Question

CALCULATE THE PH AT THE EQUIVALENCE POINT FOR THE TITRATION OF 0.250M METHYLAMINE (CH3NH2) WITH 0.25 HCL. THE Kb OF METHYLAMINE IS 5.0X10-4. (HINT: THERE IS NO VALUE FOR VOLUME IN THE PROBLEM SO I AM NOT SURE HOW TO PROCEED! PLEASE HELP.)

Answer 1

The reaction of CH3NH2 with HCl is

CH3NH2 + HCl ----- > CH₃NH₃⁺ + Cl^-

1 mol-------- 1 mol------- 1 ml ------- 1 mol

At equivalence point all of the CH3NH2 are neutralized and there will be only CH₃NH₃⁺ and Cl^-

Since equal moles of CH3NH2 and HCl react with each other, equal volume of 0.250 M HCl is required to titrate equal volume of CH3NH2.

Hence the volume of the solution doubles at equivalence point

Hence [CH₃NH₃⁺] = 0.250 *V / 2V = 0.125 M

At equivalence point, hydrolysis of CH₃NH₃⁺ will occur to form CH3NH2 and H3O+

--------------------CH₃NH₃⁺ ------- > CH3NH2 and H3O+; Ka = 10^-14 / Kb = 10^-14 / 5.0*10^-4 = 2.0*10^-11

Init.Conc(M): 0.125 --------------- 0 --------------- 0

Eqm.conc(M):0.125(1 - x)----- 0.125x --------- 0.125x

2.0*10^-11 = 0.125x*0.125x / 0.125(1 - x)

=> x = 1.265*10^-5  

Hence [CH3NH2] = 0.125x = 0.125*1.265*10^-5 = 1.58*10^-6 M

[CH₃NH₃⁺] = 0.125(1 - x) $\approx$ 0.125 M

Now applying Hendersen equation

pOH = pKb + log[CH₃NH₃⁺] / [CH3NH2] = - log(5.0*10^-4 ) + log (0.125 M / 1.58*10^-6 M)

=> pOH = 8.20

=> pH = 14 - 8.20 = 5.8 (answer)