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Chapter 12: Q31P (page 285)

How many milliliters of 0.050 0 M EDTA are required to react with 50.0 mL of 0.010 0 M Ca2+? With 50.0 mL of 0.010 0 M Al3+?

Short Answer

Expert verified

10 milliliters(mL) of 0.050 0 M EDTA are required to react with 50.0 mL of 0.010 0 M Ca2+.

10 milliliters(mL) of 0.050 0 M EDTA are required to react with 50.0 mL of 0.010 0 M Al3+..

Step by step solution

01

Given Information

Amount of sample taken= 50.0 mL

Concentration of EDTA taken = 0.05M

Amount of Ca2+ required = 50 mL of 0.010 M Ca2+

Amount of Al3+ required = 50 mL of 0.010 M Al3+

02

Determine the amount of EDTA to react with Ca2+

Number of moles of Ca2+ required

=50mL0.010MCa2+

0.5 mmol of EDTA is required to titrate with 0.5 mmol Ca2+

Volume of EDTA required

=0.5mmol0.5mmol=10mL

03

Determine the amount of EDTA to react with Al3+

Number of moles of Al3+ required

=50mL0.010MAl3+=0.5mmolAl3+

0.5 mmol of EDTA is required to titrate with 0.5 mmol Al3+

Volume of EDTA required

=0.5mmol0.05M=10mL

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Most popular questions from this chapter

According to Appendix I, Cu2+ forms two complexes with acetate:

Cu2++CH3CO2鈭�鈬�Cu(CH3CO2)+鈥夆赌夆赌夆赌夆赌夆赌夆赌�尾1(=K1)Cu2++2CH3CO2鈭�鈬�Cu(CH3CO2)2鈥夆赌夆赌夆赌夆赌夆赌夆赌�尾2

(a) Referring to Box 6-2, find K2 for the reaction

Cu(CH3CO2)++CH3CO2鈭�鈬�Cu(CH3CO2)2(aq)鈥夆赌夆赌�K2

(b) Consider 1.00 L of solution prepared by mixing 1.00 脳 10-4 mol Cu(ClO4)2 and 0.100 mol CH3CO2Na. Use Equation 12-16 to find the fraction of copper in the form Cu2+

What is meant by water hardness? Explain the difference between temporary and permanent hardness.

If back titration required 13.00 mL Zn2+, what was the original concentration of Ni2+?

A 1.000-mL sample of unknown containing Co2+ and Ni2+ was treated with 25.00 mL of 0.038 72 M EDTA. Back titration with 0.021 27 M Zn2+ at pH 5 required 23.54 mL to reach the xylenol orange end point. A 2.000-mL sample of unknown was passed through an ion-exchange column that retards Co2+ more than Ni2+. The Ni2+ that passed through the column was treated with 25.00 mL of 0.038 72 M EDTA and required 25.63 mL of 0.021 27 M Zn2+ for back titration. The Co2+ emerged from the column later. It, too, was treated with 25.00 mL of 0.038 72 M EDTA. How many milliliters of 0.021 27 M Zn2+ will be required for back titration?

A 50.0-mL sample containing Ni2+ was treated with 25.0 mL of 0.050 0 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.050 0 M Zn2+. What was the concentration of Ni2+ in the original solution?
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