Hello Dear
I hope everything gorse very well.
I did not get any respond from you.
I render my solution to the problem over CO2 aq. + H2O ...> H2CO3
how much PH can be put at for the above-equation under P = 3600 psi, T= 190 oF, 0.208 liters fresh-H2O
as we know, 1.45 grams CO2 (equal to 0.033 moles) is soluble in water at 25oC / 1 atm.
1.Using Ideal gas eq.: PV=nRT z~ 1.0
n value is calculated to be around 1.14 moles CO2 (equal to 50.16 grams).
2.Ka = [H+].[HCO3-]/[H2CO3]
Ka ~ 10^-6.367 = x * x / (0.033-x)
x = [H+]~ 7*10^-4
PH = 3.15 for solubility of 50.16 grams CO2 in 0.208 liters fresh-H2O
3.Now, my questions
firstly, is it authentic the above-solution?
secondly, how much PH would be, if we have 195 kg CO2 into 0.208 liters fresh-water?
Thank you very much.
Hello,
I apologize for the delay in responding to your message. Let's go through your solution step by step and address your questions.
1. Using the ideal gas equation (PV = nRT), you calculated the number of moles of CO2. You mentioned that the calculated value is approximately 1.14 moles, which is equal to 50.16 grams.
This calculation appears to be correct based on the given conditions (P = 3600 psi, T = 190°F). It assumes that the CO2 behaves ideally as a gas.
2. You then proceed to calculate the concentration of H+ ions using the equilibrium constant expression (Ka = [H+].[HCO3-]/[H2CO3]). The given value for Ka is approximately 10^(-6.367) and you solved for x, which represents the concentration of H+ ions.
The calculation seems reasonable, assuming you made the correct simplifying assumption by neglecting the concentration of H2CO3 compared to HCO3-. The resulting concentration of H+ ions, approximately 7 * 10^(-4) M, appears to be correct based on the given values.
3. Now, let's address your questions:
a) Is the above solution authentic?
Based on the information provided, it seems like your solution is reasonable and consistent with the calculations you performed. However, please note that I am an AI and do not have access to all the context or additional information that might be relevant. It is always recommended to double-check your calculations and consult reliable sources.
b) How much pH would it be if we have 195 kg CO2 in 0.208 liters of fresh water?
To answer this question, we need to convert the mass of CO2 (195 kg) to moles. The molar mass of CO2 is approximately 44.01 g/mol. Therefore, the number of moles of CO2 can be calculated as follows:
(195 kg / 44.01 g/mol) = approximately 4430 moles
Now we can use the same approach as before to calculate the pH:
- Calculate the number of moles of CO2 dissolved:
n = (4430 moles) * (0.208 liters / 1.45 grams)
= approximately 64.41 moles
- Solve for [H+] using the equilibrium constant expression:
Ka = x^2 / (0.033 - x)
x ≈ [H+] = 1.24 x 10^(-3) M
- Calculate the pH:
pH = -log10[H+]
= -log10(1.24 x 10^(-3))
≈ 2.91
Therefore, if you have 195 kg of CO2 dissolved in 0.208 liters of fresh water, the resulting pH would be approximately 2.91.
Please note that the above calculations assume ideal gas behavior and neglect the contribution of other ions present in the solution. Additionally, the given conditions should be taken into account for accurate calculations.
I hope this explanation helps! Let me know if you have any further questions.