How many mL of 0.20 M hydrochloric acid is required to neutralize 100 mL of 0.80 M potassium hydroxide?

Aug 11, 2016

$\text{Volume of HCl} = 400$$m L$.

Explanation:

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

Now the reaction is trivial here. It is nevertheless good practice to write the stoichiometric equation to show the $1 : 1$ equivalence.

$\text{Moles of KOH}$ $=$ $0.80 \cdot m o l \cdot {L}^{-} 1 \times 0.100 \cdot L$ $=$ $0.080 \cdot m o l$. Note that here I normalized the volumes, $1 \cdot m L$ $=$ ${10}^{-} 3 L$.

Thus $\text{volume of HCl}$ $=$ $\frac{0.080 \cdot m o l}{0.20 \cdot m o l \cdot {L}^{-} 1}$ $=$ $0.40 \cdot L$

PS Sorry for shouting in the top answer; I still haven't got the hang of this editor.