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A chemistry student weighs out 0.128g of chloroacetic acid HCH2ClCO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.

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Answer:

19.3 mL

Explanation:

First we convert 0.128 g of chloroacetic acid into moles, using its molar mass:

  • 0.128 g ÷ 94.5 g/mol = 1.354x10⁻³ mol

As this acid is monoprotic, it requires the same number of moles of NaOH to be neutralized. This means that 1.354x10⁻³ moles of NaOH are required.

Now we calculate the volume of NaOH solution, using the given concentration and calculated number of moles:

  • Molarity = moles / liters
  • liters = moles / Molarity
  • 1.354x10⁻³ mol / 0.0700 M = 0.0193 L

We can convert liters into mL:

  • 0.0193 L * 1000 = 19.3 mL

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