Consider a solution made by mixing 100 mL of 2.0 M acetic acid (Ka= 1.8e-5) with 100 mL of 0.00020 M benzoic acid
(Ka= 6.4e-5). How many moles of solid NaOH must be added to the solution so that
half of the benzoic acid is protonated (i.e. such that [HA] = [A-] for benzoic acid)?
I would add mols NaOH to neutralize all of the acetic acid and add enough to neutralize 1/2 benzoic acid.
To answer this question, we need to first understand the reaction that occurs between benzoic acid (HA) and sodium hydroxide (NaOH).
Benzoic acid is a weak acid that partially dissociates in water according to the following equation:
HA + H2O ⇌ H3O+ + A-
Sodium hydroxide is a strong base that completely ionizes in water:
NaOH → Na+ + OH-
When sodium hydroxide reacts with benzoic acid, the following reaction occurs:
HA + NaOH → NaA + H2O
Now, let's find the initial concentrations of benzoic acid and acetic acid in the mixture.
For benzoic acid:
Initial volume = 100 mL
Initial concentration = 0.00020 M
Therefore, the initial amount of benzoic acid in moles = (initial concentration) x (initial volume) = 0.00020 M x 0.100 L = 0.000020 moles
For acetic acid:
Initial volume = 100 mL
Initial concentration = 2.0 M
Therefore, the initial amount of acetic acid in moles = (initial concentration) x (initial volume) = 2.0 M x 0.100 L = 0.200 moles
Since we want to protonate half of the benzoic acid (i.e., [HA] = [A-]), the concentration of benzoic acid and its conjugate base, sodium benzoate, must be equal.
Let's assume x moles of NaOH are required to protonate half of the benzoic acid.
After the reaction, the concentration of benzoic acid (HA) will decrease by x moles, and the concentration of sodium benzoate (A-) will increase by x moles.
Now, we can determine the concentrations of benzoic acid and sodium benzoate after the reaction:
[H3O+] = [HA] - x
[A-] = x
We can use the pH equation for a weak acid with known Ka:
Ka = [H3O+][A-] / [HA]
Substituting the values, we have:
Ka = ([HA] - x)(x) / ([HA])
Now, we can solve this equation to find the value of x:
Ka = x^2 / ([HA] - x)
Substituting the values of Ka, [HA], and solving the equation, we get:
6.4e-5 = x^2 / (0.000020 - x)
Rearranging the equation, we have:
x^2 = 6.4e-5 * (0.000020 - x)
Simplifying further, we get:
x^2 = 1.28e-9 - 6.4e-5x
Rearranging this equation, we obtain:
x^2 + 6.4e-5x - 1.28e-9 = 0
Now, we can solve this quadratic equation to find the value of x. Use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
where a = 1, b = 6.4e-5, and c = -1.28e-9.
Solving this equation will give us the value of x, which represents the number of moles of NaOH needed to protonate half of the benzoic acid.