If you have added 50 mL of water to a sample of KHP instead of 30 ML, would the

Question

If you have added 50 mL of water to a sample of KHP instead of 30 ML, would the titration of that sample then have required more, less, or the same amount of base? Explain. A student weighed out 1.106 g of KHp. How many moles was that? _________ mol A titration required 18.38 mL of 0.1574 M NaOH solution. How many moles of NaOH were in this volume? _____________ mol A student weighed a sample of KHP and found it weighed 1.276 g. Titration of this KHP required 19.84 ml of base (NaOH). Calculate the molarity of the base. ____________ M Forgetful Freddy weighed his KHP sample, but forgot to bring his report sheet along, so he recorded his mosses on a paper towel. During his titration, which required 18.46 mL of base, he spilled some base on his hands. He remembered to wash his hands, but forgot about the data on the towel, and used it to dry his hands. When he went to calculate the molarity of his base, Freddy discovered that he didn't have the mass of his KHP. His kindhearted instructor told Freddy that his base was 0.2987 M. Calculate the mass of Freddy's KHP sample. ___________ g What mass of solid NaOH would be needed to make 645 mL of Freddy's NaOH solution?

Explanation / Answer

1. If the KHP solution was made with 50 ml instead of 30 ml water. The resulting KHP solution would be more dilute and thus would need less amount of base for neutralization.

2. moles of KHP = 1.106 g/204.22 g/mol = 0.0054 mols

3. moles of NaOH used = 0.1574 M x 18.38 ml = 2.893 mmol

4. moles of KHP = 1.276 g/204.22 g/mol = 0.00625 mols

molarity of base = 0.00625 mols/0.01984 L = 0.315 M

5. mass of Freddy's KHP sample = 0.2987 M x 0.01846 L x 204.22 g/mol = 1.126 g

6. Mass of solid NaOH needed = 0.2987 M x 0.645 L x 40 g/mol = 7.7065 g

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