Review for Solving ph Problems:

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1 Review for Solving ph Problems: Acid Ionization: HA H 2 O A - H 3 O CH 3 COOH H 2 O CH 3 COO - H 3 O Base Ionization: B H 2 O BH OH - 1) Strong Acid complete dissociation [H ] is equal to original [HA] 2) Strong Base complete dissociation CH 3 NH 2 H 2 O CH 3 NH 3 OH - [OH - ] is equal to original [MOH] Where M represents a cation. Bases derived from group II metals will have the formula M(OH) 2 and will yield two hydroxides. 3) Weak acid when only weak acid is initially present. Solved by examining the equilibrium of acid ionization. What is the ph of a solution of M HOCl? The K a of HOCl is 3.5 x Set up a reaction table for the following ionization: HOCl H OCl - HOCl H OCl - initial final x x x Solve by using the equilibrium expression: K a = H [OCl ] [HOCl ] The assumption is made that x << than 0.1 M due to the small K a. Solving for x: x = 5.92 x 10-5 M and the ph is x [x] = 3.5 x 10 [0.100] 4) Weak Base when only weak base is initially present Solved by examining the equilibrium of base ionization. What is the ph of a 1.0 M methylamine solution. The K b of CH 3 NH 2 is 4.38 x Set up a reaction table for the following ionization: CH 3 NH 2 H 2 O CH 3 NH 3 OH - CH 3 NH 2 CH 3 NH 3 OH - initial final 1.0 x x x Solve by using the equilibrium expression: K b = CH 3NH 3 OH x x = = 4.38 x 10 4 CH 3 NH 2 The assumption is made that x << than 1.0 M due to the small K b. Solving for x: x = M = [OH - ] and the poh is The ph is

2 5) Salts A salt is added. A salt is the conjugate base of a weak acid with a cation or a conjugate acid of a weak base with an anion. a) conjugate base and cation What is the ph of a 0.30 M NaF solution? K a (HF) = 7.2 x 10-4 K b (F - ) = 1.4 x The first thing is to determine what is in solution. In this case, it is the base, F -. Set up a reaction table for base ionization, which is the reaction of base with water: F - H 2 O HF OH - F - HF OH - initial final 0.30 x x x Solve by using the equilibrium expression: K b = HF OH F The assumption is made that x << than 0.30 M due to the small K b. Solving for x: x = 2.05 x 10-6 M = [OH - ] and the poh is The ph is 8.31 b) conjugate acid and anion = x x 0.30 = 1.40 x What is the ph of a 0.5 M CH 3 NH 3 Cl solution. K b (CH 3 NH 3 ) = 4.38 x 10-4 K a (CH 3 NH 2 ) = 2.23 x The first thing is to determine what is in solution. In this case, it is the acid, CH 3 NH 3. Set up a reaction table for acid ionization, which is the reaction of acid with water: CH 3 NH 3 H 2 O CH 3 NH 2 H 3 O CH 3 NH 3 CH 3 NH 2 H 3 O initial final 0.5 x x x Solve by using the equilibrium expression: K a = H 3O [CH 3 NH 2 ] [CH 3 NH = 2.23 x = 3 ] The assumption is made that x << than 0.5 M due to the small K a. Solving for x: x = x 10-6 M and the ph is x [x] [0.500] 6) Common ion solution, commonly known as a buffer solution Remember that if more than one acid or an acid and a base are present in the same solution. The component with the largest equilibrium constant for the ionization with water (K a or K b ) will be most important in setting the ph and should be used to determine the ph. a) a weak acid in the presence of a salt that contains the conjugate base What is the ph of a solution 0.50 M CH 3 COOH and 0.30 M NaCH 3 COOH? K a (CH 3 COOH) = 1.8 x 10-5 Set up a reaction table for the acid ionization of CH 3 COOH: CH 3 COOH H 2 O CH 3 COO - H 3 O. This time however, the concentration of conjugate base is not zero. CH 3 COOH CH 3 COO - H 3 O initial final x 0.30 x x Solve by using the equilibrium expression: K a = H 3O [CH 3 COO ] = 1.8 x 10 5 = [CH 3 COOH ] The assumption is made that x << than 0.3 M due to the small K a. Solving for x: x = 3.0 x 10-5 and the ph is 4.52 x [0.3] [0.500]

