How should I go about writing a procedure for this Chem lab? (see Details)

I am given 25mL of an aqueous solution containing the ionic compounds #HNO_3#, #Sr(NO_3)_2#, and #AgNO_3#.

I must discover the concentrations of each of the ionic compounds in the solution above using four known solutions of #KCl#(0.15M), #Na_3PO_4#(0.15M), #K_2SO_4#(0.15M), and #NaOH#(0.25M).

I don't have to use all of the solutions of I do not need them. Also, my partner is as clueless as I am, so there's not much help from that direction.

1 Answer
Truong-Son N.
May 1, 2018

Given #"25 mL"# of a solution containing #"HNO"_3#, #"Sr"("NO"_3)_2#, and #"AgNO"_3# (all water-soluble), we have #"0.15 M KCl"#, #"0.15 M Na"_3"PO"_4#, #"0.15 M K"_2"SO"_4#, and #"0.25 M NaOH"# available.

The idea is to choose reagents that react specifically with what you want to suss out. You'll want to do this in a sequence, accounting for dilutions.

This is what I would suggest as a procedure:

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[Click on the image to zoom in.]

To summarize:

  1. Measure pH to get #["HNO"_3]# right away.

  2. React the starting solution with excess #"KCl"# (I chose #"10.00 mL"# because it's the volume of a #"10.00-mL"# volumetric pipette and it's probably excess of what is needed to precipitate #"AgCl"(s)#). Relevant reagent = #"AgNO"_3#.

  3. Separate the precipitate from the supernatant, and put the supernatant to the side for step 5. Weigh a piece of filter paper.

  4. Filter the #"AgCl"# precipitate from step 2 using that filter paper, dry it, weigh it, and get the mass of filter paper+#"AgCl"#. The #["Ag"^(+)]# links #"AgCl"# to #["AgNO"_3]#, so we don't worry about excess #"Cl"^(-)# we added in.

  5. React the supernatant from step 3 (relevant reagent = #"SrCl"_2#) with excess #"Na"_3"PO"_4# (again, convenient volume) and form #"Sr"_3("PO"_4)_2(s)# precipitate.

  6. Separate the precipitate from the supernatant, and discard the supernatant! Weigh a new piece of filter paper.

  7. Filter the #"Sr"_3("PO"_4)_2# precipitate from step 5, dry it, weigh it, and get the mass of filter paper+#"Sr"_3("PO"_4)_2#. The #["Sr"^(2+)]# links #"Sr"_3("PO"_4)_2(s)# to #["SrCl"_2]# and thus #["Sr"("NO"_3)_2]#, so we don't worry about excess #"PO"_4^(3-)# we added in.

Below are all my brainstorming thoughts.

Here's my brainstorm (all given molar solubilities are in pure water at #20^@ "C"#), using #K_(sp)# values from here and here:

  • #"NaOH"# is NOT needed.

Simply measure the pH of the starting solution to find #["H"^(+)]# and thus #["HNO"_3]#.

Good method to get #["H"^(+)]#? #color(blue)(sqrt"")#

  • #"K"_2"SO"_4# will react with #"HNO"_3# to form #"HSO"_4^(-)#, with #"Sr"("NO"_3)_3# to form #"SrSO"_4(s)# (#K_(sp) = 3.44 xx 10^(-7)#), and with #"AgNO"_3# to form #"Ag"_2"SO"_4(s)# (#K_(sp) = 1.20 xx 10^(-5)#).

Both precipitates are white, unfortunately, but their corresponding molar solubilities are #5.87 xx 10^(-4) "M"# and #"0.0144 M"#, respectively.

Good method to get #["Sr"^(2+)]#? (Possibly, depends on previous steps)

  • #"Na"_3"PO"_4# would precipitate heavily to form #"Sr"_3("PO"_4)_2# (#K_(sp) = 1.0 xx 10^(-31)#) and #"Ag"_3"PO"_4# (#K_(sp) = 8.89 xx 10^(-17)#), with corresponding molar solubilities of #2.47 xx 10^(-7) "M"# and #4.26 xx 10^(-5) "M"#, respectively.

Good method to get #["Sr"^(2+)]#? (Possibly, depends on previous steps)

  • #"KCl"# would react selectively with #"AgNO"_3# to form the white #"AgCl"(s)# with no interferences, so this is a great choice to get #["Ag"^(+)]#.

#"SrCl"_2# is very soluble, no problem there, and solid #"KCl"# reacts with concentrated #"HNO"_3# but not when they're both aqueous.

Good method to get #["Ag"^(+)]#? #color(blue)(sqrt"")#

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