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Stoichiometry of Precipitation Reactions and Ion Remaining Ion Concentration

1. Determine what reaction takes place

2. Write the balanced equation for the reaction

3. Calculate the moles (or mmol) of the reactants (use V x M)

4. Determine which reactant is limiting (I use the ICE Box)

5. Calculate the moles of product(s).

6. Convert to grams or other units, as required.

Concentration of the Ions remaining

7. Find the moles of each of the ions

8. Combine the volumes used to determine the total volume.

9. Find the Molarity (moles of solute/Liters of solution) of each ion

Example

 

100.mL of 0.100M  potassium sulfate solution is added to a100.mL solution of 0.200M barium nitrate. Calculate the mass of the precipitate formed and the concentration of remaining ions in the solution.

1. Determine what reaction takes place

potassium sulfate + barium nitrate==> potassium nitrate + barium sulfate (s)

K2SO4 + Ba(NO3)2 ---> KNO3 + BaSO4(s)

2. Write the balanced equation for the reaction

K2SO4 + Ba(NO3)2 ---> 2KNO3 + BaSO4(s)

3. Calculate the moles (or mmol) of the reactants (use V x M)

K2SO4   100.mL x 0.100M= 10.0mmol       or    0.100L x0.100M= 0.0100moles

isn't mmols a nicer #?

Ba(NO3)2 100.mL x 0.200M= 20.0mmol       or    0.100L x0.200M= 0.0200moles

4. Determine which reactant is limiting
 

 K2SO4  + Ba(NO3)2==>  2KNO3 + + BaSO4(s)
I-Initial 0.0100moles0.0200moles--------------------------------
C-Change-X-X+2X+X
E-End

0

0

  

Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction.

if K2SO4  runs out ==>   0.0100moles -X =O ; X is therefore 0.0100moles

if Ba(NO3)2runs out  ==> 0.0200moles-x=O ; X is therefore 0.0200moles

Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX.

K2SO4 is therefore limiting...X= 0.0100moles

5. Calculate the moles of product(s). Apply the value for X back in the ICE box and determine the moles (mmols) of precipitate.

 K2SO4  + Ba(NO3)2==>  2KNO3 + + BaSO4(s)
I-Initial 0.0100 moles 0.0200 moles --------------------------------
C-Change-0.0100-0.0100+2(0.0100)+0.0100
E-End

0

0.0100 mol

0.0200 mol0.0100 mol

6. Convert the precipitate to grams or other units, as required. 

0.0100 mol BaSO4(s) x 233.44 g BaSO4=2.33g BaSO4
1 mol BaSO4

7. Determine the number of moles of each ion left in the solution  (click this link for further details)

0.0100mol Ba(NO3)2

Ba(NO3)2 ==>  Ba2+ + 2NO3-

Therefore 0.0100mol of Ba(NO3)2 yields 0.0100mol of Ba2+ and 0.0200 mol of NO3- (2 x 0.0100 mol)

0.0200mol KNO3

KNO3 ==> K+ + NO3-

Therefore 0.0200mol of KNO3 yields 0.0200mol of K+ and 0.0200 mol of NO3- .

Clean that up

Ba2+NO3-K+NO3-
0.0100mol0.0200mol0.0200mol0.0200mol

Combine the nitrates

Ba2+NO3-K+
0.0100mol0.0400mol0.0200mol

8. Determine the total volume by adding the volumes used. Convert to Liters.

Vol. Total= 100.mL  + 100.mL= 200.mL

Covert to Liters   200.mL x 1l/1000mL= 0.200L

9. Determine the concentration (Molarity) of each ion left in the solution

[Ba2+] =0.0100mol / 0.200L= 0.0500M

[NO3-]=0.0400mol / 0.200L =0.200M

[K+]=0.0200mol / 0.200L= 0.100M

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