How many mL of 0.10 M barium hydroxide are required to neutralize 25.0 mL of 0.50 M of nitrous acid?
To determine the amount of barium hydroxide required to neutralize nitrous acid, we need to first write and balance the chemical equation for the reaction:
HNO2 + Ba(OH)2 -> Ba(NO2)2 + 2H2O
From the balanced equation, we can see that one mole of nitrous acid reacts with two moles of barium hydroxide.
Now, we can calculate the moles of nitrous acid present in 25.0 mL of 0.50 M solution:
Moles of HNO2 = (0.50 mol/L) x (25.0 mL / 1000 mL) = 0.0125 mol
Since we know that the ratio of nitrous acid to barium hydroxide is 1:2, we can determine the moles of barium hydroxide needed to neutralize the nitrous acid:
Moles of Ba(OH)2 = 0.0125 mol x 2 = 0.025 mol
Finally, we can calculate the volume of 0.10 M barium hydroxide needed to supply 0.025 moles of Ba(OH)2:
Volume of Ba(OH)2 = Moles / Concentration
Volume of Ba(OH)2 = 0.025 mol / 0.10 mol/L = 0.25 L = 250 mL
Therefore, 250 mL of 0.10 M barium hydroxide is required to neutralize 25.0 mL of 0.50 M nitrous acid.