Question

You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A quick analysis reveals that the pH of the water is 8.0 when it should be 7.2. The pool is 5 m wide, 12 m long, and has an average depth of 2 m. What volume (mL) of 5.00 wt% sulfuric acid (SG=1.03) should be added to return the pool to the desired pH?

Answer 1

To calculate the volume of sulfuric acid needed to return the pool to the desired pH, we need to find the difference in pH, which is 8.0 - 7.2 = 0.8. The pH scale is logarithmic, so a change of 1 pH unit corresponds to a tenfold change in acidity or basicity. Therefore, the pool water is 1/10 as acidic as it needs to be.

Step-by-step explanation:

To calculate the volume of sulfuric acid needed to return the pool to the desired pH, we first need to determine the amount of acid required to lower the pH from 8.0 to 7.2. In order to do this, we first find the difference in pH, which is 8.0 - 7.2 = 0.8.

The pH scale is logarithmic, so a change of 1 pH unit corresponds to a tenfold change in acidity or basicity. Therefore, the pool water is 1/10 as acidic as it needs to be. To make it 10 times more acidic, we can use a solution of sulfuric acid with a known concentration.

To calculate the volume of 5.00 wt% sulfuric acid needed, we can use the formula:

Volume of acid (mL) = Volume of pool (m^3) x Difference in pH x Concentration of acid / (Specific gravity x 10)