Propanoic acid that, along with its sodium salt, can be used to make a buffer that has a pH of 5.25. If you have 532.9 mL of a 0.250 M solution of that acid, how many grams of the corresponding sodium salt do you have to dissolve to obtain the desired pH?
pH = pKa + log (base)/(acid)
5.25 = pKa + log (base)/(M x L)
Solve for base and since the acid is in mols, the base will be in mols.
Then mols = grams/molar mass. You know molar mass and mols, solve for grams. That will be the number of grams of the salt to add to the 532.9 mL of the acid.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the conjugate acid and conjugate base components.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
Given that the desired pH is 5.25, we can determine the pKa value for propanoic acid. The pKa value is the negative logarithm of the acid dissociation constant (Ka). The pKa value for propanoic acid is usually given as 4.87.
Next, we need to determine the concentrations of the acid ([HA]) and its conjugate base ([A-]) in the buffer solution.
Given:
Volume of 0.250 M propanoic acid = 532.9 mL = 0.5329 L
Concentration of propanoic acid ([HA]) = 0.250 M
Using the Henderson-Hasselbalch equation, we have:
5.25 = 4.87 + log ([A-] / [HA])
Rearranging the equation:
0.38 = log ([A-] / [HA])
To solve for the ratio [A-] / [HA], we need to convert the logarithmic form into exponential form:
10^0.38 = [A-] / [HA]
Now, we can calculate the ratio [A-] / [HA]:
[A-] / [HA] = 10^0.38 ≈ 2.414
This means that the concentration of the conjugate base ([A-]) should be approximately 2.414 times higher than the concentration of the acid ([HA]) in the buffer solution.
Since we know the concentration of the acid ([HA]) is 0.250 M, we can calculate the concentration of the conjugate base ([A-]) in the buffer solution:
[A-] = 2.414 * [HA]
[A-] = 2.414 * 0.250 M
[A-] ≈ 0.604 M
Now, we need to determine the mass of the sodium salt (sodium propanoate, C3H5O2Na) required to achieve the desired concentration of the conjugate base.
The molar mass of sodium propanoate is given as follows:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol
- Sodium (Na): 22.99 g/mol
Adding up the atomic masses:
12.01 g/mol + 5 * 1.01 g/mol + 2 * 16.00 g/mol + 22.99 g/mol = 96.06 g/mol
To calculate the mass required to make the desired concentration, we can use the equation:
Mass = Concentration * Volume * Molar mass
Given:
Concentration of sodium propanoate ([A-]) = 0.604 M
Volume of the sodium propanoate solution = 0.5329 L
Mass = 0.604 M * 0.5329 L * 96.06 g/mol
Mass ≈ 31.51 grams
Therefore, you would need to dissolve approximately 31.51 grams of sodium propanoate to obtain the desired pH of 5.25.