Answer
There are necessary 74.7 mL of that $NaOH$ solution to completely neutralize that $H_2SO_4$.
Work Step by Step
1. Write and balance the equation for the reaction:
$NaOH + H_2SO_4 -- \gt Salt + H_2O$
- Since each $H_2SO_4$ gives 2 protons $(H^+)$, we must put a 2 before $NaOH$ and $H_2O$:
$2NaOH + H_2SO_4 -- \gt Na_2SO_4 + 2H_2O$
** The equation is balanced.
2. Use the given information as conversion factors to calculate the necessary volume.
- The concentration of this $H_2SO_4$ solution is equal to $0.560$ mol in 1 L.
- The concentration of this NaOH solution is equal to $0.150$ mol in 1 L.
- Each mol of $H_2SO_4$ reacts with 2 moles of $NaOH$.
$10.0mL(H_2SO_4) \times \frac{1L}{1000mL} \times \frac{0.560mol(H_2SO_4)}{1L(H_2SO_4)} \times \frac{2 mol -NaOH}{1mol-H_2SO_4} \times \frac{1L(NaOH)}{0.150mol(NaOH)} \times \frac{1000mL}{1L} = 74.7 mL (NaOH)$
