I'm completely stumped. I'm trying to figure out a formula for the following...
How much of a solution that is 400 PPM per gallon needs to be added to X amount of gallons of a 0 PPM solution in order for X amount of gallons to be 2 PPM per gallon?
OR
If container A has X amount of gallons of a 0 PPM solution. And container B has a solution that is 400 PPM per gallon. How much of the solution from container B has to be added to container A to make X amount of gallons a 2 PPM per gallon solution?
I only know how to work this backward and find out how many 0 PPM gallons it would take to dilute one gallon of a 400 PPM solution to a 2 PPM per gallon solution. Which would be if I add 199 gallons that are 0 PPM to one gallon that is 400 PPM then I would have a total of 200 gallons that were 2 PPM per gallon. I'm lost though on finding a formula to basically do this the opposite way as I described at the top.
Solution A: $400$ ppm/gallon
Solution B: $2$ ppm/gallon
How much solvent ($0$ ppm/gallon) is required? Say $X$
$\frac{1 \text{gallon} \times 400 \text{ppm/gallon}}{(1 + X) \text{gallons}} = 2 \text{ppm/gallon}$
Solve for $X$.
[Answer: $199 \text{gallons}$]
$\frac{V_o\times C_o}{V_o + V_d} = C_e$ (formula)
$V_o$ = volume of original solution taken
$C_o$ = concentration of original solution
$V_o \times C_o$ = amount of solute in $V_o$ units of the original solution
$V_d$ = volume of diluent (pure solvent)
$V_o + V_d$ = total volume when you mix $V_o$ of original solution (which contains $V_o \times C_o$ of solute) with $V_d$ of solvent/diluent.
The end concentration (post-mixing) = $C_e = \frac{V_o \times C_o}{V_o + V_d}$