# How many milliliters of 0.250M NaOH are required to neutralize 30.4 mL of 0.152 M HCl?

May 12, 2016

Approx. $19 \cdot m L$. We know that

$\text{Concentration"="Amount of substance in moles"/"Volume of solution}$

#### Explanation:

$\text{Moles of HCl}$ $=$ $30.4 \times {10}^{-} 3 \cancel{L} \times 0.152 \cdot m o l \cdot \cancel{{L}^{-} 1}$ $=$ $4.62 \times {10}^{-} 3 \cdot m o l$.

We need an equivalent quantity of sodium hydroxide; $0.250 \cdot m o l \cdot {L}^{-} 1$ sodium hydroxide is available.

So, (4.62xx10^-3*cancel(mol))/(0.250*cancel(mol)*cancel(L^-1))xx10^3*mL*cancel(L^-1 $=$ ??mL

What I should have done at the beginning is to write the stoichiometric equation:

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

Why should I write this first? Because it establishes the stoichiometry; the 1:1 molar equivalence. One equiv of sodium hydroxide reacts with one equiv hydrochloric acid. Thus if I know the quantity of acid, I also know the quantity of base. There will be reactions where a different stoichiometry pertains.