Question

Swimming Pool phH Check calculations of pH and acid molarity. 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 b 7.20 The pool is 5.00 m wide, 15.0 m long, and has an average depth of 1.50m. what is the minimum (in the absence of any buffering capacity) volume (mL) of 8.00 wt% H2SO4 (SG-1.052) that should be added to return the pool to the desired pH?O mL the tolerance is +/-2% 98.1

Answer 1

3,48 mL of 8,00wt% H₂SO₄

Step-by-step explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] =
`10^(-7,2)`

Thus, you need to add:

[H⁺] =
`10^(-7,2) - 10^(-8)` = 5,31x10⁻⁸ M

The total volume of the pool is:

5,00 m × 15,0 m ×1,50 m = 112,5 m³ ≡ 112500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 112500 L = 5,97x10⁻³ moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

5,97x10⁻³ moles of H⁺ ×
`(1 H_(2)SO_(4) moles)/(2H^+ mole)` = 2,99x10⁻³ moles of H₂SO₄

These moles comes from:

2,99x10⁻³ moles of H₂SO₄ ×
`(98,1g)/(1mol)` ×
`(100 gsolution)/(8g H_(2)SO_(4) )` ×
`(1mL)/(1,052 g)` = 3,48 mL of 8,00wt% H₂SO₄

I hope it helps!