Solution Equations - Calculations with
Ions
Dr. MJ Patterson
When we write an equation showing an ionic compound dissolving in water, we try to represent each ion as it actually exists in the solution. These examples will take you through this process and will also extend using the coefficients as conversion factors.
Example 1:
Write and balance the equation to show solid magnesium iodide dissolving in
water.
Solution 1:
First, we need the formula for magnesium iodide. Magnesium ion is Mg2+,
and iodide ion is I-. We need to cross the charges to use as subscripts
to write an electrically neutral compound: MgI2.
Next, we can start to write the equation with solid MgI2 on the left.
MgI2(s) =>(H2O)
On the right, we need to show the ions as they actually exist. Magnesium will be Mg2+(aq) and iodide will be I-(aq).
MgI2(s) =>(H2O)
Mg2+(aq) + I-(aq)
However, this equation is not balanced. Magnesium is fine, but there are two iodides on the left, and only one on the right. We can fix this by putting a coefficient of 2 in front of iodide on the right.
MgI2(s) =>(H2O) Mg2+(aq)
+ 2I-(aq)
Another way to think about this process is to use the subscript from each ion in the solid on the left as a coefficient on the right.
Example 2:
Write and balance the equation to show solid magnesium nitrate dissolving in
water.
Solution 2:
This time, magnesium nitrate is Mg(NO3)2.
Magnesium has a charge of +2, and nitrate has a charge of -1. The
understood subscript of 1 on magnesium means that a single magnesium ion is
produced upon dissolution, and the explicit subscript of 2 on the nitrate means
that 2 nitrate ions will be produced. Notice that the polyatomic ion
nitrate stays as a unit - it does not break down into its component elements.
Mg(NO3)2(s) =>(H2O) Mg2+(aq) + 2NO3-(aq)
Example 3:
Write and balance the equation to show the following solids dissolving in
water.
a. sodium sulfate
b. aluminum chloride
c. calcium phosphate
Solution 3:
a. Na2SO4(s)
=>(H2O) 2Na+(aq)
+ SO42-(aq)
b. AlCl3(s) =>(H2O) Al3+(aq)
+ 3Cl-(aq)
c. Ca3(PO4)2(s)
=>(H2O) 3Ca2+(aq)
+ 2PO42-(aq)
As we have seen before, we can use the coefficients from these balanced equations to write conversion factors between reactants and products. For the calcium phosphate example above,
1 mole Ca3(PO4)2(s) = 3 moles Ca2+(aq) = 2 moles PO42-(aq)
These conversion factors are simply the coefficients from the equation for each species.
Sometimes when working with ionic compounds, you will want to know the total number of ions produced. Since 3 calcium ions and 2 phosphate ions are produced, a total of 5 ions are produced. We can tack on this conversion as well:
1 mole Ca3(PO4)2(s) = 3 moles Ca2+(aq) = 2 moles PO42-(aq) = 5 moles total ions
Example 4:
How many aluminum and chloride ions can be produced from 1.25 moles of aluminum
chloride? How many ions total are produced?
Solution 4:
From the balanced equation in Example 3, we can write the following conversion
factors:
1 mole AlCl3 = 1 mole Al3+(aq) = 3 moles Cl-(aq) = 4 moles total ions
(The 4 moles of total ions comes from 1 mole Al3+(aq) + 3 moles Cl-(aq) = 4 moles total.)
Remember the conversion plan: start by writing down the quantity you
need to convert. Multiply by a conversion factor set up so that the units
in the bottom will cancel with the original units.
|
(1.25 moles AlCl3) |
(1 mole Al3+(aq)) |
= 1.25 moles Al3+(aq) |
|
(1) |
(1 mole AlCl3) |
|
|
(1.25 moles AlCl3) |
(3 moles Cl-(aq)) |
= 3.75 moles Cl-(aq) |
|
(1) |
(1 mole AlCl3) |
|
|
(1.25 moles AlCl3) |
(4 moles total ions) |
= 5.00 moles total ions |
|
(1) |
(1 mole AlCl3) |
|