Solution dilutions

Question

I don't need just the answer to this question I would like the formula on how to solve this question. I know that the final answer will be I need to add 15ml to make it a 10mg/ml titration but I do not know how to write it in a formula how to solve this question to come up with that answer.

You are preparing a Dopamine titration and need to add fluid to dilute the solution to 10mg/mL as well as label the bag with the percent concentration.
How many mL of fluid will you add? What we know; Dopamine is supplied as a vial 200mg/5mL.

Answer 1

Let $C_w$ be the concentration we want and let $C_h$ be the concentration we have. $C_h$ must be greater than $C_w$ because we cannot increase the concentration of a solution by diluting it.

We know that the amount of the substance does not change by dilution so it is a constant. What is it equal to?

The amount of the substance = $C\cdot F$ where $F$ is the amount of fluid. Since we know the amount of the substance is constant, we set the two expressions, $C_w\cdot F_w$ and $C_h\cdot F_h$, equal to each other to get $$C_w\cdot F_w=C_h\cdot F_h$$

This gives us $$F_w=\frac{C_h\cdot F_h}{C_w}$$ where $F_w$ is the total amount of fluid we should have when we are done. To get the amount of fluid to add, we subtract the amount we started with from $F_w$ to get $$F_{add}=\frac{C_h\cdot F_h}{C_w}-F_{init}$$

But the initial fluid $F_{init}$ is equal to $F_h$ so we get $$F_{add}=\frac{C_h\cdot F_h}{C_w}-F_{h}=\left (\frac{C_h}{C_w}-1\right)F_h$$

Answer 2

We assume Dopamine has negligible volume.

We want to have a $10\frac{\text{mg}}{\text{mL}}$ solution, i.e. a solution such that $$\frac{A\,\text{mg}}{B\,\text{mL}}=10\frac{\text{mg}}{\text{mL}},$$ where $A$ is the mass of Dopamine dissolved in the solution measured in $\text{mg}$ and $B$ is the volume of the solution measured in $\text{mL}$.

The mass of Dopamine is given and is not changing, hence we set $A=200$. We also know the initial volume of the solution, so if we let us say we add $x\,\text{mL}$ of fluid, then we can write $B=5+x$, that is, we add $x\,\text{mL}$ of fluid to the initial $5\,\text{mL}$. Hence we have $$\frac{200\,\text{mg}}{(5+x)\,\text{mL}}=10\frac{\text{mg}}{\text{mL}}$$ to solve for $x$. Divide both sides by $\frac{\text{mg}}{\text{mL}}$: $$\frac{200}{5+x}=10;$$ multiply both sides by $5+x$: $$200=10(5+x);$$ expand: $$200=50+10x;$$ subtract $50$ from both sides: $$150=10x;$$ swap sides: $$10x=150;$$ divide both sides by 10: $$x=15.$$

That is, we need to add $15\,\text{mL}$ of fluid: $$\therefore15\,\text{mL}.$$