# Using the correct amount of significant figures, how many milliliters of 8.48×10^(−2) M Ba(OH)2 (aq) are required to titrate 54.90 mL of 5.38×10^(−2) M H(NO)3?

##### 1 Answer
Jul 14, 2016

$\textsf{17.4 \textcolor{w h i t e}{x} m l}$

#### Explanation:

Start with the equation:

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

So 1 mole of $\textsf{B a {\left(O H\right)}_{2}}$ will react with 2 moles of $\textsf{H N {O}_{3}}$

Concentration = amount of solute / volume of solution.

So:

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{n = c \times v}$

$\therefore$$\textsf{n H N {O}_{3} = 5.38 \times {10}^{- 2} \times \frac{54.90}{1000} = 2.953 \times {10}^{- 3}}$

$\therefore$$\textsf{n B a {\left(O H\right)}_{2} = \frac{2.953 \times {10}^{- 3}}{2} = 1.477 \times {10}^{- 3}}$

$\textsf{v = \frac{n}{c} = \frac{1.477 \times {10}^{- 3}}{8.48 \times {10}^{- 2}} = 0.01742 \text{ } L}$

$\textsf{v = 17.4 \textcolor{w h i t e}{x} m l \text{ ""to 3sf}}$.