A solution of acetic acid is prepared in water by adding 11.1 g of sodium acetate to a volumetric flask and bringing the volume to 1.0 L with water. The final pH is measured to be 5.25. What are the concentrations of acetate and acetic acid in solution? (Assume that the temperature of the solution is 25 degrees Celsius)
To find the concentrations of acetate and acetic acid in the solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH is the measured pH
pKa is the acid dissociation constant of acetic acid (4.75 at 25 degrees Celsius)
[A-] is the concentration of acetate
[HA] is the concentration of acetic acid
Rearranging the equation, we have:
[H+] = 10^(-pH)
[H+] = [A-]
[HA] = [Acetic Acid]
Substituting the given values:
[H+] = 10^(-5.25)
[H+] = 5.62 x 10^(-6) M
This concentration of H+ ions is equal to the concentration of Acetate ions since they were formed by the dissociation of sodium acetate (NaC2H3O2):
[Acetate] = 5.62 x 10^(-6) M
Now, we can use the equation:
[Acetic Acid] = [Sodium Acetate initial] - [Acetate]
[Sodium Acetate initial] = 11.1 g
First, we need to convert grams to moles:
Molar mass of Sodium Acetate = 82.03 g/mol
Number of moles = Mass / Molar mass
Number of moles = 11.1 g / 82.03 g/mol
Number of moles = 0.135 mol
Now, we can calculate the concentration of acetic acid:
[Acetic Acid] = 0.135 mol / 1.0 L
[Acetic Acid] = 0.135 M
Therefore, the concentrations of acetate and acetic acid in the solution are:
Acetate: 5.62 x 10^(-6) M
Acetic Acid: 0.135 M
To determine the concentrations of acetate and acetic acid in the solution, we need to use the Henderson-Hasselbalch equation, which relates the pH, pKa, and the concentrations of the conjugate acid-base pair.
The pKa of acetic acid is 4.74. Since the pH of the solution is 5.25, which is greater than the pKa, we can conclude that acetic acid is mostly dissociated, and the concentration of acetate will be higher than the concentration of acetic acid.
First, we need to calculate the initial concentration of acetate. To do this, we need to convert the mass of sodium acetate to moles. The molar mass of sodium acetate is:
CH3COONa: 12.01 g/mol (C) + 1.01 g/mol (H) + 3 * 16.00 g/mol (O) + 22.99 g/mol (Na) = 82.03 g/mol
The moles of sodium acetate can be calculated using the formula:
moles = mass / molar mass
moles = 11.1 g / 82.03 g/mol ≈ 0.1353 mol
Since the volume of the solution is 1.0 L, the concentration of acetate (CH3COO-) can be calculated as follows:
Concentration (acetate) = moles / volume
Concentration (acetate) = 0.1353 mol / 1.0 L = 0.1353 M
Since acetic acid is a weak acid, we can assume that the concentration of the conjugate acid (acetic acid) will be very small compared to the concentration of the conjugate base (acetate). Therefore, we can approximate the concentration of acetic acid to be negligible.
So, the concentration of acetate (CH3COO-) in the solution is approximately 0.1353 M, and the concentration of acetic acid (CH3COOH) is negligible.
NaHCO3 balance g NaHCO3 s ⇒16.4 = 11.1+ ms ⇒ms = 5.3 b g
% crystallization = 5.3 g crystallized × 100% = 32 3%
16.4 g fed