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

7.23mL of the solution you need to add

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) =

i think slash and

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )