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) please show work!

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

Explanation:

To solve this question we have to find the moles of OH- present in the basic solution. Then, using the chemical equation:

2OH- + H2CO3 → 2H2O + CO3²⁻

We can find the moles of carbonic acid and its volume using the 0.1M solution 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:

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

Volume:  

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

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