Question

The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive.

Answer 1

Answer:

The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive.

Explanation:

The given substances are:

barium chloride(BaCl2),

glucose(C6H12O6),

butanoic acid (C3H7COOH) which is a weak acid,

sodium bromide (NaBr).

The conductivity of a solution is proportional to the number of ions present in a particular solution.

1mol. of BaCl2 in water produces a total three mol. of ions.

$BaCl_2 (aq) -> Ba^2^+(aq) + 2Cl^-(aq)$

Gluocse is a covalent compound and it does not dissociate into ions in water.

So, it does not conduct electricity.

Butanoic acid is a weak acid. But due to the release of H+ ions it can conduct a very less amount of electricity.

NaBr is an ionic compound and in 1mol. of NaBr in water gives two mol. of ions.

NaBr (aq)  -> Na+ (aq)  + Br- (aq)

Hence, the order of conductivity among the given substances in aqueous solution is:

BaCl2 > NaBr > butanoic acid > glucose