Construct and calculate a theoretical titration curve, showing and discussing al

Question

Construct and calculate a theoretical titration curve, showing and discussing all key areas, for the titration of 0.100 M NaOH with 50.0 mL 0.100 M HNO_2 (K_a HNO_2 = 7.1 times 10^-4) A 30.0 L of air sample (Density of 1.20 g/L) was passed through an absorption tower containing a solution of Cd^2+, where H_2S was retained as Cds. The mixture was acidified and treated with 10.0 mL of 0.01070 M I_2. After the reaction S^2-_ + I_2 = S(s) + 2T was complete, the excess iodine was titrated with 12.85 mL of 0.01344 M thiosulfate, S_2O_3^2-. Calculate the H_2S in ppm. Mol. Mass = 34 g/mol A 6.881-g sample containing MgCl_2 and NaCl was dissolved in sufficient water to give a 500-mL volume. Analysis of the chloride content in a 50-0 mL aliquot resulted in the formation of 0.5923-g of AgCl. The magnesium in a second 50-0 mL aliquot was precipitated as MgNH_4PO_4; on ignition 0.1796-g of Mg_2P_2O_7 was found. Calculate the percentages of MgCl_2.H_2O and NaCl in the original sample. Mol. Mass: MgP_2O_7 = 222.55 g/mol, NaCl = 58.44 g/mol, MgCl_2.6H_2O = 203.30 g/mol The homogeneity of a standard chloride solution was tested by analyzing portions of the material at the top and bottom with the following found Is homogeneity indicated at the 95 % confidence level, use both t-test and f-test DERIVE and calculate the solution potential for the titration of Sn^2+ with Cr_2O_7^2- + 14 H^+ = 2Cr^3+ + 3Sn^4+ + 7H_2O The hydrogen sulfide, H_2S, in a 50-g sample of crude petroleum was removed by distillation and collected in a solution of CdCl_2. The precipitated CdS was filtered,

Explanation / Answer

The volume of air taken = 30 L

Initial moles of I2 taken = Molarity x volume of Iodine solution = 0.01070 X 10 millimoles = 0.1070 millimoles

The reaction between thiosulphate with iodine moelcules is

I2 + 2S2O32- = > 2I- + S4O62-

so with two moles of thiosulphate one mole of iodine will react

Excess moles of I2 left = 0.5 X moles of thiosulphate used

= 0.5 X molarity X volume = 0.5 X 0.01344 X 12.85 = 0.08635 millimoles

So moles of Iodine reacted with sulphide = Initial moles of iodine taken - moles of iodine left

                                                             = 0.1070 - 0.08635 = 0.02065 millimoles

So moles of sulphide ions present in air = 0.02065 millmoles

Moles of H2S present = 0.02065 millimoles

Mass of H2S present = Molecular weight X moles = 34 X 0.02065 X 10^-3 grams = 0.7021 mg

Mass of air taken = Volume X density = 30 X 1.2 = 36 grams

So mass of H2S in 36 grams = 0.7021 mg

Mass in 1 gram of H2S = 0.7021 / 36 = 0.0195 mg

So ppm = 0.0195 X 1000 = 19.5 ppm

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