2) A student performed the experiment you\'ll be performing, and obtained the fo

Question

2) A student performed the experiment you'll be performing, and obtained the following nformation The standard curve generated from his data is at the right Mass of unknown nickel salt he used to prepare 25 mL of solution: 0.2508g. y 16.943x+0.0077 R0.99983 Absorbance reading of unknown nickel salt solution (at appropriate wavelength): 0.322. concentration of nickel (II) (M) Use this information to determine the percent by mass of Nickel (II) and the percent by mass of sulfate ion in the student's unknown salt. a) 3) In this experiment, you will dissolve some of your Salt A or Salt B in 1 M H,SO, (aq). We won't know the empirical formula of your synthesized salt until you've performed the spectroscopy and tiration experiments, so for the purposes of this question, assume the empirical formula of the complex salt is: [Ni(en).(H,O) SO, 8H,O. a) Build a handheld molecular model of the complex cation in this salt and bring it to lab. If you run out of atoms of a given color when building your model, be prepared to describe what substitutions were made to complete your structure. b) What is the purpose of dissolving the salt in acid? Write a balanced chemical equation to illustrate the reaction that takes place between the salt and the acid when they combine c) d) Students observed in lab that Salt A was blue, and Salt B was lavender. (Recall that the color of the salt is impacted by which ligands are attached to the nickel (II) ion.) What color do you expect the salt dissolved in acid to be? Why?

Explanation / Answer

2a) The regression equation is used to determine the concentration of Ni(II) in the unknown salt. Plug in y = 0.322 and obtain x, the concentration of Ni(II) in the unknown.

0.322 = 16.943x + 0.0077

====> 16.943x = 0.3143

====> x = 0.3143/16.943 = 0.01855

The concentration of Ni(II) in the unknown salt is given as 0.01855 M.

Determine the number of mole(s) of Ni(II) in the unknown salt as

Mole(s) of Ni(II) = (0.01855 M)*(1 mol.L-1/1 M)*(2.5 mL)*(1 L/1000 mL) = 4.6375*10-5.

Atomic mass of Ni(II) = 58.6934 g/mol.

Mass of Ni(II) in the unknown salt = (4.6375*10-5 mole)*(58.6934 g/mol) = 2.7219*10-3 g.

Percent Ni(II) in the unknown salt = (mass of Ni(II))/(mass of unknown salt)*100 = (2.7219*10-3 g)/(0.2508 g)*100 = 1.08528% 1.085% (ans).

One mole Ni(II) combines with one mole SO42- to form one mole Ni(II) sulfate; therefore, mole(s) of sulfate = mole(s) of Ni(II) = 4.6375*10-5.

Molar mass of sulfate, SO42- = (1*32.065 + 4*15.9994) g/mol = 96.0626 g/mol.

Mass of sulfate in the unknown salt = (4.6375*10-5 mole)*(96.0626 g/mol) = 4.4549*10-3 g.

Percent sulfate in the unknown salt = (4.4549*10-3 g)/(0.2508 g)*100 = 1.776% (ans).

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