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)
[HA] = .033M
[A-] = .102M
To find the concentrations of acetate and acetic acid in the solution, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH is the measured pH
pKa is the dissociation constant for acetic acid
[A-] is the concentration of acetate ion
[HA] is the concentration of acetic acid
First, we need to find the pKa value for acetic acid. The pKa value for acetic acid is approximately 4.76.
Using the Henderson-Hasselbalch equation, we can rearrange the equation as:
log([A-]/[HA]) = pH - pKa
Now let's substitute the given values into the equation:
log([A-]/[HA]) = 5.25 - 4.76
log([A-]/[HA]) = 0.49
To solve for [A-]/[HA], we need to take the antilog of both sides of the equation:
[A-]/[HA] = 10^0.49
Using a scientific calculator or logarithmic tables, we find that 10^0.49 is approximately 3.16.
Now, since we added 11.1 g of sodium acetate to a 1.0 L volumetric flask, we know that the number of moles of sodium acetate is:
moles of sodium acetate = mass of sodium acetate / molar mass of sodium acetate
The molar mass of sodium acetate is:
sodium (Na): 22.99 g/mol
carbon (C): 12.01 g/mol
hydrogen (H): 1.01 g/mol
oxygen (O): 16.00 g/mol
So, molar mass of sodium acetate (CH3COONa) = 22.99 + 12.01 + (3 * 1.01) + 16.00 = 82.03 g/mol
Now, let's calculate the moles of sodium acetate:
moles of sodium acetate = 11.1 g / 82.03 g/mol
Next, we need to consider the stoichiometry of the reaction. Since sodium acetate is the salt of acetic acid, it completely ionizes in water:
CH3COONa -> CH3COO- + Na+
This means that one mole of sodium acetate produces one mole of acetate ion (A-) and one mole of sodium ion (Na+).
So, the concentration of acetate (A-) and sodium (Na+) ions is the same as the concentration of sodium acetate:
[A-] = [Na+] = moles of sodium acetate / volume of solution (in L)
Since we made a 1.0 L solution, the volume of the solution is 1.0 L.
Now, let's calculate the concentration of acetate:
[A-] = [Na+] = moles of sodium acetate / volume of solution
= (11.1 g / 82.03 g/mol) / 1.0 L
= 0.135 M
Since the concentration of acetate is equal to the concentration of acetic acid ([A-]/[HA] = 3.16), the concentration of acetic acid is:
[HA] = [A-] / (concentration ratio)
= 0.135 M / 3.16
= 0.043 M
Therefore, the concentrations of acetate and acetic acid in the solution are approximately 0.135 M and 0.043 M, respectively.