Answer
48.2 mL of that $KOH$ solution.
Work Step by Step
1. Write and balance the equation for the reaction between $KOH$ and $HNO_3$:
$KOH(aq) + HNO_3(aq) --\gt KNO_3(aq) + H_2O(l)$
- The equation is balanced.
2. Identify the conversion factors.
- The concentration of that $KOH$ solution is equal to 0.215 mole per liter.
$\frac{0.215 \space mole \space KOH}{1 \space L \space (soln. \space KOH)}$ and $\frac{1 \space L \space (soln. \space KOH)}{0.215 \space mole \space KOH}$
- The concentration of that $HNO_3$ solution is equal to 0.560 mole per liter.
$\frac{0.560 \space mole \space HNO_3}{1 \space L \space (soln. \space HNO_3)}$ and $\frac{1 \space L \space (soln. \space HNO_3)}{0.560 \space mole \space HNO_3}$
- According to the balanced equation, each mole of $HNO_3$ reacts with one mole of $KOH$:
$\frac{1 \space mole \space HNO_3}{1 \space mole \space KOH}$ and $\frac{1 \space mole \space KOH}{1 \space mole \space HNO_3}$
3. Calculate the required volume:
$18.5 \space mL \space (Soln. \space HNO_3) \times \frac{0.560 \space mole \space HNO_3}{1 \space L \space (soln. \space HNO_3)} \times \frac{1 \space mole \space KOH}{1 \space mole \space HNO_3} \times \frac{1 \space L \space (soln. \space KOH)}{0.215 \space mole \space KOH}= 48.2 \space mL \space (Soln. \space KOH)$
