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Titration of a Weak Acid Lab - Stephanie Warner Chem 153...

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Stephanie Warner Chem 153 Shidong Chu 23 September 2005 Titration of a Weak Acid Purpose: The purpose of this lab is to determine the amount of acid in a solid by titrating the solid with a strong base. Calculations: Standardization: (0.9479 g KHP/ 204.22 g/mol) x (1 mol OH/1 mol KHP) = 0.003173 mol OH 31.89 mL NaOH x (1L/1000 mL) = 0.03389 L NaOH 0.003173 mol OH/0.03389 L NaOH = 0.09363 M NaOH Q Test for Standardization: 0.0982, 0.09813, 0.09363, 0.09379 (0.09813-0.09379)/ (0.09813-0.9282) = 0.871 0.09813 is an outlier. Average for Standardization: 0.09282 + 0.09363 + 0.09379 = 0.28024/3 = 0.09341 M NaOH Unknown Titration: 24.06 mL x (1L/ 1000mL) = 0.02406 L NaOH (0.02406 L NaOH)(0.09341 M NaOH)x(1 mol H + / 1 mol OH - )x(1 mol KHP/ 1 mol OH - ) = 0.002247 mol KHP [(0.002247 mol KHP)(204.22 g/mol)/ 0.9236 g Unknown] x 100 = 49.68% KHP Q Test: 49.97%, 47.25%, 49.68%, 49.99%, 49.49% (49.99-49.97)/ (49.99-47.25) = 0.0072 (47.25-49.68)/ (49.99-47.25) = 0.88 This means that 47.25% is an outlier. (49.99-49.97)/ (49.99-49.49) = 0.04 (49.49-59.68)/ (49.99-49.49) = 0.38 Average: 49.49 + 49.68 + 49.99 + 49.97 = 199.13/ 4 = 49.78 % average
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Standard Deviation d 1 = (x mean value – x measured value ) = (49.78-49.87) = -0.19 d 2 = (49.78- 49.68) = 0.1 d 3 = (49.78- 49.99) = -0.21 d 4 = (49.78- 49.49) = 0.29 S = [(d 1 2 + d 2 2 + d 3 2 + d 4 2 ) / (number of measurements – 1)] ½ = [((-0.19) 2 + (0.1) 2 + (-0.21) 2 +(0.29) 2 )/ (4-1)] ½ = 0.24% Questions: 1. Molecular Equation: C 6 H 5 COOH (aq) + NaOH (aq) → C 6 H 5 COO - (aq) + Na + (aq) + H 2 O (l) Net Ionic Equation: H + (aq) + OH - (aq)
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