A 0.15 M solution of acetic acid (CH3COOH) is titrated with 0.10 M sodium hydroxide (NaOH). Calculate the pH at the halfway point of the titration (the half-equivalence point).
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At the halfway point of the titration, the number of moles of acetic acid (CH3COOH) will be equal to half of its initial amount.
Let's assume the initial volume of acetic acid solution is V liters.
The number of moles of acetic acid (CH3COOH) = 0.15 * V moles
At the halfway point, half of the acetic acid has reacted, so only 0.075 * V moles of acetic acid are left.
Since acetic acid is a weak acid, it partially dissociates in water:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
The dissociation constant (Ka) for acetic acid is 1.8 x 10^-5.
At equilibrium, the concentration of CH3COO- is equal to the concentration of H+ which can be calculated using the equation: Ka = [CH3COO-][H+]/[CH3COOH]
Since we already know the concentration of CH3COOH at the halfway point (0.075 * V), we can rearrange the equation to solve for [H+].
[H+] = Ka * [CH3COOH] / [CH3COO-]
Plugging in the values:
[H+] = (1.8 x 10^-5) * (0.075 * V) / (0.15 * V)
[H+] = 9 x 10^-6
To find the pH, we can take the negative logarithm of the concentration of H+:
pH = -log [H+]
pH = -log (9 x 10^-6)
pH = 5.05
Therefore, the pH at the halfway point of the titration is 5.05.
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