Question

How many ml of 10% can you prepare by mixing 2% of stock solution with 160ml of 20% stock solution?

Answer 1

To find out how many milliliters of a 10% solution can be prepared by mixing a 2% stock solution with 160 mL of a 20% stock solution, we can use the concept of mass balance.

Let's denote:
- \( V_1 \) = volume of the 2% solution (in mL)
- \( V_2 \) = volume of the 20% solution, given as 160 mL
- \( C_1 \) = concentration of the 2% solution = 0.02 (in decimal form)
- \( C_2 \) = concentration of the 20% solution = 0.20 (in decimal form)
- \( C_f \) = concentration of the final mixture = 0.10 (in decimal form)

The total mass of the solute in the mixture must equal the sum of the masses of the solute in each solution.

The mass of solute in the 2% solution is:
\[
\text{Mass from 2% solution} = C_1 \cdot V_1 = 0.02 \cdot V_1
\]

The mass of solute in the 20% solution is:
\[
\text{Mass from 20% solution} = C_2 \cdot V_2 = 0.20 \cdot 160 = 32 \, \text{g}
\]

The total mass of solute in the final mixture (which is a 10% solution) is:
\[
\text{Total volume} = V_1 + V_2 = V_1 + 160 \, \text{mL}
\]
\[
\text{Total mass of solute in 10% solution} = C_f \cdot \text{Total volume} = 0.10 \cdot (V_1 + 160)
\]

Now we need to set up the mass balance equation:
\[
0.02 \cdot V_1 + 32 = 0.10 \cdot (V_1 + 160)
\]

Expanding the equation gives:
\[
0.02 \cdot V_1 + 32 = 0.10 \cdot V_1 + 16
\]

Now we can rearrange the equation to isolate \( V_1 \):
\[
32 - 16 = 0.10 \cdot V_1 - 0.02 \cdot V_1
\]
\[
16 = 0.08 \cdot V_1
\]
\[
V_1 = \frac{16}{0.08} = 200 \, \text{mL}
\]

Now we can find the total volume of the prepared 10% solution:
\[
\text{Total volume} = V_1 + V_2 = 200 \, \text{mL} + 160 \, \text{mL} = 360 \, \text{mL}
\]

Therefore, you can prepare **360 mL of a 10% solution** by mixing 200 mL of the 2% stock solution with 160 mL of the 20% stock solution.