Question

You have 1 M solutions of H₃PO₄, NaH₂PO₄, Na₂HPO₄, Na₃PO₄ (no NaOH).

=> Using only two of these solutions, how would you prepare 150 mL of 1 M phosphate buffer at pH 3.0?

A)		131.4 mL of NaH2PO4 + 18.6 mL H3PO4
B)		50 mL Na2HPO4 + 100 mL Na3PO4
C)		75 mL Na2HPO4 + 75 mL Na3PO4
D)		18.6 mL of NaH2PO4 + 131.4 mL H3PO4

Answer 1

pH of buffer = 3.0

Volume of buffer = 150 mL = 0.150 L

pKa of H3PO4 = 2.15

Using Henderson-Hasselbalch equation.

pH = pKa + log [(base)/(acid)]

3.0 = 2.15 + log [NaH2PO4] / [H3PO4]

0.85 = log [NaH2PO4] / [H3PO4]

7.079 = [NaH2PO4] / [H3PO4]

ratio of [NaH2PO4] / [H3PO4]  = 7.079 means that -

[NaH2PO4] = 7.079*x M

[H3PO4] = 1*x M

Fraction of each component in buffer is given by -

NaH2PO4 = 7.079*x / ( 7.079*x + 1*x)

= 0.876

H3PO4 = 1*x / ( 7.079*x + 1*x)

= 0.124

Concentration of each component in buffer is given as -

Concentration = Molarity of buffer * Fraction of component in buffer

[NaH2PO4] = 1 * 0.876 = 0.876 M

[H3PO4] = 1 * 0.124 = 0.124 M

Moles of NaH2PO4 in buffer = 0.876 * 0.150

= 0.1314

Moles of H3PO4 in buffer = 0.124 * 0.150

= 0.0186

Now, we have molarity of the soluion = 1 M

Volume of NaH2PO4 = Moles * Molarity

= 0.1314 * 1

= 0.1314 L = 131.4 mL

Volume of H3PO4 = 0.0186 * 1

= 0.0186 L = 18.6 mL

So, 150 mL of 1 M phosphate buffer at pH 3.0 can be prepared by adding -

131.4 mL of NaH2PO4 + 18.6 mL H3PO4