1. 33.32/50 points | Previous Answers My Notes Ask Your Teacher Part I: You are

Question

1. 33.32/50 points | Previous Answers My Notes Ask Your Teacher Part I: You are instructed to produce solutions of HI and NaOH for your lab experiment. The concentration you should produce is shown in the table below. Complete the table and copy it into your lab notebook for use during your lab. You will be provided with a 3.0 M stock solution of HI and solid NaOH. You will need to make 100 mL of each solution. Compound Solution Concentration (M) Moles in 100 mL NaOH 0.82 HI 1.17 0.082 Volume HI Stock Needed in mL 39.00 Grams NaOH Needed 3.28 Part 2: You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label making it impossible to read. You know the concentration is below 1.0 M as you can make out "0.xxx" at the beginning of the label. In order to determine the concentration, you decide to precipitate out the lead in the solution as Pbl2. If you added 1.0 mL of the unknown Pb(NO3)2 to a test tube, what is the amount of HI in mL you will need to add to be sure the HI is the excess reagent? NOTE: HI is expensive so you should not use more than absolutely necessary. mL HI Needed .667 Part 3: After the reaction is complete, you centrifuge the sample, dry and weigh it. The ppt is found to weigh 0.399 grams. Based on this mass, what was the original concentration of the unknown Pb(NO3)2 solution? [Pb(NO3)2l-

Explanation / Answer

part 2

Pb(NO3)2 + 2HI = PbI2 + 2HNO3

According to reaction, 2mole of HI react with ! mole of Pb(NO3)2 to completely form ppt.

Now,1 ml of Pb(NO3)2 1M solution contain 0.001 mole of  Pb(NO3)2 . So we can be sure that atmost 0.002 mole of HI will be required to completely form ppt.

Density of HI = 2.850 gm/ml

Molecular mass of HI = 127.97 gm/mol

Hence, moles of HI = 2.850 / 127.97 mol/ml

= 0.02188 mol/ml

Now,

volume of HI required = moles of Pb(NO3)2 in 1 ml / moles of HI in 1 ml

= 0.002 / 0.02188 = 0.0914 ml

Answer= 0.0914ml

Part 3

Molecular mass of PbI2 = 461.01 gm /mol

mass of ppt = 0.399 gm ( given )

Moles of ppt form = 0.399 / 461.01 = 0.0008655 mol

Hence moles of Pb(NO3)2 =  0.0008655 mol (as all Pb is precipitated)

now concentration of Pb(NO3)2 = 0.0008655 mol/ml = 0.0008655 * 1000 mol / litre = 0.8655 M

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