An alloy contains |2.0 wt% Ni. What volume of 0.83 wt% DMG should be used to provide a 50% excess of DMG for the analysi

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An alloy contains |2.0 wt% Ni. What volume of 0.83 wt% DMG
should be used to provide a 50% excess of DMG for the analysis of
1.8 g of steel? What mass of Ni(DMG)2 precipitate is expected?
(Answer: 33 mL, 0.18 g)

An Alloy Contains 2 0 Wt Ni What Volume Of 0 83 Wt Dmg Should Be Used To Provide A 50 Excess Of Dmg For The Analysi 1 (209.81 KiB) Viewed 50 times

EXAMPLE Calculating How Much Precipitant to Use (a) To measure the nickel content in steel, the alloy is dissolved in 12 M HCl and neu- tralized in the presence of citrate ion, which maintains iron in solution. The slightly basic solution is warmed, and dimethylglyoxime (DMG) is added to precipitate the red DMG-nickel complex quantitatively. The product is filtered, washed with cold water, and dried at 110°C. OH OH Ni2+ + 2 + 2H (27-7) N OH 7...0 DMG FM 116.12 Bis(dimethylglyoximate)nickel(II) FM 288.91 FM 58.69 If the nickel content is known to be near 3 wt% and you wish to analyze 1.0 g of steel, what volume of 1.0 wt% alcoholic DMG solution should be used to give a 50% excess of DMG for the analysis? Assume that the density of the alcohol solution is 0.79 g/mL. Solution Because the Ni content is about 3%, 1.0 g of steel will contain about 0.03 g of Ni, which corresponds to 0.03 g NT = 5.11 X 10 mol Ni 58.69 g Nt/mol Ni 27-3 Examples of Gravimetric Calculations This amount of metal requires 2(5.11 X 10-met-NT)(116.12 g DMG/mel NI) = 0.119 g DMG because 1 mol of Ni2+ requires 2 mol of DMG. A 50% excess of DMG would be (1.5)(0.119 g) = 0.178 g. This much DMG is contained in 0.178 g DMG = 17.8 g solution 0.010 g-DMG/g solution which occupies a volume of 17.8 g-seltition = 23 mL 0.79 g-solution/mL (b) If 1.163 4g of steel gives 0.179 5 g of precipitate, what percentage of Ni is in the steel? Solution For each mole of Ni in the steel, 1 mol of precipitate will be formed. Therefore, 0.179 5 g of precipitate corresponds to 0.179 5 g NI(DMG): = 6.213 x 10-4 mol Ni(DMG), 288.91 g Ni (MGT3/mol Ni(DMG)2 The mass of Ni in the steel is (6.213 X 10-4 mel Ni) (58.69 g/mol Ni) = 0.036 46 g, and the wt% Ni in steel is 0.036 46 g Ni x 100 = 3.134% 1.1634 g steel A slightly simpler way to approach this problem comes from realizing that 58.69 g of Ni (1 mol) would give 288.91 g (1 mol) of product. Calling the mass of Ni in the sample X, we can write , Grams of Ni analyzed 58.69 Ni = 0.036 46 g Grams of product formed 0.1795 288.91 TEST YOURSELF An alloy contains -2.0 wt% Ni. What volume of 0.83 wt% DMG should be used to provide a 50% excess of DMG for the analysis of 1.8 g of steel? What mass of Ni(DMG), precipitate is expected? (Answer: 33 mL, 0.18 g)