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 =  

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

the atomic mass is the bottom number on the element.

-steel jelly