A titration required 18.38 mL of 0.1574 M NaOH solution. How many moles of NaOH

Question

A titration required 18.38 mL of 0.1574 M NaOH solution. How many moles of NaOH were in this volume? 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. Forgetful Freddy weighed his KHP sample, but forgot to bring his report sheet along, so he recorded the mass of KHP 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. What mass of solid NaOH would be needed to make 645 mL of Freddy's NaOH solution?

Explanation / Answer

Q3

mol = M*V

M = molarity in mol per liter

V = liter of solution

so

V = 18.39 mL = 18.38*10^-3 L

so

mol = MV = (18.38*10^-3)(0.1574) = 0.00289301 mol of NaOH

Q4

KHP + NaOH = H2O + NaKP

mol of KHP = mass/MW = 1.276/ 204.22 = 0.00624 mol of KHP

so..

ratio is 1:1

so 0.00624 mol of NAOH must be used

M = mol/V = 0.00624 / (19.84*10^-3) = 0.31451 M of NaOH

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