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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: MgI_{2}.
Next, we can start to write the equation with solid MgI_{2} on the left.
MgI_{2(s)} =>(H_{2}O)
On the right, we need to show the ions as they actually exist. Magnesium will be Mg^{2+(}_{aq}_{)} and iodide will be I^{}_{(aq)}.
MgI_{2(s) } =>(H_{2}O) Mg^{2+}_{(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.
MgI_{2}_{(s) } =>(H_{2}O) Mg^{2+}_{(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(NO_{3})_{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(NO_{3})_{2(s)} =>(H_{2}O) Mg^{2+}_{(aq) } + 2NO_{3}^{}_{(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. Na_{2}SO_{4(s)} =>(H_{2}O) 2Na^{+}_{(aq)} + SO_{4}^{2}_{(aq)}
 AlCl_{3}_{(s)} =>(H_{2}O) Al^{3+}_{(aq)} + 3Cl^{}_{(aq)}
 Ca_{3}(PO4)_{2}_{(s)} =>(H_{2}O) 3Ca^{2+}_{(aq)} + 2PO_{4}^{2}_{(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 Ca_{3}(PO4)_{2(s)} = 3 moles Ca^{2+}_{(aq)} = 2 moles PO_{4}^{2}_{(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 Ca_{3}(PO4)_{2(s)} = 3 moles Ca^{2+}_{(aq)} = 2 moles PO_{4}^{2}_{(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 AlCl_{3} = 1 mole Al^{3+(}_{aq}_{)} = 3 moles Cl^{}_{(aq)} = 4 moles total ions
(The 4 moles of total ions comes from 1 mole Al^{3+(}_{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 AlCl_{3})  (1 mole Al^{3+}_{(aq)})  = 1.25 moles Al^{3+}_{(aq)} 
(1)  (1 mole AlCl_{3}) 

(1.25 moles AlCl_{3})  (3 moles Cl^{}_{(aq)})  = 3.75 moles Cl^{}_{(aq)} 
(1)  (1 mole AlCl_{3}) 

(1.25 moles AlCl_{3})  (4 moles total ions)  = 5.00 moles total ions 
(1)  (1 mole AlCl_{3}) 
