Calculate the amount of 0.1 M acid needed to neutralize 10,000 liters of pH 8.0 water. The acid is carbonic acid (H₂CO₃).

50mL of the 0.1M acid are needed to neutralize the solution

Explanation:

To solve this question we have to calculate the moles of OH- present in the basic solution. Based on the reaction:

2OH- + H2CO3 → 2H2O + CO3²⁻

we can find the moles of carbonic acid (And its volume) required for the complete neutralization as follows:

Moles OH-:

pH = -log [H+]

10^-pH = [H+]

[H+] = 1x10⁻⁸M

As:

[OH-] = Kw / [H+]

[OH-] = 1x10⁻¹⁴ / 1x10⁻⁸

[OH⁻] = 1x10⁻⁶M

The moles in 10000L are:

10000L * (1x10⁻⁶moles OH- / L) = 0.01 moles OH-

Moles H₂CO₃ required:

0.01 moles OH- * (1mol H₂CO₃ / 2mol OH⁻) = 0.005 moles H₂CO₃

Volume:

0.005 moles H₂CO₃ * (1L / 0.1moles) = 0.05L =

it's c (engineer's do their work with mechanical stuff not on any living being)