Base Acid Equlibrium Problem Set I
By S.E. Van Bramer Widener University One University Place Chester, PA 19013 [e-mail] February 23, 1999 M mole liter
1.For a 0.1 M solution of HCl calculate the equlibrium concentration of a.H3O1+ b.Cl 1c.OH1Given the initial concentration of HCl: C HCl 0.1 .M
Since HCl is a strong acid, it undergoes complete dissociation by the reaction: HCl + H 2O --> H 3O1+ + Cl1C H3O C Cl C HCl C HCl C H3O = 0.1 M C Cl = 0.1 M
The concentration of OH1- is determined by the equlibrium reaction: 2 H2O <--> H3O1+ + OH1Kw 1.0 .10
14 .
M
2
K w C H3O .C OH C OH Kw C H3O 13 M
C OH = 1 10
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S.E. Van Bramer [e-mail]
2.For a 0.1 M solution of KOH calculate the equlibrium concentration of a.K1+ b.H3O1+ c.OH1-
Given the initial concentration of KOH: C KOH 0.1 . M
Since KOH is a strong base, it undergoes complete dissociation by the reaction: KOH --> K1+ + OH1C K = 0.1 M C OH = 0.1 M
CK C OH
C KOH C KOH
The concentration of H3O1+ is determined by the equlibrium reaction: 2 H2O <--> H3O1+ + OH1Kw 1.0 .10
14 .
M
2
K w C H3O .C OH C H3O Kw C OH 13 M
C H3O = 1 10
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S.E. Van Bramer [e-mail]
3.For a 0.1 M solution of acetic acid (Ka = 1.8*10-5) calculate the equlibrium concentration of a.CH3COOH b.CH3COO1c.H3O1+ d.OH1Given the initial concentration of acetic acid: C CH3COOH 0.1 .M
The acetic acid is in equlibrium with acetate ion in the following reaction CH3COOH + H2O <--> CH 3COO1- + H3O1+ The equlibrium expression for this is: C CH3COO. C H3O C CH3COOH
KA
At equlibrium the concentration of each of these is: C CH3COOH_eq C CH3COOH C CH3COO_eq X C H3O_eq X Ka X .X C CH3COOH X
2
X
X
5 1.8 . 10 .M
0.1 . M
X two The possible roots for X, Solved using the quadratic. Only one of these roots gives a possible solution below. 0.101 0.099 1.351 10 1.333 10 C H3O_eq = 1.351 10 1.333 10 3 M 3 3 M 3
X
.0013506709730779748818 .M .0013326709730779748818 .M
C CH3COOH_eq
C CH3COOH
X
C CH3COOH_eq =
M
C CH3COO_eq
X
C CH3COO_eq =
C H3O_eq
X
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S.E. Van Bramer [e-mail]
The concentration of OH1- is determined by the equlibrium reaction: 2 H2O <--> H3O1+ + OH1Kw 1.0 .10
14 .
M
2
K w C H3O .C OH C H3O C H3O_eq Kw C H3O 12 M
1
From the results above.
C OH
C OH = 7.504 10
4.For a 0.1 M solution of Ammonia (Kb = 1.78*10-5) calculate the equlibrium concentration of a.NH3 b.NH41+ c.H3O1+ d.OH1-
Given the initial concentration of ammonia: C NH3 0.1 . M
The ammonia is in equlibrium with ammonium ion in the following reaction NH3 + H2O <--> NH 41+ + HO1The equlibrium expression for this is: C NH4 .C OH C NH3
KB
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S.E. Van Bramer [e-mail]
At equlibrium the concentrat...