Ch18_Titrations - 1 Titrations 2 Acid-Base Titrations...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 Titrations 2 Acid-Base Titrations Acid-Base Titrations 3 Acid-Base Titrations Acid-Base Titrations pH pH Titrant volume, mL Titrant volume, mL 4 Acid-Base Titrations Acid-Base Titrations 5 QUESTION: You titrate 100. mL of a QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at equivalence point? What is the pH at equivalence point? Equivalence Equivalence point point Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point. Additional NaOH is added. pH rises as equivalence point is approached. Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly. Benzoic acid + NaOH Acid-Base Reactions Section 18.4 QUESTION: You titrate 100. mL of a 0.025 M QUESTION: solution of benzoic acid with 0.100 M NaOH solution of benzoic acid with 0.100 M to the equivalence point. What is the pH of to the equivalence point. What is the pH of the final solution? the final solution? + + Bz + H O HBz + NaOH ---> Na+ + Bz-- + H2O HBz 2 pH of solution of pH of solution of benzoic acid, a benzoic acid, a weak acid weak acid C6H5CO2H = HBz Benzoate ion = Bz- 6 Acid-Base Reactions 7 Acid-Base Reactions Section 18.4 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? STOICHIOMETRY PORTION STOICHIOMETRY PORTION 1. Calc. moles of NaOH req’d 1. Calc. Calc. (0.100 L HBz)(0.025 M) = 0.0025 mol HBz (0.100 L HBz)(0.025 This requires 0.0025 mol NaOH This requires 2. Calc. volume of NaOH req’d Calc. 2.Calc. 0.0025 mol (1 L / 0.100 mol) = 0.025 L mol) 0.0025 25 mL of NaOH req’d 25 Strategy — find the conc. of the find the conc. conjugate base Bz-- in the solution AFTER in the solution AFTER conjugate base the titration, then calculate pH. the titration, then calculate pH. This is a two-step problem This is a two-step problem Bz-- + H2O ¸ HBz + OH-Bz HBz 2 1. stoichiometry of acid-base reaction of acid-base reaction 1. + 2. equilibrium calculation 2. + Kb = 1.6 x 1.6 9 Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? The product of the titration of benzoic acid, The product of the titration of benzoic acid, the benzoate ion, Bz--,, is the conjugate base the benzoate ion, is the conjugate base of a weak acid. of a weak acid. Therefore, final solution is basic. Therefore, final solution is basic. ¸ 8 10 -10 10 Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? STOICHIOMETRY PORTION STOICHIOMETRY PORTION 25 mL of NaOH req’d 25 3. Moles of Bz-- produced = moles HBz = produced = moles 3. Moles of 0.0025 mol 0.0025 4. Calc. conc. of Bz-4. Calc. conc. Calc. There are 0.0025 mol of Bz-- in a TOTAL in a TOTAL There are 0.0025 SOLUTION VOLUME of 125 mL SOLUTION VOLUME [Bz-] = 0.0025 mol / 0.125 L = 0.020 M 11 12 Acid-Base Reactions Acid-Base Reactions Acid-Base Reactions Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point? QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point? Equivalence Point Equivalence Point Most important species in solution is benzoate Most important species in solution is benzoate ion, Bz--,, the weak conjugate base of benzoic ion, the weak conjugate base of benzoic acid, HBz. acid, HBz. Bz-- + H2O ¸ HBz + OH-Kb = 1.6 x 10-10 Bz + H2O ¸ HBz Kb = 1.6 x 10-10 [Bz--] [HBz] [OH--] [HBz] HBz] [OH ] initial 0.020 0 0 initial 0.020 0 0 change -x +x +x change -x +x +x equilib 0.020 - x x x equilib 0.020 - x x x Equivalence Point Equivalence Point Most important species in solution is benzoate ion, Bz--,, Most important species in solution is benzoate ion, Bz the weak conjugate base of benzoic acid, HBz. the weak conjugate base of benzoic acid, HBz. HBz. Bz-- + H2O ¸ HBz + OH-Kb = 1.6 x 10-10 Bz + H2O ¸ HBz + OH Kb = 1.6 x 10-10 Kb 1.6 x 10-10 = x2 0.020 - x x = [OH-] = 1.8 x 10 -6 pOH = 5.75 -----> pH = 8.25 pOH 13 13 QUESTION: You titrate 100. mL of a QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point? What is the pH at half-way point? Half-way Half-way point point 14 Acid-Base Reactions Acid-Base Reactions QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point? HBz + H2O ¸ H3O+ + Bz-HBz + H2O ¸ H3O+ + Bz Ka = 6.3 x 10 -5 Ka = 6.3 x 10 -5 Both HBz and Bz-- are present are present Both [H3 O ] = [HBz] •K a [Bz- ] At the half-way point, [ HBz] = [Bz-] HBz] [Bz Therefore, [H 3O+] = Ka = 6.3 x 10 -5 6.3 pH = 4.20 ...
View Full Document

Ask a homework question - tutors are online