In the titration of the unknown diprotic acid with NaOH assume a chemist

In the titration of the unknown diprotic acid with

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3.In the titration of the unknown diprotic acid with NaOH, assume achemist overshoots the end point and adds too much NaOH. Howwould this error affect the final calculation for the molar mass of theunknown acid? Would it be overestimated, underestimated or remainunaffected? Explain your reasoning
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4. KHP, potassium hydrogen phthalate (KHC 8 H 4 O 4 ), is often used to standardize basic solution used in titration. If a 0.855-g sample of KHP requires 31.44 mL of a KOH solution to fully neutralize it, what is the [KOH] in the solution? The reaction is KHC 8 H 4 O 4 (aq) + KOH(aq) K 2 C 8 H 4 O 4 (aq) + H 2 O(l) KHC 8 H 4 O 4 (aq) + KOH(aq) K 2 C 8 H 4 O 4 (aq) + H 2 O(l) = 4.19 moles n KHC H O 8 4 4 = 204.22 g 0.855 gKHC H O 8 4 4 0 × 1 −3 4.19 0 moles n KOH = × 1 −3 [KOH] = = 0.133 M 0.03144 L 4.19 ×10 moles −3 The concentration of KOH in the solution is 0.133 M 5. The KOH solution standardized above is used to titrate a 20.00-mL sample of sulfuric acid (H 2 SO 4 ) solution of unknown concentration. Determine [H 2 SO 4 ] for the unknown acid solution if 41.27 mL of the KOH solution is needed to fully react with it.
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V KOH = 41.27 mL = 0.04127 L M KOH = 0.133 M 2KOH(aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + H 2 O(l) = = 0.137 M M H SO 2 4 = V H SO 2 4 M . V KOH KOH 0.02000 L (0.133 M )(0.04127 L ) 2 mol KOH 1 mol H SO 2 4 The concentration of sulfuric acid is 0.137 M 6. A 25.00 mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125 ml Erlenmeyer flask and two drops of phenolphthalein are added. The above sodium hydroxide solution (the titrant) is used to titrate the nitric acid solution (the analyte). If 16.77 mL of the titrant is dispensed from a burette in causing a color change of the phenolphthalein, what is the molar concentration of the nitric acid (a monoprotic acid) solution? Express the molar concentration of nitric acid to the correct number of significant figures. V HNO3 = 25.00 mL = 0.02500 L V NaOH = 16.77 mL = 0.01677 L M NaOH = 0.133 M HNO 3
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NaOH(aq) + HNO 3 (aq) NaNO 3 (aq) + H 2 O(l) = = 0.08922 M M HNO 3 = V HNO 3 M . V NaOH NaOH 0.02500 L (0.133 M )(0.01677 L ) The molar concentration of nitric acid is 0.08922 M I. Conclusion: The concentration of base B-35 is around 0.600 M, and the concentration of base B-37 is around 0.450 M. Because the acid and bases are classified as strong acids and bases, the pH of the solution at equivalence point is 7.00. J. Sources of Error Sources of error include (a) wet buret, (b) contaminated stock and standardized solutions, (c) dirty glassware, (d) inaccurate measure equipments, (e) human error in controlling the buret knob, (f) human error in putting the solutions in right places, and (g) human error in reading the meniscus. K. Recommendation: Use a purer source of standardized and stock solutions, use a clean buret to titrate, be more patient when dropping the standardized solution into the Erlenmeyer flasks, and more accurate measure glassware would help improve the results. -----------------------THE END-----------------------
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  • Fall '15
  • Neethling,Melanie
  • Chemistry, ml, Sodium hydroxide, Erlenmeyer flask

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