Answer:
concentration of H3O+ in the solution is 10^(-3.73) = 2.20 x 10^(-4) M.
Explanation:
To solve this problem, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pH is the desired pH (4.56), pKa is the dissociation constant of benzoic acid (4.20), [A-] is the concentration of sodium benzoate (0.23 M), and [HA] is the concentration of benzoic acid (0.10 M).
First, we need to calculate the ratio of [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa) = 10^(4.56 - 4.20) = 1.78
Next, we need to use the total amount of benzoate and benzoic acid to calculate the total buffer concentration:
[buffer] = [A-] + [HA] = 0.23 M + 0.10 M = 0.33 M
Now we can use the amount of HCl added to calculate the final concentration of H3O+:
0.060 mol HCl / 1.7 L = 0.035 M HCl
Since HCl is a strong acid, it completely dissociates in water, so the concentration of H3O+ in the solution is also 0.035 M.
To calculate the new pH of the buffer solution, we need to use the Henderson-Hasselbalch equation again, but this time with the new concentrations of [A-] and [HA]:
pH = pKa + log([A-]/[HA])
pH = 4.20 + log(1.78 * 0.23 M / 0.10 M)
pH = 3.73
Therefore, the new concentration of H3O+ in the solution is 10^(-3.73) = 2.20 x 10^(-4) M.