Answer:
50mL of the 0.1M acid are needed to neutralize the solution
Step-by-step 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 =
50mL of the 0.1M acid are needed to neutralize the solution