Acid-Base Titrations End point and equivalence point

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Presentation transcript:

Acid-Base Titrations End point and equivalence point Calculating the concentration of unknown acid or base Choosing an indicator dye Titration of a strong acid by a strong base Titration of a weak acid by a strong base Calculation of the titration curve - buffer region of a titration curve Dependence of titration curve on pKa Using the H-H equation to calculate pH

Acid-Base Titrations strong acid titrated with strong base pH At the equivalence point, mol H+ = mol OH- strong acid titrated with strong base strong base titrated with strong acid

Titrations - Determination of the concentration of an unknown acid or base A burette is used to add a precise volume of acid (or base) to the unknown solution. At the end-point of the reaction, the indicator changes color. 1. The flask contains the unknown acid 2. Base is added drop by drop to the end-point. 3. The indicator (in this case phenolphthalein) turns pink at the end-point.

Calculating the Concentration of an Unknown Acid or Base end point: indicator changes color equivalence point: moles [H+] = moles [OH-] When the indicator is properly chosen: - end point = equivalence point within experimental error. Example: 0.231 M HCl titrates 50.00 mL unknown base. 45.27 mL HCl is added at the equivalence point. (Molarity * Volume)acid = (Molarity * Volume)base [base] = ( 0.321 M * 0.04527 L )/ .05000 L = 0.291 M

The Common Indicator Dyes

Titration of a Weak Acid with a Strong Base pH = pKa + log10 [base] [acid] CH3COOH = CH3COO- + H+ pKa = 4.75 When pH = 4.75, [CH3COOH] = [CH3COO-] This is the center of the buffering region.

Exercises: Calculate the Titration Profile pH = pKa + log10 [base] [acid] What is the pH when 90.0 mL of NaOH has been added? Ans: 5.70 What is the pH when 30.0 mL of NaOH has been added? Ans: 4.38

Dependence of the Titration Curve on Acid Strength On this diagram, note that the pH = pKa at the midpoint of the buffering region for each weak acid.

Titration of a Weak Base by a Strong Acid pH = pKa + log10 [base] [acid] NH4+ = NH3 + H+ pKa = 9.25 When pH = 9.25, [NH4+] = [NH3]

Problem Calculate the pH of a 500 mL solution prepared from: 0.050 mol of acetic acid and 0.020 mol sodium acetate. HAc = H+ + Ac- pKa = 4.75 0.020 / 0.50 [Ac-] pH = pKa + log10 [HAc] 4.75 0.050 / 0.50 pH = 4.35

Problem 10.45 (2) (b) Suppose that 0.010 mol NaOH is added to the buffer of part (a). What is the pH? HAc + OH- = H2O + Ac- pKa = 4.75 (0.020 + 0.010) / 0.50 [Ac-] pH = pKa + log10 [HAc] (0.050 - 0.010) / 0.50 4.63 4.75 As long as the buffer capacity is not exceeded, the change of pH is small, in this case, 4.35 to 4.63