Hi! My name's Isabow! can someone please help with this question? Thanks in advance :3 <3 <3 <3

If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the molarity of the nitric acid solution. Show all of the work used to solve this problem.

Unbalanced equation: Ba(OH)2 + HNO3 -> Ba(NO3)2 + H2O

Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

mols Ba(OH)2 = M x L = ?
mols HNO3 = 2x that (look at the coefficients).
Then M HNO3 = mols HNO3/L HNO3.

you are soo cute

Hi Isabow! I'd be happy to help you solve this problem. To find the molarity of the nitric acid solution, we can use the equation for molarity which is:

Molarity = moles of solute / volume of solution in liters

First, let's find the moles of barium hydroxide (Ba(OH)2) using its volume and molarity. We can use the formula:

moles = volume (in liters) * molarity

Given that the volume of barium hydroxide is 23.4 milliliters and the molarity is 0.65:

23.4 mL = 23.4 / 1000 = 0.0234 L (conversion from milliliters to liters)
moles of Ba(OH)2 = 0.0234 L * 0.65 mol/L = 0.01521 mol

According to the balanced equation, the stoichiometry of barium hydroxide to nitric acid is 1:2. Since the stoichiometry is 1:2, this means that one mole of barium hydroxide reacts with two moles of nitric acid.

So, if we have 0.01521 moles of barium hydroxide, we will need double that amount of moles of nitric acid. Therefore, we need:

0.01521 mol * 2 = 0.03042 mol of nitric acid

Now, we can find the molarity of nitric acid by dividing the moles by the volume of nitric acid. The volume of nitric acid given is 42.5 milliliters, we need to convert it to liters:

42.5 mL = 42.5 / 1000 = 0.0425 L

Finally, we can calculate the molarity of nitric acid:

Molarity = moles of nitric acid / volume of nitric acid

Molarity = 0.03042 mol / 0.0425 L = 0.716 M

Therefore, the molarity of the nitric acid solution is 0.716 M.

Hope that helps! Let me know if you have any other questions.