2.1. The lithium ion in a 250,00 mL sample of mineral water was precipitated with sodium tetraphenylboron:

Lit + B(C6H4)4 — LiB(CH3)4(s)

The precipitate was filtered, washed and re-dissolved in an organic

solvent. An excess of mercury (II) EDTA chelate was added:

4HgY2- + B(C6H4)4 + 4H20 =H3BO3 + 4C6H5Hg+ + 4HY3- + OH

The liberated EDTA was titrated with 29,64 mL of 0,05581 M Mg2+.

Calculate the lithium ion concentration in ppt.

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

Moles Mg²⁺:

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

Moles B(C₆H₄)₄ = Moles Lithium:

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

your answer would be “dry”

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.