Myron Stokes
Nov 03, 2024
Calculate the masses of oxygen and nitrogen that are dissolved in 6.0 L of aqueous solution in equilibrium…?
I'm having a bit of trouble with this chemistry question. I took the partial pressures of the gases and used those to find the molarity using Henry’s law. Then, I calculated the number of moles and used the molar mass to find the mass in milligrams. What would you do? Here is the entire question: Calculate the masses of oxygen and nitrogen that are dissolved in 6.0 L of aqueous solution in equilibrium with air at 25 °C and 760 Torr. Assume that air is composed of 21% oxygen and 78% nitrogen by volume. The gas constants are O2: 7.9 × 10^2 bar/M and N2: 1.6 × 10^3 bar/M.
2 Answers
Henry’s Law c = p/k, where p = partial pressure and k = Henry’s constant for the gas.. Partial pressure is proportional to volume fraction, 760 torr = 1 atm so p(O2) = 0.21 atm, p(N2) = 0.78 atm
k for O2 = 769.2 L-atm/mol
k for N2 = 1639 L-atm/mol
c(O2) = 0.21/769.2 = 2.73*10^-4 mol/L
c(N2) = 0.78/1639 = 4.76*10^-4 mol/L
There are 6.0 L of solution, so the no of moles is
n(O2) = 6* 2.73*10^-4 = 1.64*10^-3 mol
n(N2) = 6*4.76*10^-4 = 2.86*10^-3 mol
The molar mass of O2 = 32 so there are 32*1.64*10^-3 = 52 mg of O2
The molar mass of N2 = 28 so there are 28*2.86*10^-3 = 80 mg of N2
you want to envision an ICE chart (I = initial, C = replace, E = equilibrium) for a weak acid. HB = cinnamic acid on the best of the chart, think about HB –> H+ + B- The initial quantity of HB is a million.6*10^-3 The replace for HB is -x, at the same time as the replace for H+ and B- is +x The equilibrium quantity for HB is a million.6*10^-3 – x The equilibrium quantity for H+ is x. Ka = [H+]^2/[HB] So Ka = x^2/(a million.6*10^-3 – x) because x is this style of small variety, it must be removed from the denominator with out causing a lot disturbance. so that you’re left with Ka = x^2/a million.6*10^-3 After plugging in Ka and fixing for x, you get: x = 2.40 3*10^-4 = [H+] -log ([H+]) = pH pH = 3.sixty one
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