Question

How many grams of dipotassium phthalate (242.3 g/mol) must be added to 900 mL of 0.050 M potassium hydrogen phthalate to give a buffer of pH 5.00?
The K_a's for phthalic acid are 1.12 x 10^-3 and 3.90 x 10^-6.

Answer 1

pH = pka + log [conjugate base] / [ acid]

acid is KHP , conjugate base is K2HP ( dipotassium phthalatae)

pH 5 is nearer to pka2 , where pka2 = -log ka2 = -log ( 3.9x10^-6) = 5.4

pH = pka+ log [K2P]/[KHP]

5 = 5.4 + log [K2P]/[KHP]

-0.4 = log [K2P]/[KHP]

[K2P]/[KHP] = 10^ -(0.4) = 0.398

K2HP moles / KHP moles = 0.398 ...........(1)

we have KHP moles = M x V = 0.05 x ( 900/1000) = 0.045

Henc K2P moles = 0.398 x 0.045 = 0.01791        ( by eq1)

K2P mass = moles x molar mass of K2HP

      = 0.01791 x 242.3 g/mol

      = 4.34 g

hence mass of dipotassium pthalate needed = 4.34 g