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2 (Gravimetric Analysis/Volumetric Analysis) Answer Key

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2 (Gravimetric Analysis/Volumetric Analysis) Answer Key CHEM 201 Self Quiz – 2 (Gravimetric analysis/Volumetric analysis) Answer key 1. Calculate the solubility product constant for SrF given the molar concentration of the 2 -4 saturated solution is 8.6 × 10 M. 2+ " SrF2 ! Sr + 2F 2+ ! 2 Ksp = [Sr ][F ] 2+ "4 [Sr ] = [SrF2 ] = 8.6!10 ! !4 !4 [F ] = 2"[SrF2 ] = 2" 8.6"10 =17.2"10 "4 "4 2 "10 Ksp = 8.6!10 ! (17.2!10 ) =1.5!10 -11 2. Calcium fluoride, CaF , is a sparingly soluble salt with a K = 3.9 × 10 . 2 sp 2+ " CaF ! Ca + 2F 2 x x 2x (a) Calculate its molar solubility in a saturated solution. 2+ ! 2 2 3 Ksp = [Ca ][F ] = x " (2x) = 4x 3.9!10"11 4x 3 = 3.9!10"11 # x = 3 = 2.1!10"4 4 Molar solubility is 2.1×10-4 M (b) Calculate its molar solubility in a saturated aqueous solution that is also 0.050 M in fluoride ion, F-. 2 Ksp = (x)(2x + 0.05) 2x = 4.2!10"4 Assumption: 2x << 0.050 2 "11 Ksp = (x)(0.05) = 0.0025x = 3.9!10 "11 3.9!10 x = =1.6!10"8 0.0025 Molar solubility is 1.6×10-8 M. Solubility decreased. -4 3. A solution containing 250 mL of 2.00 × 10 M Ni(NO3)2 is mixed with 250 mL of a -8 solution containing 4.00 × 10 M Na2S. The solubility product constant (Ksp) for NiS is 3.00 × 10-21. Show all your work. Ni(NO3)2 + Na2S ! NiS + 2NaNO3 (a) Write the net ionic reaction that occurs. 2+ 2! "21 Ni + S " NiS# Ksp = 3.00!10 (b) What are the initial concentrations of the predominant species participating in the net ionic reaction (prior to the reaction)? Total volume is: 250 ml + 250 ml = 500 ml 250ml Ni2+ = 2.00!10"4 ! =1.00!10"4 M 500ml 250ml S 2! = 4.00"10!8 " = 2.00"10!8 M 500ml (c) Does a precipitate form? Justify your answer with the calculation that demonstrates whether, or not a precipitate forms. Q = [Ni+2][S!2] =1.00"10!4 " 2.00"10!8 = 2.00"10!12 Q > Ksp : precipitate forms 4. Distinguish between end point and equivalence point. Equivalence point: amount of titrant is equal to the amount of analyte. End point: point where is the sudden change of physico-chemical property of the analyte occurs. It is located very close to the equivalence point. 5. Consider the following reaction - + 2+ 5 H2C2O4 + 2 KMnO4 + 6 H 10 CO2 + 2 Mn + 8 H2O How many milliliters of 0.165 M KMnO4 are needed to react with 108.0 mL of 0.165 M oxalic acid? How many milliliters of 0.165M oxalic acid are required to react with 108 mL of 0.165 M KMnO4? ! ! 2" MnO4 ! n of MnO4 = 2! (108.0ml " 0.165M) = 7.128 mmoles MnO4 5" C2O4 2- volume and concentration of C2O4 7.128 mmoles = 43.20 ml of KMnO 0.165 mmoles/ml 4 How many milliliters of 0.165M oxalic acid are required to react with 108 mL of 0.165 M KMnO4? 2! 5 5 Volume of the C O = (volume of the KMnO4) = "108 ml = 270 ml 2 4 2 2 6. Why solubility of an ionic compound increases as the ionic strength of the solution increases? The net attraction between the cation with its ionic atmosphere and the anion with its ionic atmosphere is smaller than it would be between pure cation and anion in the absence of ionic atmosphere. 7. Find the activity coefficient of each ion at the indicated ionic strength: 2- (a) SO4 ( = 0.01 M) "0.51z2 µ "0.51$ 4 $ 0.01 log! 2" = = = "0.18 SO4 1+ (# µ /305) 1+ (400 0.01/305) "0.18 ! 2" =10 = 0.66 SO4 3+ (b) Sc ( = 0.005 M) "0.51z2 µ "0.51$ 9$ 0.005 log! 3+ = = = "0.27 Sc 1+ (# µ /305) 1+ (900 0.005 /305) "0.27 ! Sc 3+ =10 = 0.54 2+ 8. Calculate the concentration of Hg2 in saturated solutions of Hg2Br2 in: (a) 0.001 M KNO3 Ionic strength of the KNO3 is: 1 1 µ = c z 2 = (0.001"12 + 0.001"12) = 0.001 2 ! i i 2 The solubility value product for the Hg2Br2 is: -23 Ksp = 5.6× 10 2+ " 2 2 2 2 "23 [Hg2 ]! 2+ [Br ] ! Br" = x (0.867)(2x) (0.964) = 5.6#10 Hg2 2+ "8 x = [Hg2 ] = 2.6!10 M (b) 0.01 M KNO3 Ionic strength of the KNO3 is: 1 1 µ = c z 2 = (0.01"12 + 0.01"12) = 0.01 2 ! i i 2 2+ " 2 2 2 2 "23 [Hg2 ]! 2+ [Br ] ! Br" = x (0.660)(2x) (0.899) = 5.6#10 Hg2 2+ "8 x = [Hg2 ] = 3.0!10 M (c) 0.1 M KNO3 Ionic strength of the KNO3 is: 1 1 µ = c z 2 = (0.1"12 + 0.1"12) = 0.1 2 ! i i 2 2+ " 2 2 2 2 "23 [Hg2 ]! 2+ [Br ] ! Br" = x (0.355)(2x) (0.755) = 5.6#10 Hg2 2+ "8 x = [Hg2 ] = 4.1!10 M .
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