Question

Calculate the pH of a buffer solution that contains 0.44 M NaH2PO4 and 0.29M Na2HPO4

Calculate the change in pH if 0.100 g of solid NaOH is added to 250 mL of the solution in the problem above.

Answer 1

For the Buffer NaH₂PO₄ and Na₂HPO₄ buffer an weak acidic component H₂PO₄^- (aq) and basic conjugate componant is HPO₄^2- (aq).

pH of buffer is given by Henderson Hasselbalch equation as,

pH = pKa + log([Base]/[Acid])

For given buffer pKa of acidic componant = 7.20,

pH = 7.20 + log([Na₂HPO₄]/[NaH₂PO₄])

pH = 7.20 + log(0.29/0.44)

pH = 7.20 + (-0.18)

pH = 7.02

=================================

In 250 mL let us calculate milimoles of each component in buffer,

Milimoles of NaH₂PO₄ = Molarity x volume = 0.44 x 250 = 110

Milimoles of NaH₂PO₄ = Molarity x volume = 0.29 x 250 = 72.5

Let us calculate milimoles of naOH added to 250 mL of given buffer,

milimoles of NaOH = Mass of NaOH in mg / Molar mass of NaOH in gram per mole = 100 mg / 40 = 2.5

On addition of a strong base NaOH milimoles of Acidic componant decreases and of Basic componant increases.

New Milimoles of NaH₂PO₄ = 110 - 2.5 = 107.5

New Milimoles of Na₂HPO₄ = 72.5 + 2.5 = 75.0

With this,

pH = 7.20 + log (75.0/107.5)

pH = 7.20 + (-0.16)

pH = 7.04

===================XXXXXXX======================

pH = 7.20 +