# In a certain titration, 34.5 mL of 0.20 M NaOH is required to neutralize 15.3 mL of HCl. What is the concentration of the acid (in normality and molarity)?

Jun 22, 2016

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

#### Explanation:

The equation demonstrates 1:1 stoichiometry: moles of sodium hydroxide are equivalent to moles of hydrochloric acid.

$\text{Moles of NaOH}$

$=$ $34.5 \times {10}^{-} 3 L \times 0.20 \cdot m o l \cdot {L}^{-} 1 = 0.0069 \cdot m o l$.

Given the stoichiometry, there were an equivalent quantity of moles in the given volume of $H C l$.

And thus $\left[H C l\right] = \frac{0.0069 \cdot m o l}{15.3 \times {10}^{-} 3 L}$ $\cong$ $0.50 \cdot m o l \cdot {L}^{-} 1$.

For a monoprotic acid such as hydrochloric acid,

$\text{molarity "=" normality}$.

Capisce?