 # 8.7 Acid-Base Titration Learning Goals … … determine the pH of the solution formed in a neutralization reaction.

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8.7 Acid-Base Titration Learning Goals … … determine the pH of the solution formed in a neutralization reaction

Titrations are used to determine the concentration of an acid or a base The equivalence point is the point in the titration when the acid and the base completely react with each other If you know the volumes of both solutions at the equivalence point and the concentration of one of them, you can calculate the unknown concentration.

Titrations Involving Strong Acids and Strong Bases Calculate the pH of HNO 3 before any base is added. pH = - log [H+] = - log (0.200) = 0.699 Since HNO 3 is a strong acid, [H+] = [HNO 3 ] = 0.200 M

Calculate the pH of the solution after 50.0 mL of NaOH is added. pH = - log [H+] = - log (0.05) = 1.30 HNO 3 + NaOH  H 2 O + NaNO 3 nCVnCV 0.005 mol0.010 mol 0.200 M 0.050 L 0.100 M 0.050 L LR initial used XS 0.010 mol 0.005 mol [H+] = [HNO 3 ] = 0.005 mol 0.100 L = 0.05 M XS

What volume of NaOH will have been added at the equivalence point? What is the pH at the equivalence point? HNO 3 + NaOH  H 2 O + NaNO 3 nCVnCV 0.010 mol 0.200 M 0.050 L 0.100 M 0.100 L At the equivalence point all the acid and base have been neutralized. Since this is a titration between a strong acid and a strong base, the salt produced (NaNO 3 ) will have no effect on pH. The pH at the equivalence point will be 7.0 (neutral)

The equivalence point is the middle of the steep rise that occurs in a titration curve. The endpoint of a titration occurs when the indicator changes colour, which happens over a range of about 2 pH units. The pH changes rapidly near the equivalence point. Chemists use indicators to observe the endpoint of the titration. Some common indicators are the following: colourless  Phenolphthalein(colourless to pink) pH 8.2 – 10.0  Methyl Red (red to yellow) pH 4.3 – 6.2

Titration Curve for a Strong Acid with a Strong Base: These titrations have a pH of 7 at equivalence (neutral)

Titration of a Weak Acid with a Strong Base: Recall: The anion from the salt produced will react with water Ex 1) 100 mL of 0.1M cyanic acid (HCN) is titrated with 100 mL of 0.1M KOH. Calculate the pH at equivalence. 0.1 M HCN + KOH  H 2 O + KCN nCVnCV 0.100 L 0.01 mol 0.20 L 0.05 M CN - + H 2 O  HCN + OH - iceice 0.05 M0 0 - x+ x 0.05 - x xx

K b for CN - = K w = 1x10 -14 = 1.6x10 -5 K a 6.2x10 -10 K a for HCN = 6.2x10 -10 Can we simplify? 0.05/Kb > 100YES x 2 = 1.6x10 -5 (0.05) x 2 = 8.0x10 -7 x = 8.94x10 -4 [OH-] = x = 8.94x10 -4 pOH = - log [OH-] = 3.05 pH = 10.95 x 2 = 1.6x10 -5 (0.05-x)

These titrations have pH values that are greater than 7 at equivalence. The conjugate bases of weak acids will react with water At the equivalence point, the weak acid is neutralized by the base but the conjugate base is present, therefore increasing the pH

0.2 M NH 3 + HCl  NH 4 + + Cl - nCVnCV 0.020 L 0.004 mol 0.040 L 0.1 M Titration for a Weak Base with a Strong Acid Recall: The cation from the salt produced will react with water Ex 1) 20 mL of 0.2M NH 3 is titrated against 0.2M HCl. Calculate the pH at equivalence. NH 4 + + H 2 O  NH 3 + H 3 O + iceice 0.1 M0 0 - x+ x 0.1 - x xx

K a for NH 4 + = K w = 1x10 -14 = 5.56x10 -10 K b 1.8x10 -5 K b for NH 3 = 1.8x10 -5 Can we simplify? 0.1/Ka > 100YES x 2 = 5.56x10 -10 0.1 x = 7.45x10 -6 [H 3 O+] = x = 7.45x10 -6 pH = - log [H 3 O+] pH = 5.13 x 2 = 5.56x10 -10 0.1-x

These titrations have pH values that are less than 7 at equivalence. The conjugate acids of weak bases will react with water At the equivalence point, the weak base is neutralized by the acid but the conjugate acid is present, therefore decreasing the pH

HOMEWORK p547 #1,2 p554 #1,2 p557 #1-8 Self Check How prepared am I to start my homework? Can I … … determine the pH of the solution formed in a neutralization reaction

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