Experiment 18 Prelaboratory Assignment Potentiometric Analyses I. a. Far a weak

Question

Experiment 18 Prelaboratory Assignment Potentiometric Analyses I. a. Far a weak acid (e.g. CH,COOH) that is titrated with a strong base (e.., NaOH), what species (ions/molecules) DateLab Sec. Name Desk No. are present in the solution at the stoichiometrie point? b. For a weak acid (e.g. CH,COOH) that is titrated with a strong base (e.g. NaOH), what species (ions/molecules) are present in the solution at the halfway point in the titration toward the stoichiometric point? 2. Briefly explain how the pK, for a weak acid is determined in this experiment. 3. A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid chiometric point. Determine the molar concentration of the number of significant figures. weak acid solution. Express your answer to the correct 4. Data in the following table were obtained for the titration of a 0.297-g sample of a solid, monoprotic weak acid with a 0.150 M NaOH solution. Plot (at right) pH (ordinate) vs. Vou (abscissa). a. To determine the molar mass and the pK, of a solid, monoprotic weak acid, a titration of the weak acid with a standardized NaOH solution provided the following data in the table. Vsou added (mL) pH 1.96 0.00 2.00 4.00 7.00 10.00 12.00 14.00 2.22 2.46 2.77 3.06 3.29 3.60 4.26 11.08 11.67 12.05 16.00 17.00 18.00 20.00 25.00 12.40

Explanation / Answer

Q1 a). At stoichiometric point, species present is Na+ , CH3COO- , OH-

b) At halway point, species present is Na+ , CH3COOH, CH3COO- ,

Q2 In this experiment, a weak acid like CH3COOH is titrated with a strong base like NaOH. A graph is plotted where x-axis represents volume of NaOH added to solution and y-axis represents the pH of the solution at every point when NaOH is added. A sharp increse in pH value on graph indicates achievement of stoichiometric point. Volume of NaOH is noted of this point. Now this volume of NaOH is halved and pH at this halved volume of NaOH is found using the graph. This pH value is pKa of acid.

Q3. Moles of compound = (molarity of compound) x (volume of compound)

At stoichiometric point, moles of NaOH = moles of weak monoprotic acid

or (molarity of NaOH) x (volume of NaOH) = (molarity weak monoprotic acid) x (volume of acid)

or (0.0981 M) x (23.74 mL) = (molarity weak monoprotic acid) x (25.0 mL)

or molarity weak monoprotic acid = [(0.0981 M) x (23.74 mL)] / (25.0 mL)

or molarity weak monoprotic acid = 0.0932 M

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