Question

As a technician in a large pharmaceutical research firm, you need to produce 200. mL of 1.00 mol L−1 potassium phosphate buffer solution of pH = 7.01. The pKa of H2PO4− is 7.21.

You have the following supplies: 2.00 L of 1.00 mol L−1 KH2PO4 stock solution, 1.50 L of 1.00 mol L−1 K2HPO4 stock solution, and a carboy of pure distilled H2O.

How much 1.00 mol L−1 KH2PO4 will you need to make this solution?

Express your answer to three significant digits with the appropriate units.

Answer 1

Ans:

Stock solution-Supply

1.0 M KH2PO4 (2L)

1.0 M K2HPO4 (1.5L)

Required pH of the buffer = 7.01

pKa of (H₂PO₄)^- = 7.21

(H₂PO₄)^- (aq)+ H₂O (l) ⇋ (HPO₄)^2- (aq) + H⁺ (aq)

pH of this solution if given by Henderson–Hasselbalch equation;

pH = pKa + log([Conjugate base]/[Acid]) = pKa + log([(HPO₄)^2-]/[(H₂PO₄)^-])

7.01 = 7.21 + log ([(HPO₄)^2-]/[(H₂PO₄)^-])

[(HPO₄)^2-]/[(H₂PO₄)^-] = 0.631= [(K₂HPO₄)]/[(KH₂PO₄)]

0.631= (No. of moles of K₂HPO₄/0.2)/ (No. of moles of KH₂PO₄ /0.2)

No. of moles of K₂HPO₄ = 0.631 x No. of moles of KH₂PO₄

Volume of the buffer required = 200 mL = 0.2 L

Volume of 1.0 M KH₂PO₄ (V1) + Volume of 1.0 M K₂HPO₄ (V2) = 0.2 L ==> V1 + V2 = 0.2L

We know that volume = Mole/Molarity

V1 = no. of moles of KH₂PO₄ / 1.0

V2 = no. of moles of K₂HPO₄ / 1.0 = 0.631 x No. of moles of KH₂PO₄ / 1.0

Volume of 1.0 M KH₂PO₄ (V1) + Volume of 1.0 M K₂HPO₄ (V2) = 0.2 L ==> V1 + V2 = 0.2L

(no. of moles of KH₂PO₄ / 1.0 ) + (0.631 x No. of moles of KH₂PO₄ / 1.0) = 0.2

No. of moles of KH₂PO₄ = 0.2/1.631 = 0.1226 mol

No. of moles of K₂HPO₄ = 0.631 x No. of moles of KH₂PO₄ = 0.631 x 0.1226 = 0.0774 mol

V1 = no. of moles of KH₂PO₄ / 1.0 = 0.1226/1.0 = 0.1226 L

V2 = no. of moles of K₂HPO₄ / 1.0 = 0.631 x No. of moles of KH₂PO₄ / 1.0 = 0.0774/1.0 = 0.0774 L

122.6 mL of KH₂PO₄ (1.0 M) mixed with 77.4 mL of K₂HPO₄ (1.0 M) gives us 200 mL of buffer with pH = 7.01