A water chemist obtained a 250 ml sample from a nearby lake and fixed the oxygen on-site with alkaline solutions of MnSO4 and KI-NAN3. Returning to the laboratory, a 200 ml sample was analyzed by acidifying the sample with conc H2SO4 and then titrating with 14.4 ml of 0.0213 M Na2S2O3 solution to the starch end point.
a) Calculate the number of moles of I3- that reacted with the Na2S2O3.
b) Calculate the number of moles of Mn(OH)3 that were produced from the reduction of the dissolved oxygen.
c) Calculate the number of moles and milligrams of O2 present in the titrated sample.
d) What is the dissolved oxygen concentration in the sample, expressed in ppm O2?
In addition,
5. a) What is the procedure for preparing 250 mL of 0.0210 M Na2S2O3 for this experiment from a 100-mL volume of standard 0.106 M Na2S2O3?
b) For the preparation of the 0.0210 M solution in a 250-mL volumetric flask, only a 25.0 mL calibrated volumetric pipet is available. Explain how you would prepare the 0.0210 M Na2S2O3 solution using the 25.0 mL pipet. What would be its exact molar concentration?
thank you in advance.
M1V1=M2V2
V2= 250x0.0210/0.106 = 49.52 ml
by taking 49.52 ml of 0.106M solution dilute it up to 250 ml by adding water so tha desired you will get desired one.
b)by using 25 ml pipet first take TWO TIMES 50 ml of 0.106M solution in volumetric flask then dilute solution up to the mark by adding water carefully.
concentration of resulting solution is
MOLARITY X VOLUME =0.0210X250/1000= 0.00525 M
a.
2[S2O3]^2- + [I3]^- ==> 3I^- + [S4O6]^2-
mols [S2O3]^2- = M S2O3 x L S2O3 = ?
mols I2 = 1/2 mols S2O3
b. The equations are as follows although there is some doubt if all of the Mn is in the form of Mn^3+; it is thought some may be in Mn^4+
O2 + 2Mn^2+ ==> 2O^2- + 2Mn^3+
c. I'll leave this for you.
d. After you have mg O2, then convert to mg O2/L and that will give you ppm. Post any and all work if you want to pursue this.
2Mn^3+ + 3I- ==> 2Mn^2+ + [I3]^-
Now use the coefficients to convert from mols [S2O3]^2- to mols Mn^3+.
2 mols [S2O3]^2- = 1 mol [I3]^- = 2 mols Mn^2+.
To calculate the different values required for this problem, we need to use stoichiometry and molar relationships between the reactants and products involved.
a) To calculate the number of moles of I3- that reacted with Na2S2O3, we need to use the balanced chemical equation of the reaction between I3- and Na2S2O3.
The balanced equation is: I3- + 2S2O3^2- -> 3I- + S4O6^2-
From the equation, we can see that for every mole of I3- that reacts, it consumes 2 moles of S2O3^2-. In the given information, 14.4 ml of 0.0213 M Na2S2O3 solution was used, which is equivalent to 0.0213 moles of Na2S2O3.
Since the stoichiometric ratio is 2:1 between I3- and S2O3^2-, the number of moles of I3- can be calculated as:
Number of moles of I3- = (0.0213 moles of Na2S2O3) / 2 = 0.01065 moles of I3-
b) To calculate the number of moles of Mn(OH)3 that were produced from the reduction of dissolved oxygen, we need to consider the stoichiometry of the reaction between Mn(OH)3 and I3-.
The balanced chemical equation for the reaction is: 2Mn(OH)3 + 3I3- -> 2Mn(OH)2 + 3I- + 3IO3-
From the equation, we can see that for every mole of I3- that reacts, it produces 2 moles of Mn(OH)3. Since we have already determined that 0.01065 moles of I3- reacted, the number of moles of Mn(OH)3 can be calculated as:
Number of moles of Mn(OH)3 = 2 * 0.01065 = 0.0213 moles of Mn(OH)3
c) To calculate the number of moles and milligrams of O2 present in the titrated sample, we need to consider the stoichiometric relationship between O2 and Mn(OH)3.
The balanced chemical equation for the reaction is: 3O2 + 6OH- -> 6O2- + 6H2O
From the equation, we can see that for every 3 moles of O2, it produces 6 moles of Mn(OH)3. Therefore, the number of moles of O2 can be calculated as:
Number of moles of O2 = (0.0213 moles of Mn(OH)3) * (3 moles of O2 / 6 moles of Mn(OH)3) = 0.01065 moles of O2
To calculate the number of milligrams of O2, we need to use the molar mass of O2, which is approximately 32 g/mol.
Number of milligrams of O2 = (0.01065 moles of O2) * (32 g/mol) * (1000 mg/g) = 341.6 mg of O2
d) To calculate the dissolved oxygen concentration in the sample expressed in ppm O2, we need to know the volume of the original sample and the amount of dissolved oxygen in the water sample.
The volume of the original sample was 250 ml, and we have calculated that there were 341.6 mg of O2 present in the titrated sample.
The concentration in ppm (parts per million) is defined as: ppm = (mass of solute / volume of solution) * 10^6
Here, the solute is O2, and the volume of the solution is the volume of the original sample. Let's calculate the concentration:
Concentration of O2 in ppm = (341.6 mg / 250 ml) * (10^6) = 1366.4 ppm O2