Answer
74.0 mL of that $KOH$ solution
Work Step by Step
1. Write and balance the equation for the reaction between $KOH$ and $HCl$:
$KOH(aq) + HCl(aq) --\gt KCl(aq) + H_2O(l)$
- 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 $HCl$ solution is equal to 3.18 moles per liter.
$\frac{3.18 \space moles \space HCl}{1 \space L \space (soln. \space HCl)}$ and $\frac{1 \space L \space (soln. \space HCl)}{3.18 \space moles \space HCl}$
- According to the balanced equation, each mole of $HCl$ reacts with one mole of $KOH$:
$\frac{1 \space mole \space HCl}{1 \space mole \space KOH}$ and $\frac{1 \space mole \space KOH}{1 \space mole \space HCl}$
3. Calculate the required volume:
$5.00 \space mL \space (Soln. \space HCl) \times \frac{3.18 \space moles \space HCl}{1 \space L \space (soln. \space HCl)} \times \frac{1 \space mole \space KOH}{1 \space mole \space HCl} \times \frac{1 \space L \space (soln. \space KOH)}{0.215 \space mole \space KOH}= 74.0 \space mL \space (Soln. \space KOH)$
