Question

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

Answer 1

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