Question

0.5N titrant is used to reach the equivalence point with 800mL of an acetic acid solution with a pH of 3. What volume of titrant is needed to reach equivalence with the acetic acid solution?

Answer 1

• For the beginning: $pH = -lg[H^+] \to$ [H⁺] = 10^-3
• Because acetic acid is a weak acid, therefore this formula should be applied: $[H^+] = \sqrt (Ka•C(acid))$ , then, the concentration will be calculated: $C(acid) = [H^+]^2 / Ka$ = (10^-3)² / 1.8•10^-5 = 0.06 M
• Since the conditions don't indicate what kind of titrant was applied, normality will be equal to molarity. At the equivalence point, equality is observed for reacting substances: $C(acid) • V(acid) = C(titrant) • V(titrant)$. Then: $V(titrant) = (C(acid)•V(acid)) / C(titrant)$ = (0.06•0.8) / 0.5 = 0.096 L = 96 ml
• Answer: V(titrant)=96 ml.