Question

How many grams of dipotassium phthalate (242.3 g/mol) must be added to 200 mL of 0.150 M potassium hydrogen phthalate to give a buffer of pH 5.40? The Ka's for phthalic acid are 1.12 x 10-3 and 3.90 x 10-6.

Answer 1

we have Hederson equation

pH = pka2 + log [ C6H14CO2K2] / [C6H14CO2HK]   , pka = -log ( 3.9 x 10^ -6) = 5.4

5.4 = 5.4 + log [C6H14CO2K2]/[C6H14CO2HK]

[dipotassium pthalate ] = [Potassium hydrogen phtalate]

dipotassium phtahalate moles = potassium hydrogen pthalate moles

Potassium hydrogen pthalate moles = M x V ( in liters) = 0.15 x ( 200/1000) =0.03

hence dipotassium phtahalet moles = 0.03

its mass = Moles x molar mass of dipotassium pthalate

          = 0.03 x 242.3 = 7.27 g

hence 7.29 grams of dipotassium pthalate is required to be added to shown amount of potassium hydrogen pthalate get pH 5.4