Question

A volume of 12.95 mL of 0.1080 M NaOH solution was used to titrate a 0.596 g sample of unknown containing K₂HPO₄.

a. What is the molecular mass of K₂HPO₄? (report answers to 4 or 5 significant figures)

b. What is the percent by mass of K₂HPO₄ in the unknown?

c. In this problem what mass of sample in grams would be needed to deliver about 22.00 mL in the next trial?

d. In the second trial above, exactly 0.996 g was transferred into a flask to be titrated. If the initial buret reading is 0.10 mL, predict what the final buret reading be.

Answer 1

Hi,

A) To calculate the molecular mass of K2HPO4, first we will calculate the moles of NaOH required to titrate the unknown sample containing K2HPO4

Moles of NaOH = Volume in Litre X concentration in Molarity = 0.01295 x 0.1080 = 1.3986 X 10^-3 moles

Now, It can be assumed that mole ratio of NaOH to K2HPO4 would be 1:2 (Since K2HPO4 has 2 replaceable hydrogen ions)

Therefore, number of moles of K2HPO4 = 1.3986 X 10^-3 / 2 = 6.993 X 10^-4 moles

Now, Molecular Mass of K2HPO4 = Moles of K2HPO4 X molecular weight of K2HPO4 = 6.993 X 10^-4 mol X 174.2 g/mol = 0.12189 g

Therefore, Molecular Mass of K2HPO4 is 0.12189 g

B) Percent by mass of K2HPO4 in unknown sample = 0.12189/0.596 X 100 = 20.44 %

C) We will calculate first the molecular mass of K2HPO4 as done in solving problem (A).

Moles of NaOH = 0.022 x 0.1080 = 2.376 X 10^-3 moles

Therefore, number of moles of K2HPO4 = 2.376 X 10^-3 / 2 = 1.188 X 10^-3 moles

Now, Molecular Mass of K2HPO4 = 1.188 X 10^-3 moles X 174.2 g/mol = 0.20695 g

From A, we know in 0.596 g of unknown sample, 0.12189g of K2HPO4 is present

Therefore, 0.20695g of K2HPO4 will be present in = 0.20695 X 0.596 / 0.12189 = 1.0119 g

The mass of unknown sample is 1.0119g

D) We will solve this problem using the above equation used in problem A

Mass of K2HPO4 present in 0.996 g sample is 0.996 X 0.12189 / 0.596 = 0.20369 g

Now, Number of moles of K2HPO4 will be 0.20369g/ 174.2 = 1.1693 X 10^-3 moles

therefore the number of moles of NaOH would be 2.3386 X 10^-3 moles (1.1693 X 10^-3 molex 2)

Now, the final volume of Naoh( burette reading ) = 2.3386 X 10^-3 moles/ 0.1080 = 0.02165 L = 21.65 mL

The final burette reading would be 21.65 mL