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 8.00 when it should be 7.20. The pool is 5.00 m wide, 12.0 m long, and has an average depth of 2.50m. What is the minimum (in the absence of any buffering capacity) volume (mL) of 5.00 wt% H2SO4 (SG

Answer 1

7.23mL of the solution you need to add

Step-by-step explanation:

Assuming SG of H2SO4 = 1.080

pH is defined as -log [H⁺]

The initial [H⁺] is 10^-8 but you want a concentration of 10^-7.20

You need to add some acid to reach this concentration:

[H⁺] = 10^-7.20 - 10^-8M = 5.31x10⁻⁸M [H⁺]

To know the moles of H⁺ that we need to add to the pool we need to calculate volume of pool in Liters:

Volume pool:

5.00m ₓ 12.0m ₓ 2.50m = 150m³. 1m³ are 1000L:

150m³ * (1000L / 1m³) = 150000L.

The moles of H⁺ you need to add are:

5.31x10⁻⁸M [H⁺] * (150000L) = 7.965x10⁻³ moles of H⁺

Now, in therms of a 5.00 wt% of H2SO4, these moles are:

7.965x10⁻³ moles of H⁺ * (1 mole H2SO4 / 2 moles H⁺= 3.9825x10⁻³ moles H2SO4

3.9825x10⁻³ moles H2SO4 * (98g / mol) = 0.3903g H2SO4

0.3903g H2SO4 * (100g solution / 5.00g H2SO4) = 7.8057g of solution you need to add.

Using SG of the solution:

7.8057g * (1mL / 1.080g) =

7.23mL of the solution you need to add