Question

Preparing Buffer Solutions: Calculating the Number of Grams of Conjugate Base Needed

Use the table of K values given with this problem to choose the best weak acid to start from for making a buffer that holds the pH of the solution at 3.20. Make your selection so that you maximize the capacity of the buffer.
You select a 200.0 ml volumetric flask to which you add 54.00 mL of a 2.90 M solution of the weak acid you just selected. To finish preparing your buffer you must now add the sodium salt of the conjugate base. How many grams of the sodium salt of your conjugate base must you add so that when you finally fill the flask to the mark with deionized water the buffer will have a pH of 3.20.

Weak Acid	Ka
CH3COOH (Acetic Acid)	1.8 X 10-5
C6H5COOH (Benzoic Acid)	6.5 X 10-5
CH3CH2CH2COOH (Butanoic Acid)	1.5 X 10-5
HCOOH (Formic Acid)	1.8 X 10-4
HBrO (Hypobromous Acid)	2.8 X 10-9
HNO2 (Nitrous Acid)	4.6 X 10-4
HClO (Hypochlorous Acid)	2.9 X 10-8
CH3CH2COOH (Propanoic Acid)	1.3 X 10-5
HCN (Hydrocyanic Acid)	4.9 X 10-10

I posted this question before and they got the answer of 8.00g which is NOT CORRECT, so I was looking someone else to help me with this.

Answer 1

sodium salt of conjugate base = 7.88 g

Explanation

Given : pH of buffer = 3.20

pK_a of the acid should be closest to pH of buffer for an effective buffer

acid : HNO₂

K_a = 4.6 x 10^-4

pK_a = -log(K_a)

pK_a = -log(4.6 x 10^-4)

pK_a = 3.34

moles of HNO₂ added = (concentration of HNO₂) * (volume of HNO₂ in Liter)

moles of HNO₂ added = (2.90 M) * (0.05400 L)

moles of HNO₂ added = 0.1566 mol

According to Henderson - Hasselbalch equation,

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

pH = pK_a + log(moles of NaNO₂ / moles of HNO₂)

3.20 = 3.34 + log(moles of NaNO₂ / 0.1566 mol)

log(moles of NaNO₂ / 0.1566 mol) = 3.20 - 3.34

log(moles of NaNO₂ / 0.1566 mol) = -0.14

moles of NaNO₂ / 0.1566 mol = 10^-0.14

moles of NaNO₂ / 0.1566 mol = 0.729

moles of NaNO₂ = (0.1566 mol) * (0.729)

moles of NaNO₂ = 0.114 mol

mass of NaNO₂ = (moles of NaNO₂) * (molar mass NaNO₂)

mass of NaNO₂ = (0.114 mol) * (69.0 g/mol)

mass of NaNO₂ = 7.88 g