What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution?

Options

A. 320
B. 325
C. 316
D. 330

Explanation

To solve the problem, we need to determine the volume of a 45% acid solution that must be mixed with a 20% acid solution to achieve a final volume of 800 mL of a 29.875% acid solution.

### Step-by-Step Solution:

1. **Define Variables**:
Let \( V \) be the volume (in mL) of the 45% acid solution. Therefore, the volume of the 20% acid solution will be \( 800 - V \) mL.

2. **Set Up the Equation for Acid Content**:
The total amount of acid in the final solution can be expressed as the sum of the acid from both solutions:
\[ \text{Acid from 45% solution} + \text{Acid from 20% solution} = \text{Acid in final solution} \]
This can be written mathematically as:
\[ 0.45V + 0.20(800 - V) = 0.29875 \times 800 \]

3. **Calculate the Right Side of the Equation**:
Calculate the amount of acid in the final solution:
\[ 0.29875 \times 800 = 239 \]

4. **Expand and Rearrange the Equation**:
Substitute the calculated value into the equation:
\[ 0.45V + 160 - 0.20V = 239 \]
Combine like terms:
\[ 0.25V + 160 = 239 \]

5. **Isolate \( V \)**:
Subtract 160 from both sides:
\[ 0.25V = 239 - 160 \]
\[ 0.25V = 79 \]

6. **Solve for \( V \)**:
Divide both sides by 0.25:
\[ V = \frac{79}{0.25} = 316 \text{ mL} \]

### Final Answer:
The volume of the 45% acid solution that must be mixed is **316 mL**.

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