The butylammonium ion, C 4 H 9 NH 3 + , has a K a of 2.3 × 10 −11 . C 4 H 9 NH 3 + (aq) + H 2 O( ℓ ) ⇄ H 3 O + (aq) + C 4 H 9 NH 2 (aq) a) Calculate K b for the conjugate base, C 4 H 9 NH 2 (butyl amine ). b) Place the butylammonium ion and its conjugate base in Table 16.2. Name an acid weaker than C 4 H 9 NH 3 + and a base stronger than C 4 H 9 NH 3 . c) What is the pH of a 0.015M solution of butylammonium chloride?
The butylammonium ion, C 4 H 9 NH 3 + , has a K a of 2.3 × 10 −11 . C 4 H 9 NH 3 + (aq) + H 2 O( ℓ ) ⇄ H 3 O + (aq) + C 4 H 9 NH 2 (aq) a) Calculate K b for the conjugate base, C 4 H 9 NH 2 (butyl amine ). b) Place the butylammonium ion and its conjugate base in Table 16.2. Name an acid weaker than C 4 H 9 NH 3 + and a base stronger than C 4 H 9 NH 3 . c) What is the pH of a 0.015M solution of butylammonium chloride?
The butylammonium ion, C4H9NH3+, has a Ka of 2.3 × 10−11.
C4H9NH3+(aq) + H2O(ℓ) ⇄ H3O+(aq) + C4H9NH2(aq)
a) Calculate Kb for the conjugate base, C4H9NH2 (butyl amine).
b) Place the butylammonium ion and its conjugate base in Table 16.2. Name an acid weaker than C4H9NH3+ and a base stronger than C4H9NH3.
c) What is the pH of a 0.015M solution of butylammonium chloride?
Definition Definition Transformation of a chemical species into another chemical species. A chemical reaction consists of breaking existing bonds and forming new ones by changing the position of electrons. These reactions are best explained using a chemical equation.
(a)
Expert Solution
Interpretation Introduction
Interpretation:
Kb has to be calculated for the conjugate base (C4H9NH2, butyl amine)
Concept Introduction:
Equilibrium constants:
The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.
Ka is an acid constant for equilibrium reactions.
HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]
Kb is a base constant for equilibrium reaction.
BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]
Ion product constant for wter Kw= [H3O+][OH-] =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]
Relation between pH and pOH
pH + pOH =14
Answer to Problem 92GQ
Kb of conjugate base is 4.34 × 10-4
Explanation of Solution
According to the Brown stead –Lowry theory the compound which is accepts a proton to be considered as a base.
A weak base accepts a proton from water molecules producing hydroxide ions in water.
The Ka of 2.3× 10-11
Ka×Kb = KwKb= KwKaKw= 1.0×10-14Ka = 2.3×10-11
Substitute the values,
Kb= 1.0×10-142.3×10-11 = 4.34×10-4
Therefore, Kb of conjugate base is 4.34 × 10-4
(b)
Expert Solution
Interpretation Introduction
Interpretation:
Compound should be Named for an acid weaker than C4H9NH3+ and a base stronger than C4H9NH2.
Concept Introduction:
Equilibrium constants:
The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water .
Ka an acid constant for equilibrium reactions.
HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]
Kb is base constant for equilibrium reaction.
BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]
Ion product constant for wter Kw= [H3O+][OH-] =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]
Relation between pH and pOH
pH + pOH =14
Answer to Problem 92GQ
An acid weaker than C4H9NH3+ is HPO42-
A base stronger than C4H9NH2 is PO43-
Explanation of Solution
From the table,
Ka of C4H9NH3+ is 2.3×10-11
Kb of C4H9NH2 is 4.4×10-4
The butyl ammonium ion has Ka value is in between HPO42- (Ka=3.6×10-13) and [Ni(H2O)6]2+ (Ka=3.6×10-13)
The order of Ka is as follows.
HPO42- < C4H9NH3+ < [Ni(H2O)6]2+
Butyl amine has a Kb value in between [Ni(H2O)5OH]+ (Kb=4.0×10-4) and PO43- (Kb=2.8×10-2).
The Kb order is as follows.
[Ni(H2O)5OH]+< C4H9NH2< PO43-
Therefore, An acid weaker than C4H9NH3+ is HPO42- . A base stronger than C4H9NH2 is PO43-
(c)
Expert Solution
Interpretation Introduction
Interpretation:
The pH has be calculated for 0.015M solution of butylammonium chloride?
Concept Introduction:
Equilibrium constants:
The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water .
Ka is an acid constant for equilibrium reactions.
HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]
Kb is a base constant for equilibrium reaction.
BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]
Ion product constant for wter Kw= [H3O+][OH-] =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]
Relation between pH and pOH
pH + pOH =14
Answer to Problem 92GQ
The pH of the 0.015M solution of butylammonium chloride is 6.2.
Explanation of Solution
Butylammonium chloride dissociates into butyl ammonium ion and chloride ion. Chloride ion does not affect the solution pH. Butylammonium donate a proton to water to form hydronium ions.
The equilibrium chemical reaction of butylammonium chloride is as follows.
C4H9NH3+(aq) + H2O(l) ⇌H3O+(aq) + C4H9NH2(aq)
The equilibrium expression:
Ka= [C4H9NH2][H3O+][C4H9NH3+]
The initial concentration of ammonium chloride is 0.015M.
Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.
C4H9NH3+(aq) + H2O(l) ⇌H3O+(aq) + C4H9NH2(aq)I 0.015M -- -- --C -x -- + x + xE (0.015- x) -- x x
Hence,
Ka= [C4H9NH2][H3O+][C4H9NH3+]
Ka= 2.3×10-11
Substitute the values form the ICE table.
2.3×10-11 = (x)(x)0.015- x2.3×10-11 = (x)20.015- x(0.015-x) approximately equals to 0.0152.3×10−11 = (x)2(0.015) x2= (0.015)(2.3×10-11)
Therefore,
x = (0.015)(2.3×10-11) = 5.9×10-7[H3O+] =5.9×10−7M
Let’s calculate the pH of the solution:
pH = -log[H3O+]pH = -log[5.9×10-7]pH = 6.2
Therefore, pH of butylammonium chloride solution is 6.2.
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