Please help with both questions. The solutions in the first question are in the

Question

Please help with both questions. The solutions in the first question are in the second picture.

s th Consider a solution containing bromide ions (Br). Which of the fourteen solutions (isted on pg. 7-23 across the top of the table would cause a precipitate to form when mixed with this bromide solution? e HW3. Explain which of the following ions you would expect to be more water soluble or methyl mercury (CH3Hg). More lipid soluble? Hint: Lipids are the large organic molecules which make up cell membranes. Based on your assessment of their solubilities, which of these would you expect to be more toxic? (Reasoning is more important here than conclusions.) HW4.

Explanation / Answer

HW3 - According to solubility rules, all bromides are soluble except those of silver, mercury and lead. For example- AgBr, Hg2Br2 and PbBr2.

So, among the fourteen solutions Pb(NO3)2 and AgNO3 solutions will form a precipitate when mixed with bromide solution.

HW4 - The Hg2+ ions mixes easily with water. Due to its ionic nature, it is closer in nature to polar substances and is thus soluble in polar solvent that is water and not lipids.

Methyl mercury(CH3Hg+) is much more lipid soluble. Due to the presence of methyl group, polarity and effective nuclear charge is reduced. Therefore, CH3Hg+ will be less water-soluble.

CH3Hg+ is lipophilic and can therefore readily pass through biological membranes. Also it forms relatively stable complexes and have a tendency to accumulate, making it more toxic.

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