A sample of pure CaCO_3 weighing 0.2500 g is dissolved in hydrochloric acid and

Question

A sample of pure CaCO_3 weighing 0.2500 g is dissolved in hydrochloric acid and the solution diluted to 250.0 ml in a volumetric flask. A 25.00 ml aliquot requires 39.11 ml of an EDTA solution for titration. Calculate the Tier of the EDTA solution in terms of mg CaCO_3 per ml of EDTA A 250.0 ml of lake water containing Ca^+2 is titrated with 11.00 ml of the above EDTA solution. Calculate the hardness of the water in mg of CaCO_3 per ml of lake water. Determine the normality of a solution of KMnO_4(FW = 158.034) if 7.9935 g of primary standard K_2Cr_2O_7 (FW = 294.202) was used to prepare 0.5000 L of solution. 6e^- + 2Cr^+6 = 2Cr6+3 Calculate the percent iron (At. Wt. = 55.847 g/mole) in an unknown if a 09887 g sample required 30.05 ml of the above standard KMnO_4 solution to titrate the iron as Fe^+2 in the sample. 5e^- + MnO_4^-1 + 8H^+ = Mn^+2 + 4H_2O In the spectrophotometric determination of manganese in steel, a 0.250 gram sample of unknown steel is dissolved and the manganese oxidized to MnO_4^-. This solution is then diluted to 500 ml in a volumetric flask. The quantity 0.600 grams of a steel sample, certified to contain 0.25% Mn, is treated in exactly the same way. The Spectronic 20, set at 540 nm to match the absorption band of the MnO_4^- ion, is adjusted so that the instrument reads 0% transmission with no sample in the Spectronic 20 and 100% transmission with a distilled water blank in the instrument. The % transmission of the unknown solution (%T(x)) is found to be 58.0%, while the % transmission of the known solution (%T(k)) is found to be 39.0%. Determine a) mg Mn(x). and b) the percentage of Mn in the 0.250 gram sample of unknown steel. mg M(x) = mg Mn(k) times [(2 - log%T(x))/times (2 - log%T(k))] Theory of Neutralization Titrations

Explanation / Answer

2.

molarity of CaCO3 solution = 0.25 g/100.087 g/mol x 0.25 L = 9.99 mM

moles of CaCO3 in 25 ml aliquot = 9.99 x 10^-3 M x 25 ml = 0.25 mmol

a) EDTA = 39.11 ml

So,

Titer of EDTA = 0.25 mmol x 100.087 g/mol/39.11 ml

                       = 0.64 mg CaCO3/ml EDTA

b) Hardness of lake water = 0.25 mmol x 100.087 g/mol/11 ml

                                          = 2.275 mg CaCO3/ml EDTA              

3.

a) molarity of K2Cr2O7 = 7.9935 g/294.202 g/mol x 0.5 L

                                      = 0.054 M

normality of KMnO4 = 0.054 x 3/5 x 5 = 0.0065 N

b) moles MnO4- used = 0.0065 N x 2 x 30.05 ml = 0.391 mmol

moles Fe = 0.391 x 5 = 1.955 mmol

mass Fe present = 1.955 x 55.847/1000 = 0.11 g

%Fe in sample = 0.11 x 100/0.9887 = 11.12%

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