# How many milliliters of 0.250 M KoH will be needed to titrate 12.53 mL of 0.130 M HNO_3?

Apr 17, 2017

You will need 6.516 mL of KOH to reach the equivalence point.

#### Explanation:

Since KOH has only one $O {H}^{-}$ ion in its formula, and $H N {O}_{3}$ is monoprotic (donates only one hydrogen ion), the titration reaches an equivalence point when

Moles $H N {O}_{3}$ = moles $K O H$

And since moles (of solute) = concentration x volume, we can write

${M}_{a} \cdot {V}_{a} = {M}_{b} \cdot {V}_{b}$

(the $a$ and $b$ representing the acid and base, respectively)

We know the first three quantities in this equation. We find the fourth (${V}_{b}$) as follows:

$\left(0.130\right) \cdot \left(0.01253\right) = \left(0.250\right) {V}_{b}$

${V}_{b} = \frac{\left(0.130\right) \cdot \left(0.01253\right)}{0.250} = 0.006516 L$

or $6.516 m L$

Apr 17, 2017

Approx. $6 - 7 \cdot m L \ldots \ldots \ldots .$

#### Explanation:

We need (i) a stoichiometric equation........

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

And then (ii) equivalent quantities of potassium hydroxide and nitric acid:

$\text{Moles of nitric acid} = 12.53 \cdot m L \times {10}^{-} 3 L \cdot m {L}^{-} 1 \times 0.130 \cdot m o l \cdot {L}^{-} 1 = 1.63 \times {10}^{-} 3 m o l$.

And thus we need $\frac{1.63 \times {10}^{-} 3 m o l}{0.250 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 6.52 \cdot m L$