Question

Consider the titration of a 25.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr....

Consider the titration of a 25.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Determine each of the following.

a. the pH at one-half of the equivalence point

b. the pH at the equivalence point

c. the pH after adding 6.0 mL of acid beyond the equivalence point

i already found that the initial pH is 11.95, the volume of acid added to reach equivelance point is 29.0mL, and the pH og 6.0mL of added acid is 11.23

a)

pH of half equivalence point

is by definition...

pOH = pKb

pKb = -log(Kb) = -log(4.38*10^-4) = 3.358

since

pOH = pKb + log(CH3NH3+ / CH3NH2)

and in the half equivalence point CH3NH3 += CH3NH2

then log (1) = 0

pOH = pKb = 3.358

pH = 14-3.358 = 10.64

b)

pH in equilvalence

[CH3NH3+] = MV/(V1+V2) = 0.18*25/(25+29) = 0.0833

CH3NH3+ + H2O <-> CH3NH2 + H3O+

Ka= [CH3NH2][H3O+]/[CH3NH3+]

Ka = Kw/Kb =(10^-14)/(4.38*10^-4) = 2.283*10^-11

Ka= [CH3NH2][H3O+]/[CH3NH3+]

2.283*10^-11= x*x/(0.0833-x)

x = H+ = 1.379*10^-6

pH = -.log(H) = -log( 1.379*10^-6) = 5.8604