Question

3. The Henry's law constant for the solubility of 10-4 Matm at oxygen in water is 3.30 x 128 C and 2.85 x 104 M/atm at 22 C. Aiır is 21 mol% oxygen. 

(a) How many grams of oxygen can be dissolved in one liter of a trout stream at 128 C at an air pressure of 1.00 atm? 

(b) How many grams of oxygen can be dissolved per liter in the same trout stream at 22 ℃ at the same pressure as in (a)? 

(c) A nuclear power plant is responsible for the stream's increase in temperature. What percentage of dissolved oxygen is lost by this increase in the stream's temperature?

Answer 1

Ans 3

From the Henry's law

Molarity of solute is directly proportional to the the partial pressure of the solute.

M = H x PO2

Where H is Henry's constant

Part a

At 128 °C

H1 = 3.30 x 10^-4 M/atm

PO2 = 0.21 x 1 = 0.21 atm

M1 = H1 x PO2

= 3.30 x 10^-4 M/atm x 0.21 atm

= 0.0000693 M

Volume of solution V = 1L

Moles of O2 = molarity x volume

= M1 x V

= 0.0000693 mol/L x 1L

= 0.0000693 mol

Mass of O2 = moles x molecular weight

= 0.0000693 mol x 32 g/mol

m1 = 2.22 x 10^-3 g

Part b

At 22 °C

H2 = 2.85 x 10^-4 M/atm

PO2 = 0.21 x 1 = 0.21 atm

M2 = H2 x PO2

= 2.85 x 10^-4 M/atm x 0.21 atm

= 0.00005985 M

Volume of solution V = 1L

Moles of O2 = molarity x volume

= M2 x V

= 0.00005985 mol/L x 1L

= 0.00005985 mol

Mass of O2 = moles x molecular weight

= 0.00005985 mol x 32 g/mol

m2 = 1.92 x 10^-3 g

Part c

Dissolved oxygen lost = m2 - m1

= 2.22 x 10^-3 - 1.92 x 10^-3

= 3.00 x 10^-4 g