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.
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?
3 answers
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
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+.
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+.
6 answers