During the second week, suppose you decide to use volumetric analysis. For each of the unknown compounds on the left, choose the proper titrant (standard solution) and indicator from the drop-down box. There are three indicators for you to choose from: bromothymol blue, phenolphthalein, and methyl orange.

Hints:

HCl is a strong acid;

NaOH is a strong base;

CH3COONa, Na2C2O4 are basic salts that form weak bases in aqueous solutions;

NH4Cl is an acidic salt that forms weak acid in an aqueous solution.

CH3COONa

[ Choose ] HCl, methyl orange NaOH, phenolphthalein HCl, phenolphthalein, and then methyl orange HCl, bromothymol blue

Na2CO3

[ Choose ] HCl, methyl orange NaOH, phenolphthalein HCl, phenolphthalein, and then methyl orange HCl, bromothymol blue

Na2C2O4

[ Choose ] HCl, methyl orange NaOH, phenolphthalein HCl, phenolphthalein, and then methyl orange HCl, bromothymol blue

NH4Cl

[ Choose ] HCl, methyl orange NaOH, phenolphthalein HCl, phenolphthalein, and then methyl orange HCl, bromothymol blue

NaOH

[ Choose ] HCl, methyl orange NaOH, phenolphthalein HCl, phenolphthalein, and then methyl orange HCl, bromothymol blue

Answers provided for the reactions with explanations in parenthesis

Explanation:

We need to find the approx. equivalent pH so we can choose an indicator.

CH3COONa +strong acid HCl - methyl orange (pH will be about 4.5)

Na2CO3 base ---> strong acid HCl- phenolphthalein (pH will be 7-8) , and then methyl orange (since carbonic acid has 2 equivalence points, it needs two indicators)

Na2C2O4   HCl- methyl orange, and then methyl orange (since oxalic acid has 2 equivalence points, it needs two indicators)

NH4Cl NaOH- Phenolphthalein, since pH = 9.75 approximately

NaOH HCl-Bromothymol blue --> since pH = 7 best for a neutral indicator

the answer is minerals.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o