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 calculate the number of grams of CO2 needed to neutralize the water, we need to follow these steps:
1. Determine the amount of CO2 required to lower the pH from 11 to 7.
2. Convert the amount of CO2 from moles to grams using the molar mass of CO2.
3. Account for the mole percent of CO2 in the oil reservoir fluid.
Step 1: Determine the amount of CO2 required to lower the pH from 11 to 7.
The pH scale is logarithmic, meaning each unit represents a tenfold change in acidity or basicity. pH 7 is neutral, while pH 11 is highly basic. To neutralize the solution, we need to calculate the difference in acidity.
pH 7 - pH 11 = 4
So, we need to reduce the pH by 4 units.
Step 2: Convert the amount of CO2 from moles to grams using the molar mass of CO2.
The molar mass of CO2 is approximately 44 grams/mol.
To calculate the number of moles of CO2 required, we'll use the equation:
moles of CO2 = volume of water (in liters) × pH difference (in moles of H+/OH- ions per liter).
Since there is a 1:1 mole ratio between H+/OH- ions and CO2 produced, we can use the pH difference directly.
moles of CO2 = 5 liters × 4 moles/L = 20 moles
To convert moles to grams, multiply by the molar mass:
grams of CO2 = 20 moles × 44 grams/mol = 880 grams
Step 3: Account for the mole percent of CO2 in the oil reservoir fluid.
Given that the mole percent of CO2 in the oil reservoir fluid is 3.6%, we need to adjust the calculated amount of CO2. Multiply the grams of CO2 by the mole percent:
adjusted grams of CO2 = 880 grams × 0.036 = 31.68 grams
Therefore, approximately 31.68 grams of CO2 are needed to neutralize the 5 liters of water with a starting pH of 11 to a pH of 7, considering the given conditions.