A science teacher has a supply of 5% hydrochloric acid and a supply of 65% hydrochloric acid(HC1). How much of each solution sho
uld the teacher mix together to get 42 mL of 45% HC1 for an experiment? a. The teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution. b. The teacher will need 28 mL of the 5g solution and 14 mL of the 65% solution. c. The teacher will need 14 of mL of both solutions. d. The teacher will need 28 mL of both solutions.
Assume that the amount needed from the 5% solution is x and that the amount needed from the 65% solution is y.
We are given that, the final solution should be 42 ml, this means that: x + y = 42 ...........> equation I This can also be written as: x = 42-y .......> equation II
We are also given that the final concentration should be 45%, this means that: 5% x + 65% y = 45% (x+y) 0.05x + 0.65y = 0.45(x+y)
We have x+y = 42 from equation I, therefore: 0.05x + 0.65y = 0.45(42) 0.05x + 0.65y = 18.9 .........> equation III
Substitute with equation II in equation III as follows: 0.05x + 0.65y = 18.9 0.05(42-y) + 0.65y = 18.9 2.1 - 0.05y + 0.65y = 18.9 0.6y = 18.9 - 2.1 0.6y = 16.8 y = 28 ml
Substitute with y in equation II to get x as follows: x = 42-y x = 42 - 28 x = 14 ml
Based on the above calculations: amount of 5% solution = x = 14 ml amount of 65% solution = y = 28 ml The correct choice is: The teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.
