Chapters 6 and 8. Systematic Treatment of Equilibrium

Transcription

1 Chapters 6 and 8 Systematic Treatment of Equilibrium Equilibrium constants may be written for dissociations, associations, reactions, or distributions. Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) 1

2 Types of Chemical Equations Balanced Chemical Equation Charge Balance Equation Mass Balance Equation Charge Balance The sum of the positive charges in solution equals the sum of the negative charges in solution. n => charge, C =>concentration Cation n ici = i j n C j anion j 2

3 Mass Balance The sum of the amounts of all species in a solution containing a particular atom (or group of atoms) must equal the amount of that atom (or group) delivered to the solution. EXAMPLE: Write the mass-balance & charge balance equations for the system formed when a M NH 3 solution is saturated with AgCl. AgCl (s) <=> Ag + (aq) + Cl-1 (aq) Ag + (aq) + 2 NH 3(aq) <=> [Ag(NH 3 ) 2 + ] (aq) NH 3(aq) + H 2 O (l) <=> NH 4 + (aq) + OH - (aq) 3

4 AgCl (s) <=> Ag + (aq) + Cl- (aq) Ag + (aq) + 2 NH 3(aq) <=> [Ag(NH 3 ) 2 + ] (aq) NH 3(aq) + H 2 O (l) <=> NH + 4 (aq) + OH - (aq) Mass Balance Equations: [Ag+] + [Ag(NH 3 ) 2+ ] = [Cl - ] C NH3 = [NH 3 ] + [NH 4+ ] + 2 [Ag(NH 3 ) 2+ ] = M AgCl (s) <=> Ag + (aq) + Cl- (aq) Ag + (aq) + 2 NH 3(aq) <=> [Ag(NH 3 ) 2 + ] (aq) NH 3(aq) + H 2 O (l) <=> NH + 4 (aq) + OH - (aq) Charge Balance Equation: [Cl -1 ] + [OH -1 ] = [NH 4+ ] + [Ag(NH 3 ) 2+ ] + [Ag + ] 4

5 Systematic Approach to Equilibrium Problems 1. Balanced chemical equations 2. What quantity is being sought. 3. Equilibrium-constant expressions 4. Mass-balance expressions for the system 5. Charge balance expression 6. Count equations vs. unknowns. If more unknowns than equations, seek additional equations, or make appropriate approximations. Systematic Approach to Equilibrium Problems 7. Make suitable approximations to simplify the algebra. 8. Solve algebraic equations. 9. Check validity of assumptions. 5

6 EXAMPLE: Write the equation of mass balance for a M solution of acetic acid. EXAMPLE: Write the equations of mass balance for a 1.00X10-5 M [Ag(NH 3 ) 2 ]Cl solution. 6

7 EXAMPLE: Write the equation of charge balance for a solution of H 2 S. EXAMPLE: Write the equation of charge balance for a solution of 0.1 M Na 2 HPO 4. 7

8 EXAMPLE: Calculate the ph of a M solution of acetic acid in water. EXAMPLE: Calculate the molar solubility of Fe(OH) 2 in water. 8

9 EXAMPLE: Calculate the molar solubility of Fe(OH) 2 in water. Step 1. Fe(OH) 2(s) <=> Fe +2 (aq) + 2 OH- (aq) 2H 2 O <=> H 3 O + (aq) + OH- (aq) Step 1. Fe(OH) 2(s) <=> Fe +2 (aq) + 2 OH- (aq) 2H 2 O <=> H 3 O + (aq) + OH- (aq) Step 2. let x = molar solubility = [Fe +2 ] 9

10 Step 1. Fe(OH) 2(s) <=> Fe +2 (aq) + 2 OH- (aq) 2H 2 O <=> H 3 O + (aq) + OH- (aq) Step 2. let x = molar solubility = [Fe +2 ] Step 3. K sp = [Fe +2 ][OH - ] 2 = 8 X M 3 K w = O + ][OH - ] = 1 X M 2 Step 1. Fe(OH) 2(s) <=> Fe +2 (aq) + 2 OH- (aq) 2H 2 O <=> H 3 O + (aq) + OH- (aq) Step 2. let x = molar solubility = [Fe +2 ] Step 3. K sp = [Fe +2 ][OH - ] 2 = 8 X M 3 (1) K w = O + ][OH - ] = 1 X M 2 (2) Step 4. Mass Balance Equation [OH - ] = 2 [Fe +2 ] + O + ] (3) 10

