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From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.

(a) CH 3 CO 2 H: [ H 3 O + ] = 1.34 × 10 −3 M ;
[ CH 3 CO 2 ] = 1.34 × 10 −3 M ;

[CH 3 CO 2 H] = 9.866 × 10 −2 M ;

(b) ClO : [OH ] = 4.0 × 10 −4 M ;

[HClO] = 2.38 × 10 −5 M ;

[ClO ] = 0.273 M ;

(c) HCO 2 H: [HCO 2 H] = 0.524 M ;
[ H 3 O + ] = 9.8 × 10 −3 M ;
[ HCO 2 ] = 9.8 × 10 −3 M ;

(d) C 6 H 5 NH 3 + : [ C 6 H 5 NH 3 + ] = 0.233 M ;

[C 6 H 5 NH 2 ] = 2.3 × 10 −3 M ;
[ H 3 O + ] = 2.3 × 10 −3 M

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From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.

(a) NH 3 : [OH ] = 3.1 × 10 −3 M ;
[ NH 4 + ] = 3.1 × 10 −3 M ;

[NH 3 ] = 0.533 M ;

(b) HNO 2 : [ H 3 O + ] = 0.011 M ;
[ NO 2 ] = 0.0438 M ;

[HNO 2 ] = 1.07 M ;

(c) (CH 3 ) 3 N: [(CH 3 ) 3 N] = 0.25 M ;
[(CH 3 ) 3 NH + ] = 4.3 × 10 −3 M ;

[OH ] = 4.3 × 10 −3 M ;

(d) NH 4 + : [ NH 4 + ] = 0.100 M ;

[NH 3 ] = 7.5 × 10 −6 M ;
[H 3 O + ] = 7.5 × 10 −6 M

(a) K b = 1.8 × 10 −5 ;
(b) K a = 4.5 × 10 −4 ;
(c) K b = 7.4 × 10 −5 ;
(d) K a = 5.6 × 10 −10

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Determine K b for the nitrite ion, NO 2 . In a 0.10- M solution this base is 0.0015% ionized.

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Determine K a for hydrogen sulfate ion, HSO 4 . In a 0.10- M solution the acid is 29% ionized.

K a = 1.2 × 10 −2

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Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) F

(b) NH 4 +

(c) AsO 4 3−

(d) ( CH 3 ) 2 NH 2 +

(e) NO 2

(f) HC 2 O 4 (as a base)

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Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) HTe (as a base)

(b) ( CH 3 ) 3 NH +

(c) HAsO 4 3− (as a base)

(d) HO 2 (as a base)

(e) C 6 H 5 NH 3 +

(f) HSO 3 (as a base)

(a) K b = 4.3 × 10 −12 ;
(b) K a = 1.6 × 10 −8 ;
(c) K b = 5.9 × 10 −7 ;
(d) K b = 4.2 × 10 −3 ;
(e) K b = 2.3 × 10 −3 ;
(f) K b = 6.3 × 10 −13

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For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?

(a) 3 × 10 −8 M HNO 3

(b) 0.10 g HCl in 1.0 L of solution

(c) 0.00080 g NaOH in 0.50 L of solution

(d) 1 × 10 −7 M Ca(OH) 2

(e) 0.0245 M KNO 3

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Even though both NH 3 and C 6 H 5 NH 2 are weak bases, NH 3 is a much stronger acid than C 6 H 5 NH 2 . Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH 3 and 0.10 M in C 6 H 5 NH 2 ?

(a) [ OH ] = [ NH 4 + ]

(b) [ NH 4 + ] = [ C 6 H 5 NH 3 + ]

(c) [ OH ] = [ C 6 H 5 NH 3 + ]

(d) [NH 3 ] = [C 6 H 5 NH 2 ]

(e) both a and b are correct

(a) is the correct statement.

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Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO 2 H and 0.10 M in HClO.

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Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO 2 and 0.120 M in HBrO.

[H 3 O + ] = 7.5 × 10 −3 M
[HNO 2 ] = 0.127
[OH ] = 1.3 × 10 −12 M
[BrO ] = 4.5 × 10 −8 M
[HBrO] = 0.120 M

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Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH 3 NH 2 and 0.10 M in C 5 H 5 N ( K b = 1.7 × 10 −9 ).

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Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH 3 and 0.100 M in C 6 H 5 NH 2 .

[OH ] = [ NO 4 + ] = 0.0014 M
[NH 3 ] = 0.144 M
[H 3 O + ] = 6.9 × 10 −12 M
[ C 6 H 5 NH 3 + ] = 3.9 × 10 −8 M
[C 6 H 5 NH 2 ] = 0.100 M

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Using the K a value of 1.4 × 10 −5 , place Al ( H 2 O ) 6 3+ in the correct location in [link] .

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Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I .

