Saline Solution (Refer to Example 1.) A solution contains 3% salt. How much water should b...

Saline Solution (Refer to Example 1.) A solution contains 3% salt. How much water should be added to 20 ounces of this solution to make a 1.2% solution?

EXAMPLE 1 Diluting a saline solution

A solution contains 4% salt. How much pure water should be added to 30 ounces of the solution to dilute it to a 1.5% solution?

Solution

STEP 1: Assign a variable x as follows.

 x: ounces of pure water (0% salt solution)

 30: ounces of 4% salt solution

 x + 30: ounces of 1.5% salt solution

In Figure 1 three beakers illustrate this situation.

Figure 1 Mixing a Saline Solution

 

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STEP 2: Note that the amount of salt in the first two beakers must equal the amount of salt in the third beaker. We use Table 1 to organize our calculations. The amount of salt in a solution equals the concentration times the solution amount, as shown in the last column of the table.

Table 1 Mixing a Saline Solution

Solution Type	Concentration (as a decimal)	Solution Amount (ounces)	Salt (ounces)
Pure Water	0% = 0.00	x	0.00x
Initial Solution	4% = 0.04	30	0.04(30)
Diluted Solution	1.5% = 0.015	x + 30	0.015(x + 30)

The amount of salt in the first two beakers is

 0.00x + 0.04(30) = 0 + 1.2 = 1.2 ounces.

The amount of salt in the final beaker is

 0.015(x + 30) ounces.

Because the amounts of salt in the solutions before and after mixing must be equal, the following equation must hold.

 0.015(x + 30) = 1.2

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STEP 3: Solve the equation in Step 2.

 

Fifty ounces of water should be added.

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STEP 4: Adding 50 ounces of water will yield 50 + 30 = 80 ounces of water containing 0.04(30) = 1.2 ounces of salt. The concentration is or 1.5%, so the answer checks.