3 NOTE: A common ion solution is a buffer solution and this can be easily solved by using the Henderson-Hasselbach equation. ph = pk a log([a - ]/[HA] Substituting values in: ph = 4.74 log(0.30/0.50) = 4.52 b) a weak base in the presence of a salt that contain the conjugate acid What is the ph of a solution that contains 0.25 M NH 3 (K b = 1.8 x 10-5 ) and 0.40 M NH 4 Cl? Set up a reaction table for the base ionization of NH 3 with water : NH 3 H 2 O OH - NH 4. This time however, the concentration of conjugate acid is not zero. NH 3 NH 4 OH - initial final x Solve by using the equilibrium expression: K b = NH 4 OH x 0.4 = = 1.80 x 10 5 The assumption is made that x << than 0.25 M due to the small K b. Solving for x: x = 1.1 x 10-5 M = [OH - ] and the poh is The ph is 9.05 NH 3 This problem could also be solved using the Henderson-Hasselbach equation. Note that the pk a of the conjugate acid (5.5 x in this case) must be used and that it should still be base over acid in the log function. K a for NH 4 = K w /K b = 5.56 x ph = pk a log([a - ]/[HA] Substituting values in: ph = 9.25 log(0.25/0.0.40) = ) Additions to solution in each of these cases, there are two steps to the solution. First one considers the stoichiometric reaction. Second, one considers an equilibrium Because volumes change, amounts must be converted from concentration to moles. Values in reaction table are thus in moles. When finished, to find [OH - ] or [H ], moles must be divided by new volume. a) strong base to a weak acid What is the ph of 50.0 ml 0.1 M CH 3 COOH after 10.0 ml of 0.1 M NaOH has been added? K a = 1.8 x 10-5 CH 3 COOH OH - CH 3 COO - H 2 O Reaction: The strong acid will cause the reaction to go to completion. The limiting reagent here is the hydroxide ion. CH 3 COOH OH - CH 3 COO initial change final At this point, there is an acid and a base in solution. The K a for the acid is 1.8 x 10-5 while the K b for the conjugate base is 5.56 x Since the equilibrium constant for the acid is largest, the acid ionization with water, as opposed to the base ionization with water is used to determine the ph. CH 3 COOH H 2 O CH 3 COO - H 3 O The solution for this will be first shown using a reaction table. Values are entered as concentrations: mol/0.06 L and mol/0.06l.

4 Equilibrium with a reaction table: CH 3 COOH CH 3 COO - H 3 O initial final x x x x is small compared to acid concentration and can be dropped. K a = 1.8 x 10-5 = (x)( )/(0.0667) x = 7.20 x 10-5 = [H ] and ph is 4.14 Equilibrium using the Henderson-Hasselback equation: Since this is a mixture of an acid and the conjugate base, which is a buffer, this could be solved using the Henderson Hasselbach equation. ph = pk a log([a - ]/[HA] = 4.74 log(0.001/0.060)/(0.004/0.06) = =4.14 b) strong acid to a weak base What is the ph of 100 ml M NH 3 after 20 ml of 0.10 M HCl has been added. K b = 1.8 x 10-5 NH 3 HCl NH 4 Cl - Once again, because volumes change, amounts must be converted from concentration to moles. Values in reaction table are thus in moles. When finished, to find [OH - ] or [H ], moles must be divided by new volume. Cl - need not be included as it is a spectator ion and is the anion of a strong acid. Reaction: The strong base will cause the reaction to go to completion. In this case, the limiting reagent is the acid. NH 3 H NH 4 Cl - initial change final At this point, there is an acid and a base in solution. The K a for the conjugate acid (NH 4 ) is 5.56 x -10 while the K b for the base is 1.80 x Since the equilibrium constant for the base is largest, the base ionization with water, as opposed to the acid ionization with water is used to determine the ph. NH 3 H 2 O NH 4 OH - The solution for this will be first shown using a reaction table. Values are entered as concentrations: mol/0.120 L and mol/0.120 L. NH NH 4 OH - initial final x x x K b = 1.80 x 10-5 = (x)( )/(0.025) and x = [OH - ] = 2.7 x 10-5 poh = 4.57 and ph = 9.42 Equilibrium using the Henderson-Hasselback equation: Since this is a mixture of an acid and the conjugate base, which is a buffer, this could be solved using the Henderson Hasselbach equation. Remember to use acid terms. ph = pk a log([a - ]/[HA] = 9.25 log(0.025)/( ) = = 9.43

5 c) strong base to a buffer solution A problem of this type would have a reaction followed by an equilibrium. The reaction would be solved as described in 6 above. The equilibrium would then be solved as in 7, part a above. d) strong acid to a buffer solution A problem of this type would have a reaction followed by an equilibrium. The reaction would be solved as described in 6 above. The equilibrium would then be solved as in 7, part b above.

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