Question

# Suppose 25.00 mL of 0.250 M benzoic acid are titrated with 0.200 M NaOH. What is...

Suppose 25.00 mL of 0.250 M benzoic acid are titrated with 0.200 M NaOH. What is the pH at the equivalence point?

nof moles of benzoic acid = molarity x volume in liters

moles of benzoic acid = 0.25M x 0.025 L = 0.00625 moles

C6H5COOH + NaOH ------> C6H5COONa + H2O

that means one mole of benzoic acid required one mole of sodium hydroxide

no of moles of NaoH required = 0.00625 moles

volume of naOH required = no of moles / molarity = 0.00625 / 0.2 = 0.03125 L NaOH required

total volume = 0.03125 + 0.025 = 0.05625 L

no o fmoles of salts formed = no ofmoles of acid = 0.00625

now concentration of salt = 0.00625 / 0.05625 = 0.111M

there is a direct formula to calculate the pOH of the week base

pOH = 1/2 (pKb -logC)

where C is the concentration of salt = 0.111M

pKb = 9.8

pOH = 1/2 (9.8 - log(0.111))

pOH = 1/2(9.8 + 0.95)

pOH = 1/2 (10.75)

pOH = 5.4

pH = 14-pOH

pH = 14- 5.4

pH = 8.6