# Question #098cc

Jun 20, 2017

${V}_{\text{eq}}$ = 16.7 ml of 0.15M HCl
$N a O H + H C l \implies N a C l + {H}_{2} O$ has a 1 to 1 reaction ratio. This means that the moles of NaOH neutralized equals the moles of HCl added at the equivalence point ( or, end point ).
=> ${\left(M o l a r i t y \times V o l u m e\right)}_{\text{NaOH}}$ = ${\left(M o l a r i t y \times V o l u m e\right)}^{\text{HCl}}$