11 Step 1. Fe(OH) 2(s) <=> Fe +2 (aq) + 2 OH - (aq) 2H 2 O <=> H 3 O + (aq) + OH - (aq) Step 2. let x = molar solubility = [Fe +2 ] Step 3. K sp = [Fe +2 ][OH - ] 2 = 8 X M 3 (1) K w = O + ][OH - ] = 1 X M 2 (2) Step 4. Mass Balance Equation [OH - ] = 2 [Fe +2 ] + O + ] (3) Step 5. Charge Balance Equation 2 [Fe +2 ] + O + ] = [OH - ] (4) Step 1. Fe(OH) 2(s) <=> Fe +2 (aq) + 2 OH- (aq) 2H 2 O <=> H 3 O+ (aq) + OH- (aq) Step 2. let x = molar solubility = [Fe +2 ] Step 3. K sp = [Fe +2 ][OH - ] 2 = 8 X M 3 (1) K w = O + ][OH - ] = 1 X M 2 (2) Step 4. Mass Balance Equation [OH - ] = 2 [Fe +2 ] + O + ] (3) Step 5. Charge Balance Equation 2 [Fe +2 ] + O + ] = [OH - ] (4) Step 6. 3 unique equations 3 unknowns exact solution possible 11

12 K sp = [Fe +2 ][OH - ] 2 = 8 X M 3 (1) K w = O + ][OH - ] = 1 X M 2 (2) [OH - ] = 2 [Fe +2 ] + O + ] (3) Step 7. Simplify Algebra assume [OH - ] ~ 2 [Fe +2 ] i.e. 2 [Fe +2 ] >> O + ] K sp = [Fe +2 ][OH - ] 2 = 8 X M 3 (1) K w = O + ][OH - ] = 1 X M 2 (2) [OH - ] = 2 [Fe +2 ] + O + ] (3) assume [OH - ] ~ 2 [Fe +2 ] i.e. 2 [Fe +2 ] >> O + ] Step 8. Solve Algebraic Expressions Using equation (1) K sp = [Fe +2 ][OH - ] 2 = [Fe +2 ](2[Fe +2 ]) 2 = 8 X M 3 [Fe +2 ] = 3 (8 X /4)M 3 = 6 X 10-6 M 12

13 Step 8. Solve Algebraic Expressions [Fe +2 ] = 6 X 10-6 M Step 9. Check Assumption [OH - ]~ 2 [Fe +2 ] = 2(6 X 10-6 M) = 1.2 X 10-5 M ~ 1 X 10-5 M O + ] = K w / [OH - ] = (1 X M 2 )/(1 X 10-5 M) = 1 X 10-9 M 2 [Fe +2 ] >> O + ] Calculation of Chemical Equilibrium Concentration calculation (1) ph of Weak Acid ph 0.1 M HAc. A. Reaction equations HA +H 2 O A - +H 3 O + 2H 2 O H 3 O + +OH - 13

14 A. Reaction equations HA +H 2 O A - +H 3 O + 2H 2 O H 3 O + +OH - B. Equilibrium Constant Ka = [A ] O + ] [HA] Kw = O + ][OH ] = C. Mass Balance Equations C = [HA] + [A ] D. Charge Balance E. Conversion Known item: C O + ] = [A ] + Kw O + ] [A ] = O + ] = CKa Ka + O + ] CKa Ka + O + ] + Kw O + ] O + ] = [A ] + [OH ] O + ] = CKa Ka + O + ] + CKa > 20Kw, C /Ka > 400 O + ] = CKa Kw O + ] 14

15 Type Keq Formula Example Strong acids Keq= O + ] = Ca HCl, HNO 3, H 2 SO 4, HClO4 Strong Bases Weak Acids Keq= Ka = O + ][A] HA [OH-]=Cb [ H 3 O + ]= CKa NaOH, KOH HAc, HCOOH, HClO Weak Bases Kb = [OH ][B] [BOH ] [OH ] = CK b NH4OH Buffer Ka = O + ][A] HA O + ] = Ka C HA C NaA HAc-NaAc, NH 4+ - NH 3 H 2 O Amphiprotic salt(naha) Ka 1 = O + ][HA] H 2 A [ H 3 O + ]= Ka 1 Ka 2 NaHCO 3, NaHPO 4, NaH 2 PO 4, Polyprotic acids Ka 2 = O + ][A 2 ] HA Same as weak acid 0.1 M H 2 SO 4 Ka 2 =1.2X10-2 H 2 SO 4 HSO 4 - +H + HSO 4 - [HSO 4- ]+[SO 4 2- ]=C SO H + Ka 2 = [H + ][SO 4 2 ] [HSO 4 ] [H + ]= [HSO 4- ]+2[SO 4 2- ] [H + ]= C+ [SO 4 2- ] [HSO 4 ] = [H + ][SO 4 2 ] Ka 2 C = [H + ][SO 2 4 ] + [SO 2 4 ] Ka 2 C = ( [H + ] + 1)[SO 2 4 ] Ka 2 [SO 2 CKa 4 ] = 2 [H +] + Ka 2 CKa [H + ] = C + 2 [H + ] + Ka 2 15