# A titration neutralizes 25mL of an unknown acid using 32 mL of 6.0 M KOH. What is the molarity of the unknown acid in the original solution?

$\left[\left(M o l a r i t y A c i d\right) \left(V o l u m e A c i d\right)\right] = \left[\left(M o l a r i t y B a s e\right) \left(V o l u m e B a s e\right)\right]$
$\left(M o l a r i t y A c i d\right) \left(25 m l\right) = \left(6.0 M\right) \left(32 m l\right)$
$M o l a r i t y A c i d = \frac{\left(6.0 M\right) \left(32 m l\right)}{\left(25 m l\right)} = 7.68 M$