CHEM 1210 17. The volume for the second titration final reading - first reading

Question

CHEM 1210 17. The volume for the second titration final reading - first reading Part B: Determination of concentration of commercial vinegar 1. Fill the burette with NaOH and adjust to zero. 0.0 mL of commercial vinegar to 100 mL add 90 mL of distilled water into a 100 mL graduated cylinder then add 10 mL vinegar. Using stirring rod mix the solution well 3. Transfer 20.0 mL of this solution into a 250-mL beaker using a graduated cylinder and add 20.0 mL of water. 4. Follow steps 8 - 17 in Part A. To find the volume of base used to reach the end point. 5. Repeat steps 3 & 4. The volume for the second titration final reading - first reading Equations: molarity = moles/volume moles- molarity x volume a Acid + b Base Salt + Water

Explanation / Answer

Analysis of vinegar

Part A) Standardization of HCl solution

NaOH + HCl --> NaCl + H2O

moles NaOH = moles HCl

moles NaOH = 0.1 M x 10 ml = 1 mmol

Since volume of HCl solution is not given, we would for calcualtion purpose assume it to be 10 ml

So,

molarity of HCl solution = 1 mmol/10 ml = 0.1 M

Part B) moles NaOH used = 0.1 M x 17 ml = 1.7 mmol

HC2H3O2 + NaOH --> NaC2H3O2 + H2O

moles HC2H3O2 = moles NaOH = 1.7 mmol

molarity of HC2H3O2 = 1.7 mmol x 2/40 ml = 0.085 M

actual molarity of vinegar solution = 0.085 m x 100 ml/10 ml = 0.85 M

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