AP FREE RESPONSE QUESTIONS – ACIDS/BASES
1996 A
HOCl ↔ OCl– + H+
Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The acid–dissociation constant, Ka, for the
reaction represented above is 3.2×10–8.
(a) Calculate the [H+] of a 0.14–molar solution of HOCl.
(b) Write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is dissolved in water and calculate
the numerical value of the equilibrium constant for the reaction.
(e) Household bleach is made by dissolving chlorine gas in water, as represented below.
Cl2(g) + H2O → H+ + Cl– + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar.
Answer:
(a) Ka =
[H + ][OCl- ]
= 3.2×10–8
[HOCl]
X = amount of acid that ionizes = [OCl–] = [H+]
(0.14 – X) = [HOCl] that remains unionized
–8
3.2×10 =
(b)
2
0. 14 – X
; X = 6.7×10–5 M = [H+]
NaOCl(s) + H2O → Na+(aq) + HOCl(aq) + OH–(aq)
Kb =
(e)
X
K w 11014

= 3.1×10–7
K a 3.2 108
1 mol H+ for every 1 mole of HOCl produced
[H+] ~ [HOCl] = 0.065 M
pH = – log (0.065) = 1.2
AP FREE RESPONSE QUESTIONS – ACIDS/BASES
ACID-BASE PRACTICE PROBLEMS
1.
pH of 0.025 M HF.
HF + H2O

H3O+ + FHF
I
C
E
Ka 
[ H 3O  ][ F  ]
[ HF ]
7.2 104 
2.
[ x][ x]
[0.025]
[ x][ x]
[0.025  x]

7.2 104 

0.0042  x  [ H3O ]

H3O+
0
+x
x
0.025
-x
0.025 - x
since x is small., 0.025-x = 0.025
pH   log [ H3O ]

HF
I
C
E
3.


pOH  0.699

pOH  14  pH

[CH 3 NH 3 ][OH  ]

[CH 3 NH 2 ]
[ x][ x]
[0.15]


pOH  14  pH

4.38 104 
[ x][ x]
[0.15  x]
0.0081  x  [OH  ]
pH  11.91


Na+
0
0.20
0.20
NaOH
0.20
-0.20
0

OH0
0.20
0.20
pH  13.301
CH3NH2 + H2O
I
C
E
4.38 104 
NO30
0.025
0.025
1st: NaOH (s)  Na+(aq) + OH-(aq), 2nd: OH- + H2O  H2O + OHthis is a strong base so there is 100% dissociation
pH of 0.075 mol/ 0.500 L CH3NH2.
Ka 
H3O+
0
0.025
0.025
0.025
-0.025
0
I
C
E
4.
pH  2.38
pH  1.60
pH of 0.20 mol / 1.00 L NaOH
pOH   log [0.20]


HNO3 + H2O 
H3O+ + NO3
this is a strong acid so there is 100% dissociation
pH of 0.17 M HNO3.
pH   log [0.025]
F0
+x
x
CH3NH2
0.15
-x
0.15 - x
 CH3NH3+ +
OH-
CH3NH3+
0
+x
x
OH0
+x
x
since x is small., 0.15-x = 0.15
pOH   log [OH  ]

pOH  2.09
AP FREE RESPONSE QUESTIONS – ACIDS/BASES
AN INCOMPLETE LIST OF TOPCS
EQUILIBRIUM (chap 13)
Know the definition of equilibrium.
Write equilibrium expressions Kc, Kp, Ka, Kb,
Know when prods or reacts are favored.
Apply le Chatelier's principle – particularly it’s impact on K or the conc of a molecule after an add/loss of another
molecule or a temp change or volume/pressure changes
Be able to use H in le Chatelier's principle and K.
Be able given initial values to get equilibrium values given K
Be able to use stoichiometricr ratiosn
Be able to calculate Q and know if a reaction is at equilibrium or if it will need to shift. Know which way it will shift.
Know that a catalyst effects both rates equally so has no impact on equilibrium.
Solve I.C.E. problems. Also know how to do ICE if your given amounts for two molecules and not just one.
Interconvert between Kp and Kc mathematically Kp = Kc(RT)n
Be able to calculate K values when the coefficients in the original equation are multiplied by a constant or when the equation
is reversed.
ACID - BASE (chap 14, 15)
Types of acid/base - Bronsted-Lowry, Arrhenius, Lewis: (Remember Lewis is the opposite of Bronsted-Lowry)
Know formulas of strong acids and bases
Know that strong acid/base makes a weak base/acid
Know that strong acids/bases are fully dissociated and weak are not.
Be able to write the dissociation equations for an acid/base in water.
Be able to calculate the Ka or Kb for either a strong or weak acid/base dissociation
Be able to calculate the Ka or Kb of the conjugate acid/base given Ka or Kb of the original because KaKb = Kw
Know the math to calculate the pH, pOH, [H+] and [OH-]
Be able to predict which molecule is a stronger acid based on Lewis Structure (see the course guide)
Be able to predict which molecule is a stronger acid/base based on Ka or Kb values or the value of the conjugate.
Be able to use ICE, acid/base dissociation eq’s, mole ratios and molarities to calc the amount of a species in solution