Ionic Equations The 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
Ionic Equations
Jan 12, 2016
Ionic Equations. The 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. Ni +2 SO 4 -2. Na +1 CO 3 -2. The Reactants. sodium carbonate. - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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)
Related Documents
