Need Help with Part 3 Part I: You are instructed to produce solutions of H 2 SO

Question

Need Help with Part 3

Part I: You are instructed to produce solutions of H2SO4 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 H2SO4 and solid NaOH. You will need to make 100 mL of each solution.

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 PbSO4. If you added 1.0 mL of the unknown Pb(NO3)2 to a test tube, what is the amount of H2SO4 in mL you will need to add to be sure the H2SO4 is the excess reagent? NOTE: H2SO4 is expensive so you should not use more than absolutely necessary.
mL H2SO4 Needed =  

Part 3:
After the reaction is complete, you centrifuge the sample, dry and weigh it. The ppt is found to weigh 0.382 grams. Based on this mass, what was the original concentration of the unknown Pb(NO3)2solution?
[Pb(NO3)2] =   M

Compound H2SO4 NaOH Solution Concentration (M) 1.2 0.87 Moles in 100 mL Volume H2SO4 Stock Needed in mL --- Grams NaOH Needed ---

Explanation / Answer

Solution :-

Part 3)

Balanced reaction equation of the Pb(NO3)2 and H2SO4 is as follows

Pb(NO3)2(aq)+H2SO4(aq) --- >PbSO4(s) +2HNO3(aq)

Mass of precipitate (PbSO4)= 0.382 g

Concentration of the Pb(NO3)2 = ?

Volume of Pb(NO3)2 = 1.0 mL = 0.001 L

Lets calculate the moles of Pb(NO3)2 using the mass of PbSO4

(0.382 g PbSO4 * 1 mol / 303.26 g )*(1 mol Pb(NO3)2/1 mol PbSO4)= 0.00126 mol Pb(NO3)2

Now using the moles and volume of the Pb(NO3)2 we can find its concentration

Molarity = moles / volume in liter

Molarity of Pb(NO3)2 = 0.00126 mol / 0.001 L

                                        = 1.26 M

Therefore calculation gives that concentration of the Pb(NO3)2 = 1.26 M

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