The molarity of the acid is 0.086 00 mol/L.
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
Since 1 mol of acid reacts with 1 mol of base, we can use the formula
M₁V₁ = M₂V₂
M₁ = ?; V₁ = 50.00 mL
M₂ = 0.1000 mol/L; V₂ = 43.00 mL
M₁ = M₂ × [tex] \frac{ V_{2}}{ V_{1}} [/tex] = 0.1000 mol/L × [tex] \frac{43.00 mL}{50.00 mL} [/tex] = 0.086 00 mol/L
