± pH and Percent Ionization of a Weak Base The degree to which a weak base disso
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Question
± pH and Percent Ionization of a Weak Base
The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B
B(aq)+H2O(l)BH+(aq)+OH(aq)
this constant is given by
Kb=[BH+][OH][B]
Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value.
Percent ionization=[OH] equilibrium[B] initial×100%
Strong bases, for which Kb is very large, ionize completely (100%). For weak bases, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization.
Ammonia, NH3, is a weak base with a Kb value of 1.8×105.
Part A
What is the pH of a 0.105 M ammonia solution?
Express your answer numerically to two decimal places.
11.37
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Part B
What is the percent ionization of ammonia at this concentration?
Express your answer with the appropriate units.
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± pH and Percent Ionization of a Weak Base
The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B
B(aq)+H2O(l)BH+(aq)+OH(aq)
this constant is given by
Kb=[BH+][OH][B]
Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value.
Percent ionization=[OH] equilibrium[B] initial×100%
Strong bases, for which Kb is very large, ionize completely (100%). For weak bases, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization.
Ammonia, NH3, is a weak base with a Kb value of 1.8×105.
Part A
What is the pH of a 0.105 M ammonia solution?
Express your answer numerically to two decimal places.
pH =11.37
SubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 5 attempts remaining
Part B
What is the percent ionization of ammonia at this concentration?
Express your answer with the appropriate units.
% ionization =SubmitHintsMy AnswersGive UpReview Part
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Explanation / Answer
NH3 ---------> NH4+ + OH- Kb = 1.8x10-5
i) 0.105 0 0
eq) 0.105 - x x x
Kb = x2 / 0.105 - x -----> Kb is small so x is small too: 0.105 - x = 0.105
1.8x10-5 = x2 / 0.105 -----> x2 = 1.8x10-5 x 0.105 ------> x = 1.3748x10-3 M = [OH-]
pOH = -log(1.3748x10-3) = 2.86 -----> pH = 14 - 2.86 = 11.14
[NH3] = 0.105 - 1.3748x10-3 = 0.1036252
% = (0.1036252 / 0.105) x 100 = 98.69%
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