Example Question - acid solution

Calculating the Required Volume of a Concentrated Acid Solution to Achieve a Desired Concentration

<p>Let \( V \) be the volume (in litres) of the 30% acid solution needed.</p> <p>The amount of pure acid in the original solution is \( \frac{12}{100} \times 600 \) litres.</p> <p>The amount of pure acid in the added solution is \( \frac{30}{100} \times V \) litres.</p> <p>The total volume of the new solution is \( 600 + V \) litres.</p> <p>The concentration of the acid in the new solution must be more than 15% but less than 18%:</p> <p>\[ \frac{15}{100} < \frac{\frac{12}{100} \times 600 + \frac{30}{100} \times V}{600 + V} < \frac{18}{100} \]</p> <p>Multiply through by \( 100(600 + V) \) to clear the denominators:</p> <p>\[ 15(600 + V) < 12 \times 600 + 30V < 18(600 + V) \]</p> <p>Solve each inequality for \( V \):</p> <p>For the lower bound \( 15(600 + V) > 12 \times 600 \):</p> <p>\[ 9000 + 15V > 7200 \]</p> <p>\[ 15V > -1800 \]</p> <p>\[ V > -120 \] (Which is always true since volume cannot be negative)</p> <p>For the upper bound \( 18(600 + V) > 12 \times 600 + 30V \):</p> <p>\[ 10800 + 18V > 7200 + 30V \]</p> <p>\[ 12V < 3600 \]</p> <p>\[ V < 300 \]</p> <p>Therefore, the volume of the 30% acid solution to be added must be less than 300 litres.</p>