = - log[H3O+] = - log 3.40*10^-7 M = 6.47 (answer)