Chemical Equilibrium Calculations Lecture

Chemical equilibrium calculations Strong acid (CPS lesson) HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) Chemical equilibrium calculations HA Initial ca Revised initial initial, Ka > 1 0 Change +x Equilibrium x H3O+ 10‐7 10‐7 + ca ≈ ca –x ca – x ≈ ca A‐ 0 ca –x ca – x ≈ ca 10‐7 + ca ≈ ca because Ka > Kw ca – x ≈ ca because Ka > 1 (HA) = x = ca2/Ka Strong acid example Weak acid (CPS lesson) HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) Ka of HA is 1 x 10+6. What is the pH and (HA) in 0.001 M HA? (H3O+) = ca = 0.001 0 001  pH = 3.0 30 2 +6 2 (HA) = x = ca /Ka = (0.001) /(1 x 10 ) = 1 x 10‐12 !!! HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) HA Initial ca Change ‐x Equilibrium ca – x ≈ ca H3O+ 10‐7 +x 10‐7 + x ≈ x A‐ 0 +x x 10‐7 + x ≈ x because Ka > Kw ca – x ≈ ca because 1 > Ka (H3O+) = x = √(Ka ca) Copyright © 2009 Dan Dill | Department of Chemistry | Boston University Weak acid example Conjugate base (CPS lesson) HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) Ka of HA is 1 x 10‐7. What is the pH of 0.001 M HA? H2O(l) + A‐(aq)  HA(aq) + OH‐(aq) (H3O+) = √(Ka ca) = √(1 x 10‐10) = 1 x 10‐5 pH = 5.0 A‐ Initial cb Change ‐x Equilibrium cb – x ≈ cb HA 0 +x x OH‐ 10‐7 +x 10‐7 + x ≈ x 10‐7 + x ≈ x because Kb = Kw /Kb > Kw cb – x ≈ cb because 1 > Kb (OH‐) = x = √(Kb cb) 1 Chemical equilibrium calculations Conjugate base example Weak acid + conjugate base H2O(l) + A‐(aq)  HA(aq) + OH‐(aq) Ka of HA is 1 x 10‐7. What is the pH of 0.001 M NaA? HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) Kb = Kw/Ka = 1 x 10‐7 (OH‐) = √(Kb cb) = √(1 x 10‐10) = 1 x 10‐5 HA Initial ca Change ‐x Equilibrium ca – x ≈ ca H3O+ 10‐7 +x 10‐7 + x ≈ x A‐ cb +x cb + x ≈ cb pOH = 5.0  pH = 9.0 10‐7 + x ≈ x because Ka > Kw ca – x ≈ ca and cb + x ≈ cb because 1 > Ka (H3O+) = x = Ka (ca/cb) Weak acid + conjugate base Conjugate base HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) Ka of HA is 1 x 10‐5 Equal volumes of 0.002 M HA and 0.01 M NaA are combined. What is the pH? H2O(l) + A‐(aq)  HA(aq) + OH‐(aq) ca = 0.001 M, cb = 0.005 M (H3O+) = x = Ka (ca/cb) = 2 x 10‐6 A‐ Initial cb Change ‐x Equilibrium cb – x ≈ cb HA 0 +x x OH‐ 10‐7 +x 10‐7 + x ≈ x pH = 6 – 0.3 = 5.7 10‐7 + x ≈ x because Kb = Kw /Ka > Kw cb – x ≈ cb because 1 > Kb (OH‐) = x = √(Kb cb) Conjugate base example Weak acid + conjugate base = buffer H2O(l) + A‐(aq)  HA(aq) + OH‐(aq) Ka of HA is 1 x 10‐7. What is the pH of 0.001 M NaA? HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) (OH‐) Kb = Kw/Ka = 1 x 10‐7 = √(Kb cb) = √(1 x 10‐10) = 1 x 10‐5 pOH = 5.0  pH = 9.0 HA Initial ca Change ‐x Equilibrium ca – x ≈ ca H3O+ 10‐7 +x 10‐7 + x ≈ x A‐ cb +x cb + x ≈ cb 10‐7 + x ≈ x because Ka > Kw ca – x ≈ ca and cb + x ≈ cb because 1 > Ka (H3O+) = x = Ka (ca/cb) 2 Chemical equilibrium calculations Weak acid + conjugate base = buffer Buffers resist change in pH (demo) HA(aq) + H2O(l)  H3O+(aq) + A‐(aq) Ka of HA is 1 x 10‐5 Equal volumes of 0.002 M HA and 0.01 M NaA are combined. What is the pH? ca = 0.001 M, cb = 0.005 M (H3O+) = x = Ka (ca/cb) = 2 x 10‐6 pH = 6 – 0.3 = 5.7 Buffers resist change in pH Add strong base to buffer Added strong base (say, OH‐) is gobbled up … ↓HA(aq) + OH‐(aq)  H2O(l) + ↑A‐(aq) ca lowered, lowered cb raised, raised ca/cb lowered 1 L buffer, ca = cb = 1.00 M, Ka = 1 x 10‐5 , pH = 5.00 Add 100. mL of 0.100 M NaOH ↓HA(aq) + OH‐(aq)  H2O(l) + ↑A‐(aq) HA  1.00 mol – 0.010 mol = 0.99 mol A‐ 1.00 mol + 0.100 mol = 1.01 mol Added strong acid (say, HCl) is gobbled up … HCl(aq) + ↓A‐(aq)  ↑HA(aq) + Cl‐(aq) cb lowered, ca raised, ca/cb rasied ca = 0.99 mol/1.10 L, cb = 1.01 mol/1.10 L ca/cb  0.99/1.01, pH  5.01 (tiny change!) Add strong acid to buffer Add strong acid/base to water 1 L buffer, ca = cb = 1.00 M, Ka = 1 x 10‐5 , pH = 5.00 Add 100. mL of 0.100 M HCl HCl(aq) + ↓A‐(aq)  ↑HA(aq) + Cl‐(aq) HA  1.00 mol + 0.010 mol = 1.01 mol A‐ 1.00 mol ‐ 0.100 mol = 0.99 mol ca = 1.01 mol/1.10 L, cb = 0.99 mol/1.10 L ca/cb  1.01/0.99, pH  4.99 (tiny change!) 1 L of water, Ka = 1 x 10‐14 , pH = 14.00 Add 100. mL of 0.100 M HCl (H3O+) = 0.010 mol/1.10 L = 0.0091 pH = 2.04 (huge change!) Add 100. mL of 0.100 M NaOH (OH‐) = 0.010 mol/1.10 L = 0.0091 pOH = 2.04, pH = 11.96 (huge change!) ca = 1.01 mol/1.10 L, cb = 0.99 mol/1.10 L ca/cb  1.01/0.99, pH  4.99 3 Chemical equilibrium calculations Solubility equilibria Common ion effect MX2(s)  M2+(aq) + 2 X‐(aq) MX2(s)  M2+(aq) + 2 X‐(aq) MX2 Initial excess Change −x Equilibrium excess Initial M2+ 0 +x x X‐ 0 +2x 2x M2+ 0 +x x MX2 Initial excess Change −x Equilibrium excess X‐ cion +2x cion + 2x ≈ cion Ksp = (M2+)(X‐)2 = (x)(2x)2 = 4 x3 Ksp = (M2+)(X‐)2 = (x)(cion)2 Large cion makes x smaller Will precipitation occur? Calculate precipitation MX2(s)  M2+(aq) + 2 X‐(aq) MX2(s)  M2+(aq) + 2 X‐(aq) MX2 0 M2+ cM X‐ cX Is (M2+)(X‐)2 = (cM)(cion)2 > Ksp ? If no, no precipitation. If yes, precipitate will form. Initial Revised Change Equilibrium MX2 0 excess −x excess M2+ cM 0 +x x X‐ cX > 2 c M cX − 2 cM +2x cX − 2 cM + 2 x ≈ cX − 2 cM Ksp = (M2+)(X‐)2 = (x)(cX − 2 cM )2 4