What quantity (in mL) of a $45\%$ acid solution of a monoprotic strong acid must be mixed with a $20\%$ solution of the same acid to produce $800\ mL$ of a $29.875\%$ acid solution ?
To determine the quantity of a 45\% acid solution needed to mix with a 20\% acid solution to get 800\ \text{mL} of a 29.875\% acid solution, we can use the concept of weighted averages or a simple algebraic equation. Here is a step-by-step solution:
Let the volume of the 45\% acid solution be x\ \text{mL}.
Then the volume of the 20\% acid solution will be (800 - x)\ \text{mL} since the total volume of the mixture should be 800\ \text{mL}.
The amount of acid in the 45\% solution is 0.45x\ \text{mL} of acid.
The amount of acid in the 20\% solution is 0.2(800 - x)\ \text{mL} of acid.
The total amount of acid in the final solution must be equal to 29.875\% of 800\ \text{mL}: 0.29875 \times 800 = 239\ \text{mL} of acid.
Setting up the equation for the total acid content gives: 0.45x + 0.2(800 - x) = 239
Solving the equation:
0.45x + 160 - 0.2x = 239
0.25x + 160 = 239
0.25x = 239 - 160
0.25x = 79
x = \frac{79}{0.25} = 316\ \text{mL}
Therefore, 316\ \text{mL} of the 45\% solution should be mixed with the 20\% solution to achieve the desired concentration. Thus, the correct answer is 316.
