# A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH. How would you calculate the molarity of HCl in this solution?

Dec 3, 2015

Correctly! And according to the equation:

$H C l \left(a q\right) + N a O H \left(a q\right) \rightarrow N a C l \left(a q\right) + {H}_{2} O \left(a q\right)$

#### Explanation:

Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of $H C l$ was dissolved in a $30 \times {10}^{-} 3 L$ volume.

Volume ($L$) $\times$ concentration ($m o l \cdot {L}^{-} 1$) gives an answer in $m o l$. Note that $1$ $m L$ $=$ $1 \times {10}^{-} 3 \cdot L$

So $\frac{23 \times {10}^{-} 3 \cancel{L} \times 0.20 \cdot m o l \cdot \cancel{{L}^{-} 1}}{30 \times {10}^{-} 3 L}$ $=$ ???mol*L^-1