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| | #1 (permalink) |
| New Member Join Date: Oct 2003
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25 Oct 2005, 3:54 PM ![]() | You can hide this advertisement by registering. Could someone please help me with either of these two questions?1) Pthalic acid is a weak diprotic acid for which we can write the formual as H2(C8H4O4). Potassium hydrogen phtalate KH(C8H4)4) is a good primary standard for standardising alkali solutions. It contains one acidic hydrgoen per formual unit. 0.917g potassium hydrogen phtalate was dissolved in water and titrated with approximately 0.2 mol/L sodium hydroxide solution; 27.2 mL hydroxide solution was needed to reach the end point. Calculate the molatrity of the hydroxide solution. 2) 'Cloudy ammonia' is often used in the home as a cleaning agent. To determine the concentration of ammonia in the solution, a chemist first accurately diluted 25mL (by pipette) to 500 mL (volumetric flask) then titrated 25Ml (by pipette) of the dilute solution with 0.151 mol/L nitric acid solution; it required 27.2 mL. Calculate the molarity of the original ammonia solution. (ref: conquering chem p 155 q17 and 19) |
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| | #2 (permalink) |
| Argentous Fingers HSC: 2003 Gender: Male
Join Date: Sep 2003
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18 Sep 2005, 9:18 PM ![]() ![]() ![]() | H(C8H4O4)- + OH- -> C8H4O4 2- + H20 For KH(C8H4O4); m = 0.917 g M = 204.2 g/mol n = 0.00449 mol For NaOH; n = 1/1 * 0.00449 mol V = 0.0272 L c = 0.00449/0.0272 = 0.165 M
__________________ ![]() Last edited by Jumbo Cactuar; 7 Aug 2005 at 2:05 PM. |
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| | #4 (permalink) |
| Argentous Fingers HSC: 2003 Gender: Male
Join Date: Sep 2003
Posts: 426
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18 Sep 2005, 9:18 PM ![]() ![]() ![]() | NH3 + HNO3 -> NH4+ + NO3- For HNO3; c = 0.151 M V = 0.0272 L n = 0.151 * 0.0272 = 0.00411 mol For diluted NH3; n =1/1 * 0.00411 mol V = 0.025 L c = 0.00411/0.025 = 0.164 M dillution factor is 500/25 = 20 For undiluted NH3; c = 0.164 * 20 = 3.29 M
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