Chemical Mixture Problems Solving Chemical Mixture Problems Algebra 1H Glencoe McGraw-Hill J. Evans/C. Logan

Mixture problems were introduced earlier this year Mixture problems were introduced earlier this year. In those problems we saw different solid ingredients like prunes and apricots, each at their own price, combined together to form a mixture at a new price. + = 1st ingredient 2nd ingredient mixture + cost · amount cost · amount = cost · amount Chemical mixture problems are another type of mixture problem. Instead of a cost for each ingredient, we’ll consider the strength of the solution, measured in percents.

+ = First equation: 40 + x = y 40 x y Ex. 1: Mrs. Armstrong has 40 mL of a solution that is 50% acid. How much water should she add to make a solution that is 10% acid? 40 x y + = 50% ACID SOLUTION PURE WATER 10% ACID SOLUTION The problem asks for the amount of water Mrs. Armstrong should add. Let x = amount of water added We also need to know how much of the 10% solution she’ll end up with. Let y = amount of new solution First equation: 40 + x = y

The first equation only addressed the amount of the liquids. 40 + x = y The equation says that Mrs. Armstrong started with 40 mL of a strong acid solution, then added x mL of water and ended up with y mL of a weaker acid solution. The second equation in the system needs to address the strength of each solution (percentage of acid).

Why is the percentage on the water 0%? x y 40 + = 10% ACID SOLUTION 50% ACID SOLUTION WATER + = % · amount % · amount % · amount .50 · 40 + 0 · x = .10 · y 20 + 0 = .1y 20 = .1y Why is the percentage on the water 0%?

Mrs. Armstrong needs to add 160 mL of water. Solve the system using any method. .50(40) + 0x = .10y 20 + 0 = .1y 20 = .1y 200 = y Mrs. Armstrong needs to add 160 mL of water.

- = First equation: 50 - x = y Ex. 2: How many liters of water must Mr. Wade EVAPORATE from 50 L of a 10% salt solution to produce a 20% salt solution? 50 x y - = 10% SALT SOLUTION PURE WATER 20% SALT SOLUTION Let x = amount of water evaporated Let y = amount of new solution First equation: 50 - x = y

Why is the percentage on the water 0%? 50 x y - = 10% SALT SOLUTION Pure water 20% SALT SOLUTION - = % · amount % · amount % · amount .10 · 50 - 0 · x = .20 · y Why is the percentage on the water 0%?

25 L of water must be evaporated. Solve the system using the substitution method. 25 L of water must be evaporated.

Half-and-Half with 11% butterfat Ex. 3: Milk with 3% butterfat was mixed with cream with 27% butterfat to produce 36 L of Half-and-Half with 11% butterfat content. How much of each was used? x y 36 Milk with 3% butterfat + Cream with 27% butterfat = Half-and-Half with 11% butterfat Let x = amount of milk added Let y = amount of cream added First equation: x + y = 36 This equation says we started with x liters of milk and are adding y liters of cream to produce 36 liters of Half-and-Half.

Half-and-Half with 11% butterfat x y 36 Milk with 3% butterfat Cream with 27% butterfat Half-and-Half with 11% butterfat + = + = % · amount % · amount % · amount .03·x + .27·y = .11·36 Solve the system: 24 L of milk and 12 L of cream are needed.

Ex. 4: A chemistry experiment calls for a 30% solution of copper sulfate. Mrs. Maiorca has 40 milliliters of 25% solution. How many milliliters of 60% solution should she add to make a 30% solution? 40 x y + = 25% solution 60% solution 30% solution + = % · amount % · amount % · amount Let x = amount 60% solution Let y = amount of 30% solution 40 + x = y .25(40) + .60(x) = .30y

Mrs. Maiorca needs to add 6.67 mL of the 60% solution. 40 + x = y .25(40) + .60(x) = .30y 10 + .60x = .30(40 + x) 10 + .60x = 12 + .30x .60x = 2 + .30x .30x = 2 x ≈ 6.67 Mrs. Maiorca needs to add 6.67 mL of the 60% solution.