# What volume of a 1.0 M HCI is required to completely neutralize 25.0 ml of a 1.0 M KOH?

Apr 7, 2017

Why, an equal volume, i.e. a $25.0 \cdot m L$ volume............

#### Explanation:

We interrogate the reaction:

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

When we quote solution concentration, we refer to the quotient:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$, i.e.

$C = \frac{n}{V}$, and thus $\text{concentration}$ has units of $m o l \cdot {L}^{-} 1$

And you will habitually use this quotient to calculate to calculate solution concentrations, moles and mass of solute, and volume of solution needed.

It should not take a great deal of insight to appreciate that the given volume of $N a O H \left(a q\right)$ solution contains THE SAME AMOUNT OF SOLUTE that is in the given volume of $H C l \left(a q\right)$.

i.e. $\text{moles of KCl}$ $=$

$\text{concentration"xx"volume} = 1.0 \cdot m o l \cdot {L}^{-} 1 \times 25.0 \times {10}^{-} 3 \cdot L$

$= 25 \times {10}^{-} 3 \cdot m o l$.

What is the final concentration of $K C l \left(a q\right)$? Why is it different from the starting concentrations of $\left[K O H\right]$, and $\left[H C l\right]$?