How many grams of NaF should be added to 612 ml of 0.4 M HF (Ka = 6.8 x 10-4) to produce a buffer solution where pH = 3.10?
8.7 g NaF
To answer this question, we need to consider the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of the acid and its conjugate base.
Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA])
In this case, HF is the acid and NaF is its conjugate base (A-). We are given the pH (= 3.10) and the pKa (= -log(Ka) = -log(6.8 x 10^(-4))), and we need to find the concentration ratio of [A-]/[HA] in order to calculate how many grams of NaF are needed.
Let's break this down step by step:
1. Calculate the concentration of the acid, [HA]:
Molarity (M) = moles (mol) / volume (L)
Since we are given the Molarity (0.4 M) and the volume (612 mL), we can convert the volume to liters:
Volume (L) = 612 mL / 1000 mL/L = 0.612 L
Therefore, [HA] = 0.4 M x 0.612 L = 0.2448 mol
2. Use the Henderson-Hasselbalch equation to set up an equation for the concentration of the conjugate base, [A-]:
pH = pKa + log([A-]/[HA])
Rearrange the equation to solve for the concentration ratio, [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
3. Plug in the values for pH and pKa:
[A-]/[HA] = 10^(3.10 - (-log(6.8 x 10^(-4))))
4. Calculate the concentration ratio, [A-]/[HA]:
[A-]/[HA] = 10^(3.10 - (-log(6.8 x 10^(-4))))
[A-]/[HA] = 10^(3.10 + 3.17)
[A-]/[HA] = 10^(6.27)
[A-]/[HA] = 1.90 x 10^6
5. Since the concentration ratio is the same as the mole ratio, we know that for every 1.90 x 10^6 moles of [A-], there is 1 mole of [HA].
6. Convert the number of moles of [HA] to grams of NaF:
Molar mass of NaF = 22.99 g/mol + 19.00 g/mol = 41.99 g/mol
Grams of NaF = 0.2448 mol x 41.99 g/mol = 10.26 g
Therefore, approximately 10.26 grams of NaF should be added to 612 mL of 0.4 M HF to produce a buffer solution with pH = 3.10.