To solve this problem, we need to find out how much water should be added to the original solution to reduce the concentration of acid from 40% to 25%.

Step-by-Step Solution

1. Define the Variables:

   • Let $V$ be the volume of water to be added (in mL).
   • The initial volume of the solution is 200 mL, which is 40% acid.

2. Calculate the Amount of Acid in the Original Solution:

   Since the original solution is 40% acid: $\text{Amount of acid} = 200 \times \frac{40}{100} = 80 \, \text{mL}$

3. Set Up the Equation for the New Solution:

   After adding $V$ mL of water, the total volume of the new solution becomes $200 + V$ mL. We want the new solution to be 25% acid.

   The new concentration equation can be set up as follows: $\frac{\text{Amount of acid}}{\text{Total volume of solution}} = 25\%$

   Substituting the known values: $\frac{80}{200 + V} = \frac{25}{100}$

4. Solve for $V$:

   First, simplify the fraction: $\frac{80}{200 + V} = 0.25$

   Now, multiply both sides by $200 + V$ to eliminate the fraction: $80 = 0.25 \times (200 + V)$

   Expand the right-hand side: $80 = 50 + 0.25V$

   Subtract 50 from both sides: $30 = 0.25V$

   Divide both sides by 0.25 to solve for $V$: $V = \frac{30}{0.25} = 120$

Conclusion

The scientist should add 120 mL of water to the original solution to make a solution that is 25% acid.

Would you like more details or have any questions?

Related Questions:

1. How much acid is in the new 25% solution after adding water?
2. What would the concentration be if only 50 mL of water were added instead of 120 mL?
3. How much water should be added to dilute the original solution to 10% acid?
4. If the initial solution was 60% acid, how much water would be needed to reduce it to 25%?
5. How would you calculate the final concentration if a different substance was added instead of water?

Tip:

When diluting a solution, adding water changes the concentration but not the total amount of solute. The volume change is crucial in determining the new concentration.