# A volume of 32.45 mL of 0.122 M NaOH was required to titration 4.89 g of vinegar to a phenolphthalein end point. How many moles of NaOH were added?

Concentration ($c$) is always moles (or sometimes mass) per unit volume. If $c$ $=$ $\frac{n}{V}$, then $n$ (molar quantity) $=$ volume $\left(V\right)$ $\times$ concentration $\frac{n}{V}$.
$n$ $=$ $V \times c$ $=$ $32.45$ $\times {10}^{- 3} \cancel{L}$ $\times$ $0.122$ $m o l \cdot {\cancel{L}}^{-} 1$
$= 3.96 \times {10}^{- 3} m o l$ $N a O H$