A chemistry student weighs out 0.123 grams of chloroacetic acid (HCH2ClCO2) into a 250. milliliter volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1200 moles/liter NaOH solution.
Calculate the volume of NaOH solution the student will need to add to reach the equivalence point.
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Volume of a 1.320 M NaOH solution is required to titrate 25.00 mL of a 1.025 M H2SO4 solution
To calculate the volume of NaOH solution (in liters) that the student will need to add to reach the equivalence point, we need to use the balanced chemical equation between chloroacetic acid (HCH2ClCO2) and sodium hydroxide (NaOH):
HCH2ClCO2 + NaOH -> H2O + NaCH2ClCO2
From the balanced equation, we can see that the stoichiometric ratio between HCH2ClCO2 and NaOH is 1:1. This means that, for every 1 mole of HCH2ClCO2, we will need 1 mole of NaOH to react completely.
First, let's calculate the number of moles of chloroacetic acid using its molecular weight:
Molecular weight of HCH2ClCO2 = 1(1) + 1(2) + 1(12) + 1(35.5) + 2(16) = 94.5 g/mol
Number of moles of chloroacetic acid = mass / molecular weight
= 0.123 g / 94.5 g/mol
≈ 0.0013 moles
Since the stoichiometric ratio is 1:1, we will need the same number of moles of NaOH. Therefore, we have:
Number of moles of NaOH = 0.0013 moles
Now, we can calculate the volume of NaOH solution in liters using its concentration:
Volume of NaOH solution (in liters) = Number of moles of NaOH / Concentration of NaOH solution
Given that the concentration of NaOH solution is 0.1200 moles/liter, we can substitute the values into the equation:
Volume of NaOH solution (in liters) = 0.0013 moles / 0.1200 moles/liter
= 0.01083 liters
≈ 10.83 milliliters
Therefore, the student will need to add approximately 10.83 milliliters of the NaOH solution to reach the equivalence point.