Question

A water sample obtained from the neigborhood pond was analyzed for dissolved oxygen content. A volume of 300 mL, with temperature 21 degrees Celsius, was analyzed using the Wrinkler method shown in the simulated lab material. It took 15.0 mL of 0.01 mol/L Na2S2O3 solution to reach titration end point.

*Note: Na2S2O3 solution is the source of S2O32- ions for the titration reaction.

In the succeeding questions, provide the following values required. Take note of the units

a.What is the mass of the water sample?

b. milimoles of S2O32- reacted (1 mole = 1000 milimoles)

c. milimoles of Mn(OH)3

d. milimoles of dissolved oxygen

e. mg of dissolved oxygen, O2

f. ppm dissolved oxygen (mg O2 / kg water)

g. Is the water in the pond good enough for aquatic life? Why or why not?

h. Based on DO content, do you consider the water in the pond polluted?

Answer 1

Volume of water sample

= 300 mL x (1L/1000 mL) x (1m3/1000L)

= 300*10^-6 m3

Temperature T = 21°C

Density of water = 997.992 kg/m3

Part a

Mass of water sample = volume x density

= 300*10^-6 m3 x 997.992 kg/m3

= 0.2993976 kg x (1000g/kg) x (1000mg/g)

= 2.99 x 10^5 mg

Part b

milimoles of S2O3 2- reacted = milimoles of Na2S2O3 reacted

= volume x molarity

= (15 mL) x (1L/1000 mL) x (0.01 mol/L) x (1000milimoles/mol)

= 0.15 milimoles

Part c

The reactions in Winkler method

2Mn(OH)2 + 0.5O2 + H2O = 2Mn(OH)3

2Mn(OH)3 + 2I- + 6H+ = 2Mn2+ + I2 + 6H2O

I2 + I- = I3-

I3- + 2S2O3 2- = 3I- + S4O6 2-

From the stoichiometry of above reactions

milimoles of I3- reacted = 0.15/2 = 0.075 milimoles

milimoles of I2 reacted = 0.075 milimoles

milimoles of Mn(OH)3 reacted = 2*0.075 = 0.15 milimoles

Part d

milimoles of dissolved Oxygen = 0.5*0.15/2

= 0.0375 milimoles

Part e

mass of dissolved Oxygen

= 0.0375 millimoles x 32 mg/millimoles

= 1.20 mg