During the second week, suppose you decide to use volumetric analysis. For each
ID: 483421 • Letter: D
Question
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
Explanation / Answer
CH3COONa will hydrolyse to form CH3COO- ions which then
CH3COO- + H2O <-> CH3COOH + OH-
pKb = 4.75 approx so we need --> this will be basic i.e. 4.75
Choose --> HCl, methyl orange
Na2CO3
will form:
2Na+ + CO3-2
and
CO3-2 + H2O <--> HCO3- + OH- which is pretty basic
so, we need HCl in order to titrate it
pH will be pretty low so choose methyl organe + HCL
Na2C2O4
C2O4-2 + H2O <-> HC2O4- + OH-
this is not that strong, wo we could use HCl, bromothymol blue
NH4Cl --> NH4+ + H2O --> NH# + H3O
this is acidic,so pH = 4.75, use methyl orange + NAOH
NaOH--> strong base so we use HCl strong acid
pH = near 7 so choose bromothymol blue
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