What mass of Na2A should you add to NaHA solution to create a buffer with pH = 5.82?

Question

Sodium hydrogen succinate (NaHA) is a weak acid with pK_a of 5.61. You have 10.0 mL of 0.283 M NaHA solution. How much Na2A should you add in grams to the NaHA solution to create a buffer with pH = 5.82?

Answer 1

NaHA immediately dissociates into HA^-, which is given to be our weak acid. We can write an acid/base reaction:

HA^-+ H₂O ⇌ A₂^- + H₃O⁺

Using Henderson-Hasselbalch:

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

Rearrange to isolate [conjugate base]:

pH - pKa = log([conjugate base]/[acid])

e^{pH - pKa} = [conjugate base]/[acid]

[conjugate base] = [acid] e^{pH - pKa}

Finally, let's substitute in numbers to solve for amount of conjugate base needed:

[conjugate base] = [A](1.6218) = (0.283M)(1.6218) = 0.45897M

0.45897M = x mol/(10 mL) = x mol/(10/1000 L)

x mol = 0.45897M(10/1000 L) = 0.0045897 mol

Na₂A has molar mass 162.05 g/mol:

0.0045897 mol * 162.05 g/mol = 0.74 g of Na2A.