Calculate the pH of a solution at the endpoint of a titration involving 25.00 mL of CH3COOH at unknown concentration, and 21.40 mL of 0.100 M NaOH
CH3COOH + NaOH ==> CH3COONa + H2O
So the pH at the equivalence point will be determined by the hydrolysis of the salt.
CH3COO^- + HOH ==> CH3COOH + OH^-
Kb = (KwKa) = (OH^-)(CH3COOH)/(CH3COO^-)
Set up an ICE chart, substitute into the Kb expression, and solve for x, then convert to pOH, then to pH. The problem doesn't tell you the concn of the CH3COONa at the end but it gives you a way to calculate it.
moles NaOH = M x L
moles CH3COOH will be the same at the equivalence point since the titration is 1:1. Then (CH3COONa) = moles NaOH/L soln or (21.4*0.1/(21.4+25) = ??
To calculate the pH at the endpoint of the titration, you need to determine the concentration of CH3COOH at the endpoint. The endpoint of a titration occurs when the moles of the acid and base are stoichiometrically equal, meaning they have reacted in a 1:1 ratio.
First, let's determine the number of moles of NaOH used in the titration:
moles of NaOH = volume of NaOH (L) × concentration of NaOH (M)
= 0.02140 L × 0.100 M
= 0.00214 moles
Since the acid and base react in a 1:1 ratio, the number of moles of CH3COOH at the endpoint will also be 0.00214 moles.
Next, you need to find the original concentration of CH3COOH using the volume and concentration information given. In this case, we have 25.00 mL of CH3COOH at an unknown concentration.
If we assume that the final volume of the solution at the endpoint is the sum of the volumes added (25.00 mL + 21.40 mL = 46.40 mL = 0.04640 L), we can use the equation:
moles of CH3COOH = volume of CH3COOH (L) × concentration of CH3COOH (M)
Rearranging the formula, we can solve for the concentration of CH3COOH:
concentration of CH3COOH = moles of CH3COOH / volume of CH3COOH (L)
= 0.00214 moles / 0.02500 L
= 0.0856 M
Now that we know the concentration of CH3COOH, we can calculate the pH at the endpoint using the dissociation of CH3COOH.
CH3COOH is a weak acid that partially dissociates in water, so we can use the equilibrium expression for the dissociation:
CH3COOH + H2O ⇌ CH3COO^- + H3O^+
The equilibrium constant expression for this reaction is:
Ka = [CH3COO^-] × [H3O^+] / [CH3COOH]
Since the concentration of H3O^+ is equal to the concentration of CH3COOH at equilibrium in this case, we need to calculate the concentration of H3O^+.
Using the expression for the acid dissociation constant (Ka = [CH3COO^-] × [H3O^+] / [CH3COOH]), we can rearrange to solve for [H3O^+]:
[H3O^+] = (Ka × [CH3COOH]) / [CH3COO^-]
= (1.8 × 10^-5) × 0.0856 M / 0.0856 M
= 1.8 × 10^-5 M
Finally, we can calculate the pH using the equation:
pH = -log[H3O^+]
= -log(1.8 × 10^-5)
= 4.74
Therefore, the pH of the solution at the endpoint of the titration is 4.74.