What quantity of 75 per cent acid solution must be mixed with a 30 per cent solution to produce 945 mL of a 50 per cent solution?
What quantity of 75 per cent acid solution must be mixed with a 30 per cent solution to produce 945 mL of a 50 per cent solution?
Let x and y, be the quantities in milliliters, for the two acid solutions.
Mixing them, the percent acidity of the mixed acid
(75x + 30y) / (x + y) = 50.
We can simplify this
75x + 30y = 50x + 50y
25x - 20y = 0.
5x - 4y = 0
Also, the combined amount of the solution is x+y = 945.
So we have a system of two linear equations 5x - 4y = 0 and x+y = 945.
Adding four times the second equation to the first gives 9x = 3780.
Thus x = 3780/9 = 420, and y = 945 - x = 945 - 420 = 525.
So you need 420 ml of the 75% solution and 525 ml of the 30% solution.
Method I. Standard way:
x = the amount of 75% acid solution
Thus, 945-x = the amount of 30% acid solution
Balance: .75x + .3(945-x) = .5*945
Solve for x,
x = 420 mL
Answer: 420 mL of 75% acid solution is required.
Method II. Compare to the final equilibrium.
75-50 = 25
50-30 = 20
By inverse proportion, the amount of 75% acid solution = (20/(20+25))(945) = 420 mL
What quantity of 75 per cent acid solution must be mixed with a 30 per cent solution to produce 945 mL of a 50 per cent solution?
Let x = amount (mL) of 75% solution
then
945-x = amount (mL) of 30% solution
.75x + .30(945-x) = .50(945)
.75x + 283.5-.30x = 472.5
.45x + 283.5 = 472.5
.45x = 189
x = 420 mL (of 75% solution)
30% solution:
945-x = 945-420 = 525 mL (of 30% solution)