A mixture having a volume of 10.00 mL and containing 0.1000 M Ag+ and 0.04500 M

Question

A mixture having a volume of 10.00 mL and containing 0.1000 M Ag+ and 0.04500 M Hg22+ was titrated with 0.0800 M KCN to precipitate Hg2(CN)2 and AgCN. (a) Explain which salt will precipitate first and why. (b) Calculate volume of KCN required to reach first and second equivalence points and sketch by hand the expected titration curve. (c) Calculate pCN- at 8.00 mL KCN added.
A mixture having a volume of 10.00 mL and containing 0.1000 M Ag+ and 0.04500 M Hg22+ was titrated with 0.0800 M KCN to precipitate Hg2(CN)2 and AgCN. (a) Explain which salt will precipitate first and why. (b) Calculate volume of KCN required to reach first and second equivalence points and sketch by hand the expected titration curve. (c) Calculate pCN- at 8.00 mL KCN added.
A mixture having a volume of 10.00 mL and containing 0.1000 M Ag+ and 0.04500 M Hg22+ was titrated with 0.0800 M KCN to precipitate Hg2(CN)2 and AgCN. (a) Explain which salt will precipitate first and why. (b) Calculate volume of KCN required to reach first and second equivalence points and sketch by hand the expected titration curve. (c) Calculate pCN- at 8.00 mL KCN added.

Explanation / Answer

Answer:

a). AgCN will precipitate first because of its higher concentration.

b). concentration of KCN (m1) = 0.0800 M

Volume of KCN (V1) = V

Mixture of volume (V2) = 10 mL

Concentration of mixture = 0.1000+ 0.04500

According to M1V1 = M2V2

V = M2V2/M1

= (0.1000)*10/0.0800

= 12.5 mL

First reach of titration = 12.5 mL

V = (0.1*0.045)*10/0.0800

=0.05625

second reach of titration = 12.5+0.5625 = 13.0625

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