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) = ??
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
1 answer
1 answer