How many grams of CO2 are needed to neutralize a 5 liters water with a pH of 11 to 7? under the following conditions:

P: 4000 psi, T:230 oF
Note: CO2 is a one of oil reservoir fluid components with mole percent 0f 3.6%.

thanks a million

To determine the number of grams of CO2 needed to neutralize the water, we need to calculate the moles of CO2 required and then convert it to grams using the molar mass of CO2.

First, let's calculate the moles of H+ ions in the water to be neutralized. Since the pH decreases from 11 to 7, it means the concentration of H+ ions increases by a factor of 10,000.

The concentration of H+ ions at pH 11 can be calculated using the formula: [H+] = 10^(-pH). Therefore, [H+] = 10^(-11).
The concentration of H+ ions at pH 7 is then [H+] = 10^(-7).

So, the moles of H+ ions in a 5L water solution at pH 11 is 5 * 10^(-11) moles/L * 5 L = 2.5 * 10^(-10) moles.
The moles of H+ ions in a 5L water solution at pH 7 is 5 * 10^(-7) moles/L * 5 L = 2.5 * 10^(-6) moles.

Since CO2 reacts with H+ ions in water to form H2CO3, we need an equal number of moles of CO2 to neutralize the H+ ions. Therefore, the moles of CO2 required is 2.5 * 10^(-6) - 2.5 * 10^(-10) = 2.49999975 * 10^(-6) moles.

Now, let's determine the mass of CO2 required. To do this, we need to convert moles to grams using the molar mass of CO2. The molar mass of CO2 is approximately 44 g/mol.

Mass of CO2 = moles of CO2 * molar mass of CO2
Mass of CO2 = (2.49999975 * 10^(-6) moles) * (44 g/mol)
Mass of CO2 = 0.00010999998675 grams

Therefore, approximately 0.00011 grams (or 0.11 milligrams) of CO2 is needed to neutralize the 5 liters of water from pH 11 to pH 7, under the given conditions.