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During a chemistry class, David was asked to add pure acid to 6 liters of 45% acid solution, to get a solution which is 80% acid. How much water should he add?

Respuesta :

ANSWER:

David should add 10.5 liter acid solution to get a solution which is 80% acid

SOLUTION:

First, Set up a table. Fill in the unknown with variable x. The table is attached below

From this, we can easily set up the two equations.  

Sum of values of two acids = Value of mixture  

x+2.7=0.8(x + 6)

For convenience, we'll multiply the entire equation by 10,

10x + 27  = 8(x + 6)  

10x + 27 = 8x + 48

10x – 8x = 48 – 27

2x = 21

x = [tex]\frac{21}{2}[/tex]

x = 10.5 L

We can conclude that, 10.5 liters of pure acid should be added to 6 liters of 45% acid to get 80% acid.

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The amount of water that should be added to obtain a solution with 80% acid is 10.5 litres.

Let the amount water added = w

Total percentage of solution = 80%

Percentage of acid solution = 45%

Amount of acid = 6 Litres

Representing the information as a system of equation :

[(Amount of acid × percentage of acid) + amount of water] = total percentage of solution×(amount of acid + amount water)

[(6 × 0.45) + w] = 0.80(w + 6)

2.7 + w = 0.80w + 4.8

Collect like terms

w - 0.8w = 4.8 - 2.7

0.2w = 2.1

w = (2.1 ÷ 0.2)

w = 10.5

Therefore, the amount of water that should be added to the mixture is 10.5 litres

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