Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C₆H₅CO₂H, in enough water to make 1.00 ✕ 10² mL of solution and then titrate the solution with 0.128 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

Number of moles of benzoic acid = mass/mol.wt. = 0.235 g/(122.12 g/mol) = 0.001924 mol Volume of benzoic acid = 100 mL At equilibrium, moles of NaOH = moles of benzoic acid Volume of NaOH needed to reach equivalence point = 0.001924 mol/0.128 M = 0.015034 L = 15.03 mL At this point, there is no benzoic acid is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate base. moles [CH,000"]= 0.001924 mmol Total volume (0.100+ 0.015034) mL -= 0.01673M Kw 1x10-14 K, for CH COO 10-5 = 1.5873x10-10 Ka 6.3x10-5 -

benzoate ion in water is given by 1 Kb M CH COO + H2O = CH,COOH + Initial 0.01673 Change Equillibrium 0.01673 - x -=1.5873x10-10 0.01673-X x is small, so 0.01673 - x=0.01673 _ x.x <=1.5873x10-10 0.01673 x? = 2.66x10-12 x=1.63x10M :: [OH-]=1.63x10M pOH =-log(1.63x10^)=5.79 pH = 14 – POH = 14 – 5.79 = 8.21 [Na+) = moles of NaOH/total volume = 0.0167 M [OH-] = 1.63 x 10-6 M. [H3O+] = 1x 10-14/(1.63 x 10-6 M) = 6.14 x 10-9 M [C6H5COO-] = 0.01673 - x = 0.0167 M.