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 =

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

what does the table look like