A student is asked to standardize a solution of barium hydroxide. He weighs out 0.945 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid).

It requires 33.9 mL of barium hydroxide to reach the endpoint.

A. What is the molarity of the barium hydroxide solution?

This barium hydroxide solution is then used to titrate an unknown solution of nitric acid.

B. If 19.8 mL of the barium hydroxide solution is required to neutralize 20.4 mL of nitric acid, what is the molarity of the nitric acid solution.

I have no idea how to do this. Thank you for your help!

These are essentially 4-step problems.

1. Write and balance the equation. If we call potassium hydrogen phthalate just KHP, then
Ba(OH)2 + 2KHP ==> 2H2O + BaP + K2P

2. Convert grams to moles. moles KHP = 0.945/molar mass KHC8H4O4 = 0.945/204 = approximately 0.0046

3. Using the coefficients in the balanced equation, convert moles KHP to moles Ba(OH)2
0.0046moles KHP x (1 mole Ba(OH)2/2 moles KHP) = 0.0046 x (1/2) = 0.0023 moles Ba(OH)2.

4. M Ba(OH)2 = moles Ba(OH)2/L Ba(OH)2
M Ba(OH)2 = 0.0023/0.0339 = approximately 0.068 M

Part B is done the same way. Now that you have the Ba(OH)2 standardized, use that, with the same process, to determine the molarity of the HNO3.
Remember that moles = M x L.

A. To find the molarity of the barium hydroxide solution, we first need to calculate the number of moles of potassium hydrogen phthalate used. The molar mass of KHC8H4O4 is:

1 mol K + 1 mol H + 8 mol C + 4 mol H + 4 mol O = 204.22 g/mol

So, 0.945 g of KHC8H4O4 is approximately:

0.945 g / 204.22 g/mol = 0.00463 mol

Since the balanced chemical equation for the reaction is 2 KHC8H4O4 + Ba(OH)2 -> Ba(C8H4O4)2 + 2 H2O, we can see that the stoichiometry is 2:1 between KHC8H4O4 and Ba(OH)2.

Therefore, the number of moles of Ba(OH)2 is:

0.00463 mol KHC8H4O4 * (1 mol Ba(OH)2 / 2 mol KHC8H4O4) = 0.00232 mol

Now, we can find the molarity (M) of the barium hydroxide solution using the equation:

M = moles / volume(L)

M = 0.00232 mol / 0.0339 L = 0.0683 M

So, the molarity of the barium hydroxide solution is approximately 0.0683 M.

B. The stoichiometry between barium hydroxide (Ba(OH)2) and nitric acid (HNO3) is 2:2, meaning they react in a 1:1 ratio.

Using the molarity of the barium hydroxide solution (0.0683 M) and its volume (19.8 mL), we can calculate the number of moles of Ba(OH)2 used:

moles Ba(OH)2 = M * V(L)
moles Ba(OH)2 = 0.0683 M * 0.0198 L = 0.00135 mol

Since the stoichiometry is 1:1, the number of moles of nitric acid (HNO3) is also 0.00135 mol.

To find the molarity of the nitric acid solution, we use the equation:

M = moles / volume(L)

M = 0.00135 mol / 0.0204 L = 0.0662 M

So, the molarity of the nitric acid solution is approximately 0.0662 M.

I hope that helped! If you have any more questions, feel free to ask.

To determine the molarity of the barium hydroxide solution, we first need to calculate the number of moles of potassium hydrogen phthalate (KHP) used in the titration.

The balanced chemical equation for the reaction between KHP and barium hydroxide (Ba(OH)2) is:

KHP + Ba(OH)2 -> K+(aq) + Ba2+(aq) + H2O + HPhthalate2-

We know that the molar mass of KHP is 204.23 g/mol and the mass of the KHP used is 0.945 g. Using the formula:

moles of KHP = mass of KHP / molar mass of KHP

Substituting the values:

moles of KHP = 0.945 g / 204.23 g/mol ≈ 0.00463 mol

Since the reaction between KHP and barium hydroxide has a 1:1 stoichiometric ratio, the number of moles of barium hydroxide used is also 0.00463 mol.

Now, to calculate the molarity of the barium hydroxide solution, we use the formula:

Molarity (M) = moles of solute / volume of solution in liters

Substituting the values:

Molarity of barium hydroxide = 0.00463 mol / 0.0339 L ≈ 0.1364 M

Therefore, the molarity of the barium hydroxide solution is approximately 0.1364 M.

Moving on to the second part:

To determine the molarity of the nitric acid solution, we can use the same concept of titration. From the balanced equation given below:

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

We can see that the stoichiometric ratio between nitric acid (HNO3) and barium hydroxide (Ba(OH)2) is 2:1. This means that for every mole of Ba(OH)2, 2 moles of HNO3 are required.

Given that 19.8 mL of the barium hydroxide solution is required to neutralize 20.4 mL of nitric acid, we can set up the following equation based on the stoichiometry:

Molarity of Ba(OH)2 × volume of Ba(OH)2 = Molarity of HNO3 × volume of HNO3 × 2

Rearranging the equation:

Molarity of HNO3 = (Molarity of Ba(OH)2 × volume of Ba(OH)2) / (volume of HNO3 × 2)

Substituting the given values:

Molarity of HNO3 = (0.1364 M × 19.8 mL) / (20.4 mL × 2) = 0.0663 M

Therefore, the molarity of the nitric acid solution is approximately 0.0663 M.

Barium hydroxide is standardising usinig pottassium hydrogen phthalate..phenolphthalein is the indicator.(.02 N BaOH standardising with .02 N PHT)