a. A 0.6740-g sample of impure Ca(OH) 2 is dissolved in enough water to make 57.

Question

a. A 0.6740-g sample of impure Ca(OH)2 is dissolved in enough water to make 57.70 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.2546-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 17.83 mL of the acid to reach the endpoint?

b. What is the iodide ion concentration in a solution if the addition of an excess of 0.100 M Pb(NO3)2 to 33.8 mL of the solution produces 565.2 mg of PbI2?

c. All of the iron in 1.608 g of an iron sample is converted into Fe2+, which was then titrated with dichromate:

14H1+ + Cr2O72- + 6Fe2+ -> 6Fe3+ + 2Cr3+ + 7H2O

What is the mass percent iron in the sample if 34.58 mL of 0.11050 M K2Cr2O7 were required to reach the equivalence point?

Explanation / Answer

ANSWER

Dear candidate you have asked three questions. As per guidelines one question one time. Please send the other two separatly.

(a)Ca(OH)2 + 2HCl -----> CaCl2 + H2O

One mole of Ca(OH)2 requires two moles of HCl

Moles of HCl consumed = Volume of HCl solution in liters X molarity of HCl

Volume of HCl solution in liters = 17.83 / 1000 = 0.01783L

Moles of HCl consumed = 0.01783 X 0.2546 = 0.00454 moles

Hence moles of Ca(OH)2 present = 2 X 0.00454 = 0.00908 Moles

Because 2 moles of HCl are required for one mole of Ca(OH)2

Hence mass of Ca(OH)2 present in the impure sample = no. of moles X Molar mass

= 0.00908 X 74.093

= 0.6727g

Total amount of impure sample = 0.6740g

% purity = Amount of Ca(OH)2 / Total amount X 100 = 0.6727 / 0.6740 X 100 = 99.8%

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