I have a 1L solution of 0.02M sodium acetate and 0.02M acetic acid. What is the pH after the addition of 0.05 moles of hydrochloric acid?
I know this is a buffer question but I have no idea how I am supposed to go about getting the answer... I know Ka values for acetic acid, but I still can't get anywhere.
What happens in a buffered solution when HCl is added? The acetate ion combines with H of the strong acid to form acetic acid, thus making a weak acid out of a strong acid. If we let Ac^- stand for acetate and HAc acetic acid, the equation you have is
Ac^- + H^+ ==> HAc
So look at the moles H^+ added from HCl. The moles HAc will be increased by the moles H^+ added (from the HCl) and the moles of Ac^- will be decreased by that same amount. Then plug in these new moles for base and acid in the Henderson-Hasselbalch equation and you have it.
To find the pH after the addition of hydrochloric acid to the buffer solution, you need to consider the equilibrium reaction involved in the buffer system and apply the principles of the Henderson-Hasselbalch equation. Here's the step-by-step process to solve this problem:
Step 1: Write the equilibrium reaction for the buffer system. In this case, the system is composed of acetic acid (CH3COOH) and its conjugate base, sodium acetate (CH3COONa). The equilibrium reaction is as follows:
CH3COOH + H2O ⇌ CH3COO- + H3O+
Step 2: Calculate the initial concentration of the weak acid (acetic acid) and its conjugate base (sodium acetate). We have a 1L solution containing 0.02M of both acetic acid and sodium acetate. Therefore, the initial concentration of both species is 0.02M.
Step 3: Determine the concentration of acetic acid and sodium acetate after the addition of HCl. Since 0.05 moles of HCl is added, the total volume of the solution is now 1L + (0.05 moles HCl/1L) = 1.05L. However, since the HCl is a strong acid, it will completely dissociate, and the concentration of H3O+ ions is directly increased by 0.05M.
HCl → H+ + Cl-
[H3O+] = 0.05M
Step 4: Next, you need to calculate the new concentrations of acetic acid and acetate ions. The change in their concentrations is determined by the ratio of the volumes of the original buffer solution to the final solution.
Initially, the acetic acid concentration is 0.02M in 1L, so the moles of acetic acid in the original solution are 0.02M × 1L = 0.02 moles.
Considering the stoichiometry of the reaction (1:1 ratio), the concentration of acetic acid in the final solution is:
[(0.02 moles acetic acid) / (1.05L)] ≈ 0.01905M
Since sodium acetate is the conjugate base of acetic acid, it reacts with the added H3O+ ions to form acetic acid. Therefore, the concentration of acetate ions is also reduced by the same amount. The new concentration of acetate ions is:
[(0.02 moles acetate ions) / (1.05L)] ≈ 0.01905M
Step 5: Calculate the new concentration of H3O+ ions. It is equal to the initial concentration (0.05M) plus the concentration resulting from the ionization of acetic acid in the buffer solution (from Step 4).
[H3O+] = 0.05M + 0.01905M ≈ 0.06905M
Step 6: Use the Henderson-Hasselbalch equation to calculate the pH of the solution:
pH = pKa + log ([conjugate base] / [weak acid])
The pKa value of acetic acid is given as an indicator of its acidic strength, typically around 4.75.
pH = 4.75 + log (0.01905M / 0.06905M)
pH ≈ 4.75 + (-0.758)
pH ≈ 3.99
Therefore, the pH after the addition of 0.05 moles of hydrochloric acid is approximately 3.99.