Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 8.00 ✕ 102 mL of solution and then titrate the solution with 0.158 M NaOH.      C6H5CO2H(aq) + OH-(aq)--> C6H5CO2-(aq) + H2O(ℓ)

What are the concentrations of the following ions at the equivalence point? M-Na+=___

M-H3O+=____

M-OH-=____

M-C6H5CO2-=_____

What is the pH of the solution? _____

Answer 1

First we calculate the number of moles and molarity of the weak acid:
Molar mass = 122.12g/mol

moles = 0.235 g / 122.12 g/mol = 1.924 mmol

Molarity = 1.924 mmol / 800 ml = 0.0024 M

At equivalence point: M1*V1 = M2*V2, or

the moles of acid = the moles of base =1.924mmol

1.924 = 0.158 * V2

V₂ =12.18 mL

I: 1.924mmol                    1.924mmol                       0

C: -1.924mmol                 -1.924mmol                     +1.924mmol

E:   0                                  0                                    1.924mmol

All we left over is 1.924mmol conjugated base of the weak acid.

[Na+]=0,

[C₆H₅CO₂^-]= 1.924 / (800 ml + 12.18 ml) = 0.0024 M     

I: 0.0024M                                            0                     0

C: -x                                                   +x                     +x

E: 0.0024-x                                          x                      x

Now we need Kb of the conj base:

Kb = Kw / Ka = 10^-14 / 6.3*10^-5 = 1.587*10^-10

Kb = x² / (0.0024 - x)

or ~=

1.587*10^-10 = x² / 0.0024

solve for x ----> x=6.17*10^-7 M = [OH-]

Also,

[H₃O⁺] = 10^-14 / [OH-] = 10^-14 / 6.17*10^-7 = 1.62*10^-8 M

Finally: What is the pH of the solution at the equivalence point?

pOH=- log([OH^-])=-log(6.17*10-7)= 6.21

pH=14-pOH=14 - 6.23 = 7.79