# How many milliliters of .45 M HCl will neutralize 25.0 mL of 1.00 M KOH?

Jun 14, 2016

Approx. $60 \cdot m L$ of hydrochloric acid are required.

#### Explanation:

With all problems of this type, a BALANCED chemical equation is required to establish the equivalence:

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

Given the 1:1 equivalence:

$\text{Moles of KOH} = 25 \times {10}^{-} 3 \cdot L \times 1.00 \cdot m o l \cdot {L}^{-} 1 = 25 \times {10}^{-} 3 \cdot m o l$

Thus we require an equivalent quantity of $0.45 \cdot m o l \cdot {L}^{-} 1$ hydrochloric acid:

$=$ $\frac{25 \times {10}^{-} 3 \cdot m o l}{0.45 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1$ $=$ ??mL