100mL of a solution contains 30 mL of acid; this is called a 30% acid solution because
part 30
---- = ---- = 0.30.
whole 100
How many milliliters of acid x would need to be added to the solution to turn it into a 50% solution ? (HInt: Adding acid increases both the volume of acid and the volume of the complete solution.)
(30+V)/(100+V)=.5
30+V=50+.5V
solve for V
This is a really bad problem, chemically speaking, because "acid" is already a solution. Very rarely are acids in pure form.
To solve this problem, we can set up an equation using the concept of concentration.
Let's assume that x represents the amount (in mL) of acid that needs to be added to the solution.
Originally, we have 30 mL of acid in a total volume of 100 mL, which gives us a concentration of 30%.
After adding x mL of acid, the total volume of the solution will increase to 100 + x mL, and the amount of acid will increase to 30 + x mL.
The concentration of the final solution is given as 50%. We can use the equation:
(30 + x)/(100 + x) = 0.50
Now, let's solve for x:
First, let's cross multiply the equation:
(30 + x) * 1 = 0.50 * (100 + x)
Simplifying, we have:
30 + x = 0.50 * (100 + x)
Expanding the right side of the equation, we get:
30 + x = 50 + 0.50x
Now, let's simplify and isolate x:
x - 0.50x = 50 - 30
0.50x = 20
To solve for x, we divide both sides of the equation by 0.50:
x = 20 / 0.50
x = 40 mL
Therefore, we would need to add 40 mL of acid to the solution to turn it into a 50% solution.