Analysis of Silver in an Alloy

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Analysis of Silver in an Alloy

Introduction

In this experiment an alloy of silver will be analyzed to determine its silver content. The silver-copper alloy will be dissolved in nitric acid, the silver will be precipitated as silver chloride, and the silver chloride will be filtered, washed, dried and its mass determined. From the mass of the silver chloride formed and the mass of the original sample, you will be able to calculate the percent of silver in the alloy. Because the results are based on the mass of a product, this procedure is classified as a gravimetric analysis.

Silver and copper are very nonreactive metals. Neither will dissolve in hydrochloric acid or sulfuric acid. The "oxidizing" acid nitric acid, HN03, is required. In acidic solutions the nitrate ion is an excellent oxidizer, and it will oxidize Ag(s) to Ag+(aq) and Cu(s) to Cu2+(aq). The reduction product is the gas NO. As the colorless nitrogen monoxide gas forms, it immediately reacts with the oxygen in the air to produce the orange-brown gas N02. The half-reactions for the oxidation of silver and copper by nitric acid are as follows: Ag(s) → Ag+(aq) + e-

Cu(s) → Cu2+(aq) + 2e-
4H+(aq) + NO3- (aq) + 3e- → NO(g) + 2H2O(l)

Once the silver and copper ions are in solution, they can be separated from each other by precipitating the silver ions as silver chloride. Silver chloride (AgCl) is very insoluble in water, while copper(II) chloride (CuCl2) is soluble. The addition of chloride ions to the solution will precipitate essentially all of the silver and none of the copper. The silver chloride precipitate is then filtered from the solution.

Experimental Methods

Follow protocol as listed for Laboratory Experiment #1 in “Experiments for Advanced Placement Chemistry” by Sally Ann Vonderbrink, Ph. D. With these modifications: instead of using a Gooch Crucible and fiber glass pad, we used a Buchner Funnel and filter paper.

Theoretical Methods

Balance out the half reactions of Ag and NO3-

3( Ag(s) → Ag+(aq) + e- )
4H+(aq) + NO3- (aq) + 3e- → NO(g) + 2H2O(l)

3Ag(s) + 4H+(aq) + NO3- (aq) → 3Ag+(aq) + NO(g) + 2H2O(l)

Calculate amount of NaCl needed to precipitate

NaCl → Na+ + Cl-
Ag + Cl → AgCl(s)
.3015 g Ag 1 mol Ag 1 mol Cl 1 mol NaCl 58.44 g NaCl 107.87g Ag 1 mol Ag 1 mol Cl 1 mol NaCl = .1635g NaCl

multiply needed NaCl x 2

.1635 x 2 = .3269 g NaCl

Calculate the percent of Ag in the Alloy

.3555 g AgCl 1 mol AgCl 1 mol Ag 107.87 g Ag = .2675 g Ag 143.23 g AgCl 1 mol AgCl 1 mol Ag

% Ag = .2675 g Ag x 100 = 75.25%
.3555 g AgCl
Calculate the percent error between percents of Ag in an Alloy

% error = actual – experimental x 100 =
actual
= 90.08 – 75.25 x 100 = 16.46 % 90.08

Results

Experimental Results:
We first weighed our original sample of silver alloy, as shown in table 1. Then added 10 ml of nitric acid to the silver alloy which dissolved, by heating, the alloy into silver and copper ions as nitrite gas escaped in an orange cloud. After the alloy was completely dissolved we added a solution of sodium chloride dissolved in distilled water. When calculating out sodium chloride we doubled the necessary amount to make sure that a full and complete reaction occurred when once again heating. After letting that sit over night covered in para-film to form large precipitate particles of silver chloride we filtered the particles from the solution with the buchner funnel. We used a diluted nitric acid as our wash because it helped to keep the precipitate from forming to small particles. We then heated the sample so we could weigh out the final product as shown in table 1, and then calculated the percent of silver as shown...
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