Question

How many mL of 0.200 M NaOH must be added to 100.0 mL of a 0.0500 F solution of fumaric acid (transbutenedioic acid) to make a buffer of pH 4.50? Hint: What species do you need in solution to achieve this pH?

Answer 1

Suppose moles of NaOH that should be added = x moles

this x mole NaOH will react with fumeric acid to form fumerate salt. The salt and the weak acid together will produce the buffer system.

Concentration of the Fumeric acid = 1F = 0.05 Formula Mass in 1 L

                                                          = (0.05 * 116.07)gm/ L

                                                          = 5.804 gm/L

                                                          = 0.05 moles/ L

Moles of Fumeric acid present in 100mL = 0.05 * 0.1 L = 0.005 moles

Ka1 of Fumeric acid = 8.85x10^-4   pKa = 3.05

Ka2 of Fumeric acid = 3.21x10^-5

According to Hinderson hasselbalch equation

pH = pKa +log [salt]/[acid]

[salt] = amount of NaOH added = x mol

[acid] = (0.05-x)

Substitute the values in the above equiation:

4.5 = 3.05- log [x/0.05-x]

or, 0.035 = x/0.05-x

or, x = (0.035 *0.05)- 0.035x

or, x = 0.0017 moles

Moles of NaOH added = 0.0017 moles

Volume of NaOH added = moles/Molarity = 0.0017 /0.2 = 0.0085L = 8.5mL

So, 8.5mL of NaOH has to be added to make the pH of the solution as 4.5