Question

Design a buffer that has a pH of 6.55 using one of the weak acid/conjugate base systems shown below.

Weak Acid Conjugate Base Ka pKa
HC2O4- C2O42- 6.4 × 10-5 4.19
H2PO4- HPO42- 6.2 × 10-8 7.21
HCO3- CO32- 4.8 × 10-11 10.32

How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams sodium salt of weak acid =

grams sodium salt of conjugate base =

Answer 1

Ans. #1. Part 1: Choice of weak acid- conjugate base pair: A buffer has maximum buffering capacity when pH of the buffer is closest to the pKa of weak acid.

Among the given options, the pKa of weak acid H₂PO₄^- (7.21) is closest to the given buffer pH of 6.55. So, Phosphate buffer is the best choice to prepare this buffer.

Part 2: Preparation of Phosphate Buffer: One mole of each salt dissociates in water to yield 1 mol of their respective phosphate ions as follow-

Na₂HPO4 ------> HPO₄^2-(aq) + 2Na⁺(aq)             - monoprotic form, Conju. Base

NaH₂PO4 ------> H₂PO₄^-(aq) + Na⁺(aq)               - diprotic form, Acid

The diprotic phosphate H₂PO­^2- acts as weak acid (pKa = 7.21). It can donate a proton to become HPO4^2- as follow-

            H₂PO4^-(aq) <---------> HPO4^2-(aq) + H⁺

Now, using Henderson- Hasselbalch equation

            pH = pKa + log ([A^-] / [AH])                    - equation 1

                        where, [A^-] = [HPO4^2-] = [Na₂HPO₄]                      ;

[AH] = [H₂PO₄^-] = [NaH₂PO₄]

Now,

            Putting the values in equation 1-

            6.55 = 7.21 + log ([Na₂HPO₄] / [NaH₂PO₄])

            Or, 6.55 – 7.21 = log ([Na₂HPO₄] / [NaH₂PO₄])

            Or, antilog (-0.66) = [Na₂HPO₄] / [NaH₂PO₄]

            Or, [Na₂HPO₄] / [NaH₂PO₄] = 0.219

            Or, [Na₂HPO₄] = 0.219 [NaH₂PO₄]                    - equation 2

#. Given, concentration of weak base, [Na₂HPO₄] = 1.0 M

Putting the value of [Na₂HPO₄] in equation 2-

1. M = 0.219 [NaH₂PO₄]

Or, [NaH₂PO₄] = 1.0 M / 0.219

Hence, [NaH₂PO₄] = 4.566 M

# Given, Total volume of buffer solution (culture medium) = 1.00 L

Now,

Mass of [Na₂HPO₄] required = (Molarity of Na₂HPO₄ x vol. of solution in L) x Molar mass

                                                            = (1.0 M x 1.00 L) x (141.958838 g/mol)

                                                            = 141.959 g

Mass of [NaH₂PO₄] required = (Molarity of KH₂PO₄ x vol. of solution in L) x Molar mass

                                                            = (04.566 M x 1.0 L) x (119.97701 g/mol)

                                                            = 547.815 g