Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough w

Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.00*10^2 102 mL of solution and then titrate the solution with 0.148 M NaOH C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O() 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?

Explanation / Answer

Assume you dissolve 0.235 g of a weak acid, benzoic acid ( C6H5CO2H ), in enough water to make 1.00 x 10^2 mL of solution and then titrate the solution with 0.108 M NaOH: C6H5CO2H (aq) + OH- (aq) ----> C6H5CO2- (aq) + H2O (l). What are the concentrations of the following ions at equilibrium: Na+, H+, OH-, and C6H5CO2-? What is the pH of the solution? moles benzoic acid = 0.235 g / 122.12 g/mol=0.00192 moles NaOH required = 0.00192 volume of NaOH required = 0.00192 volume of NaOH = 0.00192 / 0.108 M=0.0178 L total volume = 0.0178 L + 0.100 L = 0.118 L moles benzoate formed = 0.00192 concentration benzoate = 0.00192 / 0.118 L=0.0163 M C6H5COO- + H2O C6H5COOH + OH- Kb = Kw/Ka = 1.0 x 10^-14 / 6.3 x 10^-5 = 1.6 x 10^-10 M 1.6 x 10^-10 = x^2 / 0.0163-x x = [OH-]= [C6H5COOH]= 1.61 x 10^-6 M [H+]= 1.0 x 10^-14 / 1.61 x 10^-6=6.21 x 10^-9 pH = 8.21 [Na+]= 0.00192 / 0.118 L=0.0163 M

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