What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.84 and has a freezing point of -2.0 ∘C? (Assume complete dissociation of sodium benzoate and a density of 1.01 g/mL for the solution.)
To be honest I've not seen a problem but here is what I would do. Check my thinking.
You need to know the reatio base/acid and you get that from the Henderson-Hasselbalch equation.
pH = pKa benzoic acid + log (base)/(acid). The base is sodium benzoate and the acid is benzoic acid.. I think the pKa for benzoic aic dis 4.20 but you should confirm that.
4.84 = 4.20 + log b/a
and solve for the ratio b/a. I get b = about 5a
From the freezing point data you get
2 = i*Kf*m
I would use 3 for i (2 for C6H5COONa) and 1 for C6H5COOH). 1.86 is Kf for water Solve for molality. But you still don't know grams each. I get approx 0.36 for molality.
Then with mm = molar mass i have
(b/mm base) + (a/mm acid) = 0.36 mols
(5a/mm b) + (a/mm a) = 0.36 where a = mols a and b = mols b.
That should give you mols acid and mols b (convert to grams).
If anyone else has an idea don't be shy.
Your train of thought is right because thats how i was figuring it. using the buffer solution equation and then following it with the freezing point dep. There may be a shortcut but I cant remember what/how because you have a weak acid with the salt of its conjugate base.
To determine the molar concentrations of benzoic acid and sodium benzoate in a buffered solution, we need to consider the Henderson-Hasselbalch equation for a buffer solution:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH of the buffer solution, pKa is the acid dissociation constant of benzoic acid, [A-] is the concentration of the conjugate base (sodium benzoate), and [HA] is the concentration of the acid (benzoic acid).
1. Calculate the pKa of benzoic acid:
Given that the pH of the buffer solution is 4.84, we can use the equation to solve for pKa:
4.84 = pKa + log([A-]/[HA])
Since we want to know the molar concentrations of benzoic acid and sodium benzoate, we can assume equal volumes of both components in the buffer solution ([A-] = [HA]). This simplifies the equation to:
4.84 = pKa + log(1)
Therefore, pKa = 4.84.
2. Calculate the concentration of sodium benzoate ([A-]):
Since sodium benzoate fully dissociates in water, its concentration will be expressed as molarity (M). We can use the freezing point depression formula to relate the concentration of sodium benzoate to the freezing point depression:
ΔT = i · Kf · molality
Where ΔT is the freezing point depression, i is the vant Hoff factor (which is the number of particles formed from one molecule of the compound), Kf is the cryoscopic constant for water, and molality is the molal concentration (moles of solute per kilogram of solvent).
Given that the freezing point depression (ΔT) is -2.0 ∘C and the density of the solution is 1.01 g/mL, we can convert this to molality:
molality = -ΔT / (i · Kf)
The vant Hoff factor (i) for sodium benzoate is 2 (since it dissociates into sodium ions and benzoate ions), and the cryoscopic constant for water (Kf) is 1.86 ∘C·kg/mol.
Substituting the values into the equation:
molality = -(-2.0 ∘C) / (2 · 1.86 ∘C·kg/mol) = 0.537 mol/kg
Since the density of the solution is 1.01 g/mL, this is equivalent to 0.537 mol/L.
Therefore, the concentration of sodium benzoate ([A-]) in the buffer solution is 0.537 M.
3. Calculate the concentration of benzoic acid ([HA]):
Using the Henderson-Hasselbalch equation:
4.84 = 4.84 + log(0.537/[HA])
Simplifying the equation:
log(0.537/[HA]) = 0
Taking the anti-log of both sides:
0.537/[HA] = 1
Therefore, [HA] = 0.537 M.
In summary, the molar concentration of sodium benzoate ([A-]) in the buffer solution should be 0.537 M, and the molar concentration of benzoic acid ([HA]) should also be 0.537 M.
To determine the molar concentrations of benzoic acid and sodium benzoate in the solution, we can use the Henderson-Hasselbalch equation and the freezing point depression equation.
1. Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the ratio of the concentrations of its acid and its conjugate base. The equation is as follows:
pH = pKa + log([A-]/[HA])
where:
- pH is the desired pH of the solution (4.84 in this case)
- pKa is the acid dissociation constant of benzoic acid (4.20)
- [A-] is the concentration of the conjugate base (sodium benzoate)
- [HA] is the concentration of the acid (benzoic acid)
2. Freezing point depression equation:
The freezing point depression equation relates the freezing point depression to the molality of the solute in a solution. The equation is as follows:
ΔT = Kf * m
where:
- ΔT is the freezing point depression (-2.0 ∘C in this case)
- Kf is the freezing point depression constant for the solvent (-1.86 ∘C/m for water)
- m is the molality of the solute
Since the density of the solution is given as 1.01 g/mL, we can convert it to kg/L by dividing by 1000.
Now, let's solve for the molar concentrations of benzoic acid and sodium benzoate:
1. Solving the Henderson-Hasselbalch equation:
4.84 = 4.20 + log([A-]/[HA])
Rearranging the equation, we have:
log([A-]/[HA]) = 4.84 - 4.20
log([A-]/[HA]) = 0.64
Taking the antilog of both sides, we get:
[A-]/[HA] = 10^0.64
[A-]/[HA] = 4.28
2. Solving the freezing point depression equation:
ΔT = Kf * m
Rearranging the equation, we have:
m = ΔT / Kf
m = (-2.0 ∘C) / (-1.86 ∘C/m)
m = 1.08 molal
3. Calculating the molar concentrations:
Given that the density of the solution is 1.01 g/mL, we can calculate the molar concentration using the following formula:
Molar concentration (M) = Molality (m) * Density (kg/L) / Molar mass (g/mol)
For benzoic acid:
Molar concentration of benzoic acid = 1.08 molal * (1.01 g/mL / 1000 g/L) / 122.12 g/mol
Molar concentration of benzoic acid ≈ 8.9 × 10^-3 M
For sodium benzoate (assuming complete dissociation):
Molar concentration of sodium benzoate = 4.28 * 8.9 × 10^-3 M
Molar concentration of sodium benzoate ≈ 3.8 × 10^-2 M
To summarize, in a solution buffered at a pH of 4.84 and with a freezing point of -2.0 ∘C, the molar concentration of benzoic acid should be approximately 8.9 × 10^-3 M, and the molar concentration of sodium benzoate should be approximately 3.8 × 10^-2 M.