A buffer with a pH of 4.30 contains 0.25 M of sodium benzoate and 0.20 M of benzoic acid. What is the concentration of [H ] in the solution after the addition of 0.052 mol of HCl to a final volume of 1.5 L?
I know that Ka = 6.3x10-5
do I need to use an ice table or is the pH=pKa+log(A/HA) enough?
You need to use an ICE chart. Post what you can and explain what you don't understand and I can help you through it if you get stuck.
To determine the concentration of [H+] in the solution after the addition of HCl, you can use the Henderson-Hasselbalch equation, which is pH = pKa + log(A-/HA). In this equation, pKa is the negative logarithm of the acid dissociation constant (Ka), and A- and HA represent the concentrations of the conjugate base and the acid, respectively.
In this case, you have the pKa value (6.3x10-5) and the concentrations of sodium benzoate (A-, 0.25 M) and benzoic acid (HA, 0.20 M) in the buffer solution.
First, you need to calculate the moles of HCl added to the solution:
moles of HCl = 0.052 mol
Next, you need to calculate the concentration (in moles per liter) of HCl in the final solution:
concentration of HCl = moles of HCl / volume of solution = 0.052 mol / 1.5 L = 0.03467 M
Since HCl is a strong acid, it will completely dissociate in water, resulting in the same concentration of [H+].
Now, you can use the Henderson-Hasselbalch equation to find the new pH value:
pH = pKa + log(A-/HA)
Substituting the given values:
pH = 6.3x10-5 + log(0.25/0.20)
Calculating the logarithm:
pH = 6.3x10-5 + log(1.25)
pH = 6.3x10-5 + 0.0976
pH ≈ 6.3976
Therefore, the concentration of [H+] in the solution after the addition of 0.052 mol of HCl is approximately 6.3976.