lab 10 acid base titration - 4.3 Lab 10 Acid Base Titration...

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4.3 Lab 10 Acid Base Titration Jennie Chang (odd) Period 4 10/26/11 EXTRA CREDIT
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Data Table: Part A Volume of HCl (ml) Volume of NaOH (ml) Trial 1 Initial 0.0 0.0 Final 10.0 11.9 Trial 2 Initial 10.0 11.9 Final 19.9 24.3 Trial 3 Initial 19.9 24.3 Final 29.9 36.7 Trial 4 Initial 29.9 36.7 Final 38.9 47.9 Calculations: 1. Write a balanced equation for the reaction in part A. Balanced Equation: HCl + NaOH H 2 O + NaCl 2. Calculate the molarity of the NaOH from part A. a) Use MaVa = MbVb Equation: MaVa= MbVb Substitution: (.500 M)(10.00 mL) = Mb(12.4) Solve: 0.403 M b) For the volume of the base (Vb) average trials 2 and 3 only. Trial one was an estimate. Equation: Vb= (trial 2 + trial 3) / 2 Substitution: (12.4 + 12.4) / 2 Solve: 12.4 mL 3. Calculate the percent error for part A, using the actual value provided by your teacher. Equation: % error= [(experimental value- accepted value)/ accepted value] x 100 Substitution: % error= [(.403 M – 0.400 M) / 0.400 M] x 100] Solve: % error=
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lab 10 acid base titration - 4.3 Lab 10 Acid Base Titration...

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