Calculate pH of a solution by mixing 15.0 mL of .50 M NaOH and 30.0 mL of .50 benzoic acid solution (benzoic acid is monoprotic, its dissociation constant is 6.5 x 10^ -5).

I am not sure whether I am supposed to do it by acid-base titration way or to use the formula -log (Ka)+ log [acid/base]

Because both give me totally different answers. So i wasn't sure which way I am supposed to do this.

Let HA = benzoic acid

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moles NaOH(initial) = (0.015L)(0.50 mol/L) = 0.0075 mole
moles HA(initial) = (0.0300L)(0.50 mol/L) = 0.015 mol
OH^- + HA --> H2O + A-
moles of A- formed = 0.0075 (same as initial OH-)
moles HA remaining = 0.0150- 0.0075 =0.0075 mol
[A-] = moles/liters = 0.0075mol/(0.045L) = 0.167 M at equilibrium
[HA] = 0.167 M (same math as above) at equilibrium
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In the end, you have a solution with equal concentrations of weak acid, HA, and weak base, A-. Use the Henderson-Hasselbalch equation to get your pH:
pH = pKa + log(base/acid)

To calculate the pH of a solution, you need to consider the concentration of H+ ions in the solution, which is determined by the equilibrium between the acid and its conjugate base. In this case, we need to find the pH of a solution formed by mixing NaOH and benzoic acid.

Since NaOH is a strong base, it completely dissociates in water to form Na+ and OH- ions. Therefore, we can exclude NaOH as a source of H+ ions for calculating the pH. Instead, the pH will be determined by the concentration of H+ ions that arise from the dissociation of benzoic acid, which is a weak acid.

To calculate the pH, we can use the formula:

pH = -log[H+]

However, since benzoic acid is a monoprotic acid and does not completely dissociate, we need to consider the equilibrium expression:

Ka = [C6H5COO-][H+]/[C6H5COOH]

where Ka is the acid dissociation constant for benzoic acid.

Given that the dissociation constant (Ka) is 6.5 x 10^-5 and the initial concentration of benzoic acid is 0.50 M, we can set up an ICE table to determine the equilibrium concentrations:

[C6H5COOH] ⇌ [C6H5COO-] + [H+]
0.50 M ⇌ 0 M + [H+]

Let x be the concentration of H+ ions formed. Since benzoic acid is a weak acid, we can assume that x is small compared to the initial concentration of benzoic acid.

Using the equilibrium expression, we have:

Ka = [x][x]/[0.50 - x]

Since x is small compared to 0.50, we can approximate 0.50 - x as 0.50:

6.5 x 10^-5 = x^2/0.50

Solving this quadratic equation, we find x ≈ 9.04 x 10^-3 M.

Now, to calculate the pH, we can substitute the value of [H+] into the pH formula:

pH = -log(9.04 x 10^-3)
≈ 2.04

Therefore, the pH of the solution formed by mixing 15.0 mL of 0.50 M NaOH and 30.0 mL of 0.50 M benzoic acid is approximately 2.04.

Note: The acid-base titration method would be an alternative approach to determine the pH. However, in this case, since we are given the dissociation constant (Ka) of benzoic acid, it is more appropriate to use the equilibrium expression to calculate the pH.