Calculate the masses of oxygen and nitrogen that are dissolved in 6.0 L of aqueous solution in equilibrium…?

I’m having a little trouble with this chem. question… I took the partial pressures, then used those to find the molarity using Henry’s law, then found the number of moles. then I used the molar mass to find the mass in mg… 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 21% oxygen and 78% nitrogen by volume.

Gas constants are O2 7.9×10^2 bar/M

N2 1.6x 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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