lec18_whc

# lec18_whc - Exam Feb. 26; Room Assignments Room 1051 SLB...

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Exam Feb. 26; Room Assignments ALL OTHER SECTIONS ( B01, B03, B05, B08, B10, B11, B13, B18, B19, BR0 ) Meet in Sciences Lecture Hall Edna B06 2076 SLB Edna B02 2068 SLB James B15 2059 SLB James B09 2051 SLB Mateo B16 1076 SLB Mateo B12 1068 SLB Troy B20 1075 SLB Troy B04 1067 SLB Kelly B17 1059 SLB Kelly B07 1051 SLB TA Section Room

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Last Lecture • pH calculations for weak acids and bases in water • Relative strengths of acids/bases • multiprotic acids and pH • percent ionized
Calculate the pH of a 0.4 M solution of phosphoric acid (H 3 PO 4 ). pK a1 = 2.21; pK a2 = 7.21; pK a3 = 12.86 Example 16E– pH of Phosphoric Acid

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H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4 - Initial 0.4 mol/L 0 0 After 0.4 mol/L -x +x x [H 3 O + ] = 0.047 mol/L pH = - log[H 3 O + ] = 1.30 Solution 16E – pH of Phosphoric Acid [H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] K a1 = x 2 (0.4 – x ) = = 10 -2.21 = 7.1 × 10 -3 Æ x = 0.047 M As we have seen, H 3 PO 4 is effectively a monoprotic acid! No approx should be made here!
Sulfuric Acid - A diprotic acid. H 2 SO 4 + H 2 O H 3 O + + HSO 4 - HSO 4 - + H 2 O H 3 O + + SO 4 2- K a1 = very large K a2 = 1.1x10 -2 • No need for a double equilibrium calculation because can simplify when K a1 is very large and K a2 small! • i.e. complete first ionization (see example 16-10) the first dissociation is a strong-acid equilibria; the second dissociation must be treated as a weak-acid equilibrium.

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What is the concentration of sulfuric acid required to give a pH of 0.68? K a1 = large; K a2 = 1.2 x 10 -2 H 2 SO 4 + H 2 O H 3 O + + HSO 4 ¯ Example 16F – Sulfuric Acid HSO 4 ¯ + H 2 O H 3 O + + SO 4 2-
Solution 17F – Sulfuric Acid HSO 4 ¯ + H 2 O H 3 O + + SO 4 2- Initial x x 0 Change -y +y +y Equil x-y x+y y Then, the second step starts from 'x' concentration: H 2 SO 4 + H 2 O H 3 O + + HSO 4 ¯ 00 x Unknown initial conc = x, so… strong Init +x +x -x Change xx 0 After 2 () 1.2 10 xy y + =⋅ [H 3 O + ] [SO 4 2- ] [HSO 4 - ] K a2 = = It is a strong acid .

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x = 0.209-y = 0.196 mol/L [H 3 O + ] = 10 -pH = 10 -0.68 = 0.209 = x+y so 0.209 - y = x or: 0.209 - 2y = x-y (agreed?) Insert x-y = 0.209 – 2y into the equilibrium expression: Solution 17F – Sulfuric Acid 2 2 10 2 . 1 2 209 . 0 209 . 0 = = y y K a 2 10 266 . 1 × = y 2 () 1.2 10 xy y + =⋅ [H 3 O + ] [SO 4 2- ] [HSO 4 - ] K a2 = =
Degree of Ionization HA + H 2 O H 3 O + + A - Degree of ionization = [H 3 O + ] from HA [HA] originally Percent ionization = [H 3 O + ] from HA [HA] originally × 100% K a = [H 3 O + ][A - ] [HA] K a = n H 3 O + A - n HA n 1 V

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An aqueous solution of acetic acid has pH of 2.87. From pK a =4.74 calculate the percent ionization! Example 16D – Percent Ionization HC 2 H 3 O 2 + H 2 OH 3 O + + C 2 H 3 O 2 ¯ Percent ionization = [H 3 O + ] [HC 2 H 3 O 2 ] originally × 100% Before we start: pH= -log[H 3 O + ] = 2.87 Æ [H 3 O + ] = 10 -2.87 = 1.35 × 10 -3 mol/L pK a = -log K a = 4.74 Æ K a = 10 -4.74 = 1.8 × 10 -5 Similar example text p 681