Question

What weight of sodium lactate should be added to make 750 mL of a pH 4.00 buffer solution that is 0.80 M in lactic acid? The Ka for lactic acid is 1.38×10-4.

Answer 1

Answer – We are given, volume = 750 mL, pH = 4.00 , [lactic acid] = 0.80 M

Ka = 1.38*10^-4

We know

pKa = -log Ka

       = -log 1.38*10^-4

        = 3.86

We know Henderson-Hasselbalch equation

pH = pKa + log [lactate] / [lactic acid]

4.00 = 3.86 + log [lactate] / [lactic acid]

So, log [lactate] / [lactic acid] = 4.00-3.86

                                           = 0.14

Taking antilog from both side

[lactate] / [lactic acid] = 1.38

[lactate] / 0.80 = 1.38

So, [lactate] = 1.38 *0.80

                    = 1.10 M

So, moles of lactate = 1.10 M * 0.750 L

                               = 0.828 moles

So, moles of lactate = moles of sodium lactate = 0.828 moles

Mass of sodium lactate = 0.828 moles *112.06 g/mol

                                   = 92.81 g

92.81 g of weight of sodium lactate should be added to make 750 mL of a pH 4.00 buffer solution that is 0.80 M in lactic acid