Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 4.00 ✕ 102 mL of solution and then titrate the solution with 0.163 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-

Answer 1

mol of acid = mass/MW = 0.235/122.123 = 0.0019242 mol of benzoic acid

V = 4*10^2 = 400 mL = 0.4 L

M = mol/V = 0.0019242 /0.4 = 0.00481 M of acid

initially

for the base

mol of acid = mol of base

mol of base 0.0019242

M = mol/V

V = mol/M = 0.0019242 /0.163 = 0.0118049 Liters of base = 0.0118049*10^3 = 11.8 mL required for neutralization

Vtotal = V1+V2 = 400+11.8 = 411.8 mL

Benzoate ion = 0.0019242 /(411.8*10^-3) = 0.0046726 M

so...

A- + H2O <-> HA + OH-

Kb = [HA][OH-]/[A-]

Kb = Kw/Ka = (10^-14)/( 6.46*10^-5) = 1.547*10^-10

so

[HA]= x

[OH-] = x

[A-] = M-x = 0.0046726 -x

substitute

Kb = [HA][OH-]/[A-]

1.547*10^-10 = x*x(0.0046726 -x)

x= OH- = 8.5*10^-7

so...

Na+ = Mbase*Vbase / total V = 0.163*11.8/411.8 = 0.0046707 M of Na+

[OH-] = 8.5*10^-7

H+ = (10^-14)/( 8.5*10^-7) = 1.176*10^-8