Final answer:
To find how much 0.010 M HNO3 is needed to neutralize 20 mL of 0.0050 M Ba(OH)2, the balanced reaction shows that 2 moles of HNO3 are required per mole of Ba(OH)2. Calculations lead to the conclusion that 20 mL of HNO3 is required for neutralization.
Step-by-step explanation:
To answer the question of how many milliliters of 0.010 M HNO3 will neutralize 20 mL of 0.0050 M Ba(OH)2, we need to use the concept of titration and the stoichiometry of the reaction between an acid and a base.
The balanced chemical equation for the reaction is:
Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
First, calculate the moles of Ba(OH)2 using its concentration and volume:
Moles of Ba(OH)2 = 0.0050 M × 0.020 L = 0.00010 mol
Since each mole of Ba(OH)2 requires two moles of HNO3 to neutralize it, the moles of HNO3 needed are:
Moles of HNO3 = 2 × 0.00010 mol = 0.00020 mol
Then, calculate the volume of 0.010 M HNO3 using its concentration and the moles required:
Volume of HNO3 = Moles of HNO3 / Concentration of HNO3
= 0.00020 mol / 0.010 M
= 0.020 L or 20 mL
The correct answer is that 20 mL of 0.010 M HNO3 will neutralize 20 mL of 0.0050 M Ba(OH)2.