Air is 21 mole percent oxygen, that is, 0.21 atm oxygen at 1 atm. Suppose the pH

Question

Air is 21 mole percent oxygen, that is, 0.21 atm oxygen at 1 atm. Suppose the pH of the water is 3.88 and that the concentration of iron(II) in the water is 2.30x 10-5 M, what is the potential of the corrosion reaction under the above conditions at 298 K?
Air is 21 mole percent oxygen, that is, 0.21 atm oxygen at 1 atm. Suppose the pH of the water is 3.88 and that the concentration of iron(II) in the water is 2.30x 10-5 M, what is the potential of the corrosion reaction under the above conditions at 298 K?

Explanation / Answer

Yes, it is right. You use ideal gas law.
I'll assume you know most of the variables and what they mean.

PV=nRT
n = # of moles = mass(m)/molar mass (M)
PM= (m/V)RT
you know (m/V) = density = p

so PM/(RT)= p

to solve for molar mass, you use mass percentage for oxygen and nitrogen. Oxygen and nitrogen can only exist as diatomic, so you multiply by 2..

Molar mass= 2(.21)(16) + 2(.79)(14.01) = 28.8558

back to original form.

1(28.8558)/(.08206 * (273+30)) = 1.16 g/L

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.