How many milliliters of 0.201 M Ba(OH)2 solution are needed to completely neutralize 12.8 mL of 0.369 M HNO3 solution? The reaction is

Ba(OH)2(aq) + 2HNO3(aq) → Ba(NO3)2(aq) + 2H2O(l)

How many moles HNO3 do you have? That is M x L = moles HNO3.

Use the coefficients in the balanced equation to convert moles HNO3 to moles Ba(OH)2.
Then M Ba(OH)2 = moles Ba(OH)2/L Ba(OH)2. You know moles and M, solve for L and convert to mL.

To determine how many milliliters of the Ba(OH)2 solution are needed to neutralize the HNO3 solution, we can use the concept of stoichiometry and the given reaction balanced equation. Here's how you can calculate it step by step:

Step 1: Write down the balanced equation for the reaction:
Ba(OH)2(aq) + 2HNO3(aq) → Ba(NO3)2(aq) + 2H2O(l)

Step 2: Determine the stoichiometric ratio between Ba(OH)2 and HNO3 from the balanced equation. In this case, it's 1:2. This means that one mole of Ba(OH)2 reacts with 2 moles of HNO3.

Step 3: Convert the volume of the HNO3 solution (12.8 mL) to moles using its molarity:
moles of HNO3 = volume (L) × concentration (M)
moles of HNO3 = 12.8 mL × (1 L / 1000 mL) × 0.369 M
moles of HNO3 = 0.0047288 mol

Step 4: Use the stoichiometric ratio to determine the moles of Ba(OH)2:
moles of Ba(OH)2 = (moles of HNO3) / 2
moles of Ba(OH)2 = 0.0047288 mol / 2
moles of Ba(OH)2 = 0.0023644 mol

Step 5: Convert the moles of Ba(OH)2 to volume in milliliters using its molarity:
volume (mL) = moles × (1000 mL / 1 L) × (1 M / 0.201 mol)
volume (mL) = 0.0023644 mol × 1000 mL / 1 L × 1 M / 0.201 mol
volume (mL) = 11.725 mL

Therefore, approximately 11.725 milliliters of the 0.201 M Ba(OH)2 solution are needed to completely neutralize 12.8 mL of the 0.369 M HNO3 solution.