(a) To calculate the hydrogen ion concentration of the buffer solution, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, the acid is acetic acid (HA) and its conjugate base is acetate (A-).
pKa of acetic acid = 4.74
[HA] = 2.0 M x 0.1 L = 0.20 mol
[A-] = 1.0 M x 0.1 L = 0.10 mol
pH = 4.74 + log(0.10/0.20) = 4.74 - 0.301 = 4.44
Therefore, the hydrogen ion concentration of the buffer solution is 10^-4.44 = 3.98 x 10^-5 M.
(b) When hydrochloric acid is added to the buffer solution, it reacts with the acetate ion to form acetic acid and chloride ion:
HCl + A- → HA + Cl-
The resulting solution will still be a buffer solution, but with a changed ratio of acid to base due to the addition of more acid.
[HA] = 0.20 mol (unchanged)
[A-] = 0.10 mol - 0.040 L x 1.0 M = 0.06 mol
pH = 4.74 + log(0.06/0.20) = 4.74 - 0.522 = 4.22
Therefore, the hydrogen ion concentration of the resulting solution is 10^-4.22 = 6.31 x 10^-5 M.
(a) A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid solution to 0.1 liter of a 1.0 molar sodium hydroxide solution. Compute the hydrogen ion concentration of the buffer solution.
(b) Suppose that 0.10 liter of 0.50 molar hydrochloric acid is added to 0.040 liter of the buffer
prepared in (a). Compute the hydrogen ion concentration of the resulting solution.
3 answers
What is the pH of a 2.0 molar solution of acetic acid. K, acetic acid = 1.8x10 (b) A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid
(a) To calculate the pH of a 2.0 molar solution of acetic acid, we need to use the dissociation constant (Ka) of acetic acid and the equilibrium expression:
Ka = [H3O+][CH3COO-]/[CH3COOH]
Assuming that at equilibrium, x moles of acetic acid has dissociated to form x moles of H3O+ and x moles of acetate (CH3COO-), the equilibrium concentrations are:
[H3O+] = [CH3COO-] = x mol/L
[CH3COOH] = (2.0 - x) mol/L
Substituting these values into the equilibrium expression:
1.8 x 10^-5 = x^2 / (2.0 - x)
Solving for x using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
where a = 1, b = 2, and c = -3.6 x 10^-5
x = 0.00177 or 0.0382
Since the concentration of H3O+ is negligible compared to the initial concentration of acetic acid, we can assume that x = 0.00177 mol/L.
pH = -log[H3O+] = -log(0.00177) = 2.75
Therefore, the pH of the 2.0 molar acetic acid solution is 2.75.
(b) The buffer solution prepared in part (a) has a pH of 4.44. The buffer capacity is greatest when [HA] = [A-], so the buffer should be able to resist changes in pH when small amounts of acid or base are added.
Assuming that the volume of the buffer solution remains constant at 0.1 L, the initial concentrations of acetic acid and acetate are:
[HA] = 2.0 M x 0.1 L = 0.20 mol/L
[A-] = 1.0 M x 0.1 L = 0.10 mol/L
When 0.10 L of 0.50 M HCl is added, the amount of acid and base present in the buffer solution changes. The moles of HCl added is:
0.50 mol/L x 0.10 L = 0.05 mol
The reaction between HCl and acetate goes to completion, so all the acetate ions react with HCl to form acetic acid:
HCl + CH3COO- → CH3COOH + Cl-
The number of moles of acetate ions in the buffer solution is reduced by:
0.10 L x 0.05 mol/L = 0.005 mol
The new concentration of acetate ions is:
[A-] = (0.10 mol - 0.005 mol) / 0.1 L = 0.095 mol/L
Since the initial moles of acetic acid (0.20 mol) are much larger than the amount of HCl added (0.05 mol), we can assume that the change in the concentration of acetic acid is negligible. Therefore, the new concentrations of acetic acid and H3O+ are:
[HA] ≈ 0.20 mol/L (unchanged)
[H3O+] = (0.05 mol / 0.1 L) + (1.8 x 10^-5) ≈ 5.0 x 10^-5 mol/L
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 4.74 + log(0.095/0.20) ≈ 4.46
Therefore, the pH of the resulting solution is approximately 4.46.
Ka = [H3O+][CH3COO-]/[CH3COOH]
Assuming that at equilibrium, x moles of acetic acid has dissociated to form x moles of H3O+ and x moles of acetate (CH3COO-), the equilibrium concentrations are:
[H3O+] = [CH3COO-] = x mol/L
[CH3COOH] = (2.0 - x) mol/L
Substituting these values into the equilibrium expression:
1.8 x 10^-5 = x^2 / (2.0 - x)
Solving for x using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
where a = 1, b = 2, and c = -3.6 x 10^-5
x = 0.00177 or 0.0382
Since the concentration of H3O+ is negligible compared to the initial concentration of acetic acid, we can assume that x = 0.00177 mol/L.
pH = -log[H3O+] = -log(0.00177) = 2.75
Therefore, the pH of the 2.0 molar acetic acid solution is 2.75.
(b) The buffer solution prepared in part (a) has a pH of 4.44. The buffer capacity is greatest when [HA] = [A-], so the buffer should be able to resist changes in pH when small amounts of acid or base are added.
Assuming that the volume of the buffer solution remains constant at 0.1 L, the initial concentrations of acetic acid and acetate are:
[HA] = 2.0 M x 0.1 L = 0.20 mol/L
[A-] = 1.0 M x 0.1 L = 0.10 mol/L
When 0.10 L of 0.50 M HCl is added, the amount of acid and base present in the buffer solution changes. The moles of HCl added is:
0.50 mol/L x 0.10 L = 0.05 mol
The reaction between HCl and acetate goes to completion, so all the acetate ions react with HCl to form acetic acid:
HCl + CH3COO- → CH3COOH + Cl-
The number of moles of acetate ions in the buffer solution is reduced by:
0.10 L x 0.05 mol/L = 0.005 mol
The new concentration of acetate ions is:
[A-] = (0.10 mol - 0.005 mol) / 0.1 L = 0.095 mol/L
Since the initial moles of acetic acid (0.20 mol) are much larger than the amount of HCl added (0.05 mol), we can assume that the change in the concentration of acetic acid is negligible. Therefore, the new concentrations of acetic acid and H3O+ are:
[HA] ≈ 0.20 mol/L (unchanged)
[H3O+] = (0.05 mol / 0.1 L) + (1.8 x 10^-5) ≈ 5.0 x 10^-5 mol/L
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 4.74 + log(0.095/0.20) ≈ 4.46
Therefore, the pH of the resulting solution is approximately 4.46.