3600_Ch7_probs - 7-2 7-3 7-4 7-5 7-7 7-8 7-9 The...

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Unformatted text preview: 7-2. 7 -3. 7-4. 7-5. 7-7. 7-8. 7 -9. The equivalence point occurs when the exact stoichiometric quantities of reagents have been mixed. The end point, which comes near the equivalence point, is marked by a sudden change in a physical property brought about by the disappearance of a reactant or appearance of a product. In a blank titration, the quantity of titrant required to reach the end point in the absence of analyte is measured. By subtracting this quantity from the amount of titrant needed in the presence of analyte, we reduce the systematic error. In a direct titration, titrant reacts directly with analyte. In a back titration, a known excess of reagent that reacts with analyte is used. The excess is then measured with a second titrant. Primary standards are purer than reagent grade chemicals. The assay of a primary standard must be very close to the nominal value (such as 99.95—100.05%), whereas the assay on a reagent chemical might be only 99%. Primary standards must have very long shelf lives. 40.0 mL of 0.040 0 M Hg2(N03)2 = 1.60 mmol of Hg22+, which will require 3.20 . . . . 3.20 mmol mmol of KI. This 15 contained in volume = W = 32.0 mL 108.0 mL of 0.165 0 M oxalic acid = 17.82 mmol, which requires 5molH2C204 (7- 1110 H2 2 4)—. m0 0 n 4_ 7.128 mmol / (0.165 0 mmol/mL) = 43.20 mL of KMnO4. An easy way to see this is to note that the reagents are both 0.165 0 M. Therefore, Volume of MnO; = % (volume of oxalic acid). For the second part of the question, volume of oxalic acid = 2 (volume of MnOgl) = 270.0 mL. 1.69 mg of NH3 = 0.099 2 mmol of NH3. This will react with 3(0099 2) = 0.149 mmol of OBr‘. The molarity of OBr' is 0.149 mmol/ 1.00 mL = 0.149 M 7-11. HCl added to powder = (10.00 mL)(1.396 M) = 13.96 mmol NaOH required = (39.96 mL)(0.1004 M) = 4.012 mmol HCl consumed by carbonate = 13.96 — 4.012 = 9.943 mmol mol CaC03 = % mol HCl consumed = 4.974 mmol = 0.4973 g CaCOg 0.4978 g CaCO3 0.541 3 g limestone X 100 = 92'0 Wt% wt% CaCO3 = T-IIL 3fl.lfl mL Hf Ni2+ Med with 39.35 ml. of fl.fl]3 I]? M EDTA. Thad-”um the 1"~I13+ molaritjr 15-. [NW] 2 39.35 mggrnbnlllfln? mnli'L} = m“ 1r {)9 M. lit") I'I'IL. {If N124" cuntflimi 0.417 2 ml Bf Nil'h 13.15 ITIL DIEDTJ‘L = (3.132? mmul uf EUTA. The amount «fr-1i2+ which mus-t have reactad with CH" was 0.41? 1 — 11132 T = [1.294 5 ml. The cyanidr: which mind with 1411+ "lust haw: been {4]{1329‘45} = I. ITS mnmL [CH'] = LITE. mmlfl'll'fl- IIIL = 0.1192 54 M. ...
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