# 0.013*dm^3 of 0.100*mol*L^-1 NaOH reacts completely with 0.010*dm^3 HCl; what is the concentration of [HCl](aq)?

Apr 20, 2017

The stoichiometric equation is...........

#### Explanation:

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

Sodium hydroxide and hydrogen chloride react with a 1:1 stoichiometry. And so we simply have to assess the molarity of the acid solution.

$\text{Moles of base}$ $=$ $0.100 \cdot m o l \cdot {\mathrm{dm}}^{-} 3 \times 0.013 \cdot {\mathrm{dm}}^{3} = 1.30 \times {10}^{-} 3 \cdot m o l$.

Now this molar quantity was equimolar with the $0.010 \cdot {\mathrm{dm}}^{3}$ volume of $H C l \left(a q\right)$.

$\text{Concentration of HCl} = \frac{1.30 \times {10}^{-} 3 \cdot m o l}{0.010 \cdot {\mathrm{dm}}^{-} 3} = 0.130 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$

Note that $1 \cdot {\mathrm{dm}}^{3} = 1 \times {\left({10}^{-} 1 \cdot m\right)}^{3} = {10}^{-} 3 \cdot {m}^{3} = 1 \cdot L$ as required. A $\text{cubic metre}$ is a huge volume, and is equivalent to $1000 \cdot L$.