# Equivalence Point

This chapter is part of Pharmaceutical Chemistry (PHR 1601).

What is the pH at equivalence point when 75 mL of a 0.3M solution of lactic acid is titrated against 0.1M NaOH? Ka for lactic acid is 1.37 * 10-4

At equivalence point the moles of acid = moles of base.

Moles of acid

0.3 moles = 1000mL

? = 75mL

0.0225 moles

Volume of NaOH

moles = 1000mL

0.0225 moles = ?

225mL

Total volume

Total volume in solution is the amount of acid used and the amount of base, which would be 300mL

At equivalence

At equivalence, there is no acid or base, just the salt, as the number of moles of acids is equal to the number of moles of base.

This would create a situation where the equilibrium is not from the acid to the salt, but rather from the salt to the acid.

CH₃CH(OH)COO + H2O == CH₃CH(OH)COOH + OH

The number of moles of the lactate (salt) will be equal to the initial number of lactic acid, in this case 0.0225 moles.

The salt is a base, and therefore kb is needed to find the concentration of OH.

Kw = Ka * kb

Kb = 1*10­-14 / 1.37 * 10-4

= 7.30*10-11

Using the Kequation the concentration of the OH–  can be found, which in turn can be used to find the pOH and the, therefore, the pH at equivalence.

x = 2.34*10-6

pOH = – log[OH]

pOH = 5.63

pH = pkw – pOH

= 14 – 5.63

= 8.11