This chapter is part of Pharmaceutical Chemistry (PHR 1601).
Determine the volumes of 0.3M lactic acid solution and 0.1M NaOH solution that must be mixed to give 500cm3 of a buffer of pH 4.2. Ka for lactic acid is 1.37 * 10-4.
Let’s establish that V is the volume of NaOH.
Therefore the volume of lactic acid would be 500 – V.
Moles of NaOH
moles = 1000mL
? = V
(500 – V)
Moles of acid
0.3 moles = 1000mL
? = (500 – V)mL
CH₃CH(OH)COOH + NaOH –> CH₃CH(OH)COO–Na+ + H2O
The number of acid remaining in solution would be equal to the number of moles of acid added – number of moles of NaOH.
Moles of salt
The moles of salt is equivalent to the number of moles of base (as this is the limiting factor in this reaction).
Total volume of the solution will be 500mL or 0.5dm3. (Imp to find the concentration of the salt and the acid.
Inputting all the numbers in the equation will give:
Removing logs by having both sides to the power of 10.
Volume of base is 336.42mL
Volume of acid is 163.58mL (500mL – V)
More information on acids and bases.