Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 3.00 ✕ 102 mL of solution and then titrate the solution with 0.138 M NaOH.

C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(ℓ)

What are the concentrations of the following ions at the equivalence point?

Na+, H3O+ ,OH-

C6H5CO2-

M Na+

M H3O+

M OH-

M C6H5CO2-

What is the pH of the solution?

Answer 1

m = 0.235 g of benzoic acid

MW of acid = 122.12

mol = mass/MW = 0.235/122.12 = 0.0019243 mol of B.Acid

V = 3*10^2 = 300 mL = 0.3 L

Macid = mol/V = 0.0019243 /0.3 = 0.0064143 M

M = 0.138 M of NaOH

in equivalence point:

mmol of acid = mmol of base

so

M1V1 = M2V2

0.0064143*0.3 = 0.138*V2

V2 = 0.0064143*0.3/0.138 = 0.0139 L = 13.9 mL

then..

Vtotal in equivalence point = 300+13.9 = 313.9 mL

calcualte concentration of acid:

then..

expect benzoate formation

[Benzoate] = 0.0019243 / (313.9 *10^-3) = 0.00613 M

then...

B- + H2O <--> HB + OH-

Kb = [HB][OH-]/[B-]

Kb = Kw/Ka = (10^-14)/(6.46*10^-5) = 1.54798*10^-10

Kb = [HB][OH-]/[B-]

1.54798*10^-10 = x*x/(0.00613 -x)

x = [HB] = [OH-] = 9.64*10^-7

[H3O+] = (10^-14)/(9.64*10^-7) = 1.0373*10^-8 M

[Na+] = mol of Na+ / total V = Mb*Vb/VT = 0.138*13.9/313.9 = 0.00611086M

[Benzoate] = M-x = 0.00613 - 9.64*10^-7 = 0.00612 M

pH = -log(H) = -log( 1.0373*10^-8) = 7.984