Answer
8.41 mL of that 0.558 M $HNO_3$ solution.
Work Step by Step
1. The concentration of the $Ba(OH)_2$ solution is equal to 0.0515 M.
2. The concentration of the $HNO_3$ solution is equal to 0.558 M.
3. According to the balance coefficients, each mol of $Ba(OH)_2$ reacts with 2 moles of $HNO_3$.
4. 1000 mL = 1 L
- Use this information as conversion factors to find the volume of $HNO_3$ solution we should use to react completely with 45.55 mL of that $Ba(OH)_2$ solution.
$45.55mL (Ba(OH)_2) \times \frac{1L}{1000mL} \times \frac{0.0515mol(Ba(OH)_2)}{1L(Ba(OH)_2)} \times \frac{2 mol (HNO_3)}{1mol(Ba(OH)_2)} \times \frac{1L(HNO_3)}{0.558mol(HNO_3)} = 0.00841 L (HNO_3)$
$0.00841L \times \frac{1000mL}{1L} = 8.41mL$
