Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 41 mL of 0.25 M C6H5COOH(aq) with 0.27 M NaOH(aq). The Ka of benzoic acid is 6.5 x 10-5.

Answer 1

equivelent point means all the benzoic acid converted in to the sodium benzoate

C6H5COOH + NaOH ----> C6H5COONa + H2O

no o fmoles of sodium benzoate = no of moles of Benzoic acid

no of moles of Benzoic acid = Molarity of C6H5COOH x volume of the solution in liters

= 0.25 M x 0.041 L

= 0.01025 moles

from the balanced equation no of moles of NaOH = no of moles of C6H5COOH

so moles of NaOH = 0.01025

volume of NaOH = moles of NaOH / molarity of NaOH = 0.01025 / 0.27 = 0.038 L = 38 mL

total volume = 41 + 38 = 79 mL = 0.079L

concentration of C6H5COONa = moles of CH3COONa / total volume = 0.01025 mol / 0.079 L = 0.1297 M

Construct the ICE table

convert Ka in to Pkb

ka to pKa

pKa = -log(Ka) = -log(6.5 x 10^-5) = 4.2

pKa + pKb = 14

pKb = 14-4.2 = 9.8

now there is a direct formula

pOH = 1/2 (pKb - logC)

where C is the concentration of the salt

pOH = 1/2 (9.8 - log(0.1297))

pOh = 1/2 (9.8 - (-0.89))

pOH = 5.34

now pH + pOH = 14

pH = 14-pOH = 14 - 5.34 = 8.65