# Neutralization occurs when 15.0mL of KOH react with 25.0 mL of HNO3. If the molarity of HNO3 is 0.750 M, what is the molarity of the KOH?

Jun 3, 2018

$\left[K O H\right] = 1.25 \cdot m o l \cdot {L}^{-} 1. \ldots$

#### Explanation:

We interrogate the reaction....

$K O H \left(a q\right) + H N {O}_{3} \left(a q\right) \rightarrow K N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$

$\text{Moles of nitric acid} = 25.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.750 \cdot m o l \cdot {L}^{-} 1 = 0.01875 \cdot m o l$...

And an EQUIVALENT molar quantity of $\text{potassium hydroxide}$ was present in the $15.0 \cdot m L$ volume...and thus a concentration of...

(0.01875*mol)/(15.0*mLxx10^-3*L*mL^-1)=??*mol*L^-1