Ionic Equations

Ionic EquationsThe reaction of lead nitrate and potassium iodide to form bright

yellow lead (II) iodide was dramatic.

Let’s look at another example of this type of reaction in more

detail.

The Reactants

sodium carbonate

nickel (II) sulfate

Ni +2 SO4

-2

= NiSO4

Na+1 CO3

-2

= Na2CO3

Visualize the Reactants

sodium carbonate nickel (II) sulfate

Ni +2

Na+1

CO3-2

SO4-2

Na+1

Thereare ions

everywhere!

NiSO4 (aq)Na2CO3 (aq)

Are they really Ionic?

water Na2CO3 (aq)NiSO4 (aq)

Predict the Products

NiSO4 (aq) + Na2CO3 (aq) -->

Ni +2 and CO3-2 =

NiCO3 This is insoluble by rule #5.

(s)

Na+1 and SO4

-2 = Na2SO4 This is soluble by rules #1 & 4.

(aq)

Balance the Equation

NiSO4 (aq) + Na2CO3 (aq) -->NiCO3 (s) + Na2SO4 (aq)

Its balanced as written!

Ionic & Net Equations

NiSO4 (aq) + Na2CO3 (aq) -->NiCO3 (s) + Na2SO4 (aq)

Write the ionic equation by showing all soluble compounds as separated ions.

Ni+2 + CO3-2 --> NiCO3 (s)

Identify the spectator ions.Remove the spectator ions. What’s left is the net ionic equation.

Ni+2 + SO4-2 + 2Na+1 + CO3

-2 -->

NiCO3 (s) + 2Na+1 +

SO4-2

Watch the Reaction

No, Beaker, not that reaction!

Watch the Reaction

QuickTime™ and aH.264 decompressor

are needed to see this picture.

Nature of the Products

The liquid above the solid conducts a current and must contain ions in solution.

The material at the bottom can be filtered out: it’s a solid.

Visualize the Products

For the ions calculate molarity.

NiCO3 (s)

Na+1Na+1

SO4-2

For the solid calculate mass.

THAT’S what is in the Beaker ?!

Calculate Ion Concentrations

The Na+1 ions came from the Na2CO3 (aq).

There was 100. mL of this 0.60M solution.

0.100 L x 0.60 mol. Na2CO3 x 2 Na+1

= 0.12 moles Na+1

1 liter solution

1 Na2CO3

The SO4-2 ions came from the NiSO4 (aq).

There was 300. mL of this 0.20M solution.

0.300 L x 0.20 mol. NiSO4 x 1 SO4-2 =

0.060 moles SO4-2

1 liter solution

1 NiSO4

Alternate for Calculating Ion Concentrations

The Na+1 ions came from the Na2CO3 (aq).

There was 100. mL of this 0.60M solution.

0.60 mol. Na2CO3 = X moles X = .060 moles Na2CO3

0.10 liter soln.

0.060 mol Na2CO3 x 2Na+1 = 0.12 moles Na+1The SO4

-2 ions came from the NiSO4 (aq).

There was 300. mL of this 0.20M solution.

0.20 mol. NiSO4 = X moles X = 0.060 moles NiSO4

0.300 liter soln

0.060 mol. NiSO4 x 1 SO4-2 = 0.060 moles SO4

-2

Combining 100 mL + 300 mL creates a solution of 400 mL.

For Na+1: M = 0.12 moles = 0.30M 0.400 L

For SO4-2: M = 0.060 moles =

0.15M 0.400 L

Super BeakerSolves theproblem!

CalculatePpt. Mass

The CO3 in NiCO3 came from the Na2CO3 (aq).There was 100. mL of this 0.60M solution.

0.100 L x 0.60 mol. Na2CO3 x 1 CO3 = 0.060 moles CO3

1 liter solution 1 Na2CO3

The Ni in NiCO3 came from the NiSO4 (aq).There was 300. mL of this 0.20M solution.

0.300 L x 0.20 mol. NiSO4 x 1 Ni = 0.060 moles Ni

1 liter solution 1 NiSO4

3.60 g + 3.5 g = 7.1 grams NiCO3

0.060 moles CO3 x 60.0 grams = 3.60 grams CO3

1 mole

0.060 moles Ni x 58.7 grams = 3.5 grams Ni

1 mole

Ppt. Mass: alternate

You can also solve this as a stoichiometry problem from the balanced equation:

0.100 L x 0.60 mol. Na2CO3 x 1 NiCO3 x

118.7 g NiCO3 =

1 liter solution

1 Na2CO3 1 mole NiCO37.1 grams NiCO3

NiSO4 (aq) + Na2CO3 (aq) -->NiCO3 (s) + Na2SO4 (aq)