Question

Consider the titration of a 28.0 mL sample of 0.175M CH3NH2 with a 0.145M HBr. Determine each of the following.

I already got initial pH: 11.94 and the volume of added acid required to reach equivalence point: 33.8 mL.

I still need help with these:

1. The pH of 4.0 mL of added acid.

2. The pH at one-half of the equivalence point.

3. The pH of the equivalence point.

4. The pH after adding 4.0 mL of acid beyond the equivalence point.

Answer 1

CH3NH2 + H2O <------> CH3NH3+ + OH-, Kb = 4.365x10^-4

Kb = x2/(0.175-x) => x = 0.00874 M

Therefore, initial pH = 14 -(-log0.00874) = 11.94

1) Moles of CH3NH2 = 0.028*0.175 = 0.0049

Moles of HBr in 4mL = 0.145*0.004 = 0.00058

[CH3NH2] = 0.0049/0.032 = 0.153M and [HBr] = 0.0181 M

Kb = x(x-0.0181)/(0.153-x), solving for x,

x = 0.0241

Therefore, [OH-] = 0.006 M => pH = 11.78

2) Volume of HBr added at equivalence point is 33.8mL and volume of HBr at half equivalence is 16.9 mL

[CH3NH2] at equivalence = 0.0049/(0.028+0.0338) = 0.0793M and at half equivalence = 0.1091 M

[HBr] at equivalence = 0.145*33.8/61.8 = 0.0793 and at half equivalnece = 0.0546

pH at half equivalence:

Kb = x(x-0.0546)/(0.1091-x), solving for x, x = 0.0551

[OH-] = 0.0551-0.0546 = 0.0005 = > pH = 10.7

3) pH at equivalence:

Kb = x(x-0.0793)/(0.0793-x) => Kb = -x, the negative sign indicates the solution has become acidic

Therefore, the reaction is CH3NH2 + H3O+ <------> CH3NH3+ + H2O, Ka = Kw/Kb = 2.291x10^-11

2.291x10^-11 = x/(0.0793-x)² => x = 2.291x10^-11

Therefore, [H3O+] = ~0.0793 and pH = 1.1