1. A sample of pure zinc metal weighing 816.0 mg is carefully dissolved in HCl a

Question

1.    A sample of pure zinc metal weighing 816.0 mg is carefully dissolved in HCl and then diluted to exactly 500.00 mL in a volumetric flask.  The titration of the 50-mL aliquot required 22.84 mL of the edta solution.  Calculate the (a) molarity of the EDTA solution, (b) Zn titer  and (c)  Ca titer.  

2.   A 50.00 mL aliquot of a hard water sample is titrated with 15.00 mL of 0.01280 M EDTA. A second 50.00 mL aliquot portion is made sufficiently alkaline with NaOH to precipitate the Mg ions.  The precipitate was filtered and the filtrate was titrated with 10.00 mL of 0.01280 M edta.  Calculate the ppm Mg and Ca in the water sample.

3.    A 50.00 mL solution (density = 1.046 g/mL) containing both Cd2+ and Pb2+ ions requires 38.40 mL of 0.01925 M EDTA solution for titration of both metals.  A second 50.00 mL aliquot portion is treated with KCN to mask the Cd and then titrated with 22.40 mL of the same EDTA solution.  Calculate the mg Cd per kg of solution and mg Pb/kg of solution.

4.  Calamine, a common remedy for skin irritation, is a remedy that is a mixture of zinc and iron oxides.  A 1.0540 g sample of the calamine was dissolved in acid and diluted to 250.00 mL.  Potassium fluoride was added to a 10.00 mL aliquot of the dilute solution to mask the iron.  After suitable adjustment of the pH, the Zn2+ consumed 38.24 mL of 0.01240 M EDTA to reach the end point.  Calculate the %ZnO in the sample.

Please be as detailed as possible, I am very confused.

Explanation / Answer

I'm Just Entering University At Chemistry Or Should I Say Chemical Engineering So I'll Try Help Out But Dont Fully Go With Me I Have A Different Method From Most Other People Ill Show You How I Done Titration Equations At School.

Okay This Is A Question From A Past Paper (Higher Chemistry 2004 PastPaper)

In a titration, a student found that an average of 16.7cm3 of Iron(II) Sulphate Solutation was needed to react completely with 25.0cm3 of 0.20mol-1 potassium permanganate solution.

The equation for the reaction is:

5Fe2+(aq) +MnO4-(aq) + 8H+(aq) ---> 5Fe3+(aq) +Mn2+(aq) + 4H20(l)

Now The Awnser To This Is

5Fe2+(aq) +MnO4-(aq) + 8H+(aq) ---> 5Fe3+(aq) +Mn2+(aq) + 4H20(l)

5mol 1mol

Then You Can Say That:

VpMp (Permanganate)

--------- = 1 / 5

VsMs (Iron(II) Sulphate)

You Now Cross-Multiply These And Get:

VpMp (Permanganate)

--------- = 5 / 1

VsMs (Iron(II) Sulphate)

You Can Arrange The Titration Equation To Give Ms (The Thing Your Looking For 5VpMp

------------ = 5(25/1000x0.20) / 2(16.7/1000))

2Vs

Thats: 0.025 / 0.0167 which gives 1.5mol-1 which is the correct awnser on my awnser sheet (the working i got from a book just to make sure i didnt make any mistakes)

this is just an example to try help you solve your actual problem and its the only method i've ever been taught how to tackle problems like these i hope it at least helps give maybe a possible understand on how to solve your question =]

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