Please check these thanks.

1. In a titration, 33.21 mL 0.3020M rubidium hydroxide solution is required to exactly neutralize 20.00 mL hydrofluroic acid solution. What is the molarity of the hydrofluroic acid solution?

Answer: 0.502HF

2. A 35.00 mL-sample of NaOH solution is titrated to an endpoint by 14.76 mL 0.4122M HBr solution. What is the molarity of the NaOH solution.

Answer: 0.174M NaOH

Why did you throw away some of the answer?

You have 4 places in each of these problems but you rounded your answer to 3. You threw away a perfectly good number in each problem. Otherwise, both look ok to me.

what is the molarity of a solution prepared by diluting 500ml of a 6.0 M solution to 750 ml?

how would you prepare 60.0 ml of 0.2 M HNO3 from a stock solution of 4.00 M HNO3?

To check the answers for these titration problems, we need to use the concept of stoichiometry and the balanced chemical equation. I'll explain the steps involved in solving each problem, so you can verify the provided answers:

1. To find the molarity of the hydrofluroic acid (HF) solution, we will use the concept of stoichiometry:

Step 1: Write the balanced chemical equation for the reaction between rubidium hydroxide (RbOH) and hydrofluroic acid (HF):
RbOH + HF -> RbF + H2O

From the balanced equation, we can see that the ratio between RbOH and HF is 1:1.

Step 2: Calculate the number of moles of RbOH used in the titration:
Given: volume of RbOH solution = 33.21 mL = 0.03321 L
Given: molarity of RbOH solution = 0.3020 M

moles of RbOH = volume of RbOH solution * molarity of RbOH solution
moles of RbOH = 0.03321 L * 0.3020 M

Step 3: Since the stoichiometric ratio between RbOH and HF is 1:1, the number of moles of HF is equal to the number of moles of RbOH used in the titration.

Step 4: Calculate the molarity of the HF solution:
Given: volume of HF solution = 20.00 mL = 0.02000 L

molarity of HF = moles of HF / volume of HF solution
molarity of HF = moles of RbOH / volume of HF solution

Substituting the values:
molarity of HF = (0.03321 L * 0.3020 M) / 0.02000 L

By calculating this, we find that the molarity of the hydrofluroic acid solution is approximately 0.502 HF.

2. Similarly, for the second problem:

Step 1: Write the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and hydrobromic acid (HBr):
NaOH + HBr -> NaBr + H2O

From the balanced equation, we can see that the ratio between NaOH and HBr is also 1:1.

Step 2: Calculate the number of moles of HBr used in the titration:
Given: volume of HBr solution = 14.76 mL = 0.01476 L
Given: molarity of HBr solution = 0.4122 M

moles of HBr = volume of HBr solution * molarity of HBr solution
moles of HBr = 0.01476 L * 0.4122 M

Step 3: Since the stoichiometric ratio between NaOH and HBr is 1:1, the number of moles of NaOH is equal to the number of moles of HBr used in the titration.

Step 4: Calculate the molarity of the NaOH solution:
Given: volume of NaOH solution = 35.00 mL = 0.03500 L

molarity of NaOH = moles of NaOH / volume of NaOH solution
molarity of NaOH = moles of HBr / volume of NaOH solution

Substituting the values:
molarity of NaOH = (0.01476 L * 0.4122 M) / 0.03500 L

By calculating this, we find that the molarity of the sodium hydroxide solution is approximately 0.174 M NaOH.

You can now check if the provided answers match the calculated molarities.