(a) 0.0092 M HClO, a weak acid

(b) 0.0784 M C 6 H 5 NH 2 , a weak base

(c) 0.0810 M HCN, a weak acid

(d) 0.11 M (CH 3 ) 3 N, a weak base

(e) 0.120 M Fe ( H 2 O ) 6 2+ a weak acid, K a = 1.6 × 10 −7

(a) [ H 3 O + ] [ ClO ] [ HClO ] = ( x ) ( x ) ( 0.0092 x ) ( x ) ( x ) 0.0092 = 2.9 × 10 −8
Solving for x gives 1.63 × 10 −5 M . This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [ClO]= 5.8 × 10 −5 M
[HClO] = 0.00092 M
[OH ] = 6.1 × 10 −10 M ;
(b) [ C 6 H 5 NH 3 + ] [ OH ] [ C 6 H 5 NH 2 ] = ( x ) ( x ) ( 0.0784 x ) ( x ) ( x ) 0.0784 = 4.3 × 10 −10
Solving for x gives 5.81 × 10 −6 M . This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[ CH 3 CO 2 ] = [OH ] = 5.8 × 10 −6 M
[C 6 H 5 NH 2 ] = 0.00784
[H 3 O + ] = 1.7 × 10 −9 M ;
(c) [ H 3 O + ] [ CN ] [ HCN ] = ( x ) ( x ) ( 0.0810 x ) ( x ) ( x ) 0.0810 = 4.9 × 10 −10
Solving for x gives 6.30 × 10 −6 M . This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [CN ] = 6.3 × 10 −6 M
[HCN] = 0.0810 M
[OH ] = 1.6 × 10 −9 M ;
(d) [ ( CH 3 ) 3 NH + ] [ OH ] [ ( CH 3 ) 3 N ] = ( x ) ( x ) ( 0.11 x ) ( x ) ( x ) 0.11 = 6.3 × 10 −5
Solving for x gives 2.63 × 10 −3 M . This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[(CH 3 ) 3 NH + ] = [OH ] = 2.6 × 10 −3 M
[(CH 3 ) 3 N] = 0.11 M
[H 3 O + ] = 3.8 × 10 −12 M ;
(e) [ Fe ( H 2 O ) 5 ( OH ) + ] [ H 3 O + ] [ Fe ( H 2 O ) 6 2+ ] = ( x ) ( x ) ( 0.120 x ) ( x ) ( x ) 0.120 = 1.6 × 10 −7
Solving for x gives 1.39 × 10 −4 M . This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[Fe(H 2 O) 5 (OH) + ] = [H 3 O + ] = 1.4 × 10 −4 M
[ Fe ( H 2 O ) 6 2+ ] = 0.120 M
[OH ] = 7.2 × 10 −11 M

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Propionic acid, C 2 H 5 CO 2 H ( K a = 1.34 × 10 −5 ), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698- M solution of C 2 H 5 CO 2 H?

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White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm 3 , what is the pH?

pH = 2.41

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The ionization constant of lactic acid, CH 3 CH(OH)CO 2 H, an acid found in the blood after strenuous exercise, is 1.36 × 10 −4 . If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?

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Nicotine, C 10 H 14 N 2 , is a base that will accept two protons ( K 1 = 7 × 10 −7 , K 2 = 1.4 × 10 −11 ). What is the concentration of each species present in a 0.050- M solution of nicotine?

[C 10 H 14 N 2 ] = 0.049 M
[C 10 H 14 N 2 H + ] = 1.9 × 10 −4 M
[ C 10 H 14 N 2 H 2 2+ ] = 1.4 × 10 −11 M
[OH ] = 1.9 × 10 −4 M
[H 3 O + ] = 5.3 × 10 −11 M

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The pH of a 0.20- M solution of HF is 1.92. Determine K a for HF from these data.

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The pH of a 0.15- M solution of HSO 4 is 1.43. Determine K a for HSO 4 from these data.

K a = 1.2 × 10 −2

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The pH of a 0.10- M solution of caffeine is 11.16. Determine K b for caffeine from these data:
C 8 H 10 N 4 O 2 ( a q ) + H 2 O ( l ) C 8 H 10 N 4 O 2 H + ( a q ) + OH ( a q )

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The pH of a solution of household ammonia, a 0.950 M solution of NH 3, is 11.612. Determine K b for NH 3 from these data.

K b = 1.77 × 10 −5

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Questions & Answers

How can ionic bonds dissociate in aqueous solution
Andrew Reply
Because of the polarity of both ionic compounds and water the ionic compound will dissolve as "like dissolves like", and the molecule forms bonds with the water.
Claud
I wanna understand more about isomers
Emmanuel Reply
what is catenation
Oladuji Reply
The property of carbon to form long chain with other atom!
Lareb
hydrocarbons can be classified as..1.Aliphatic compounds 2.cyclic compounds.under aliphatic compounds there are two types saturated hydrocarbons(alkanes) and unsaturated hydrocarbons(alkenes and alkynes).
Niroshan Reply
thanks but i have also heard of aromatic hydrocarbons
emmanuel
so am kinda confused
emmanuel
how
Emmanuel
hello i have big problems in understanding organic chemistry
emmanuel Reply
what are the main types of hydrocarbon
emmanuel
how many elements are in the periodic table
emmanuel Reply
118
daniel
please what are the main types of hydrocarbons
emmanuel
why Rutherford uses the gold foil instead of other metals?
Lareb Reply
Rutherford chose gold was because its extremely malleable. One can stretch gold foil until it is only a few atoms thick in places, which is not possible with aluminum. If the foil were too thick, there would be no transmission of particles at all; the whole point was to demonstrate that most alpha
daniel
wjat does Rutherford mean?
Asali
Ernest Rutherford was the scientist that preformed the experiment.
daniel
although other metals are also present which are more melleable!?so
Lareb
gold
daniel
what is a balanced equation 4 trioxonitrate (V)acid and sodium hydroxide?
Marcel Reply
proved ur Worth: If A is a of trioxonitrate(V)acid,HNO3' of unknown concentration .B is a standard solution of sodium hydroxide containing 4.00g per dm cube of solution.25cm cube portions solution B required an average of 24.00cm cube of solution A for neutralization,using 2 drops of methyl orange.
Marcel
calculate the concentration of solution B in moles per dm cube
Marcel
calculate the concentration of solution A and B in moles per DM cube
Marcel
finally calculate the concentration in g/dm cube of HNO3 in solution A (H=1,N=14,O=16,Na=23)
Marcel
calculate the standard enthalpy of formation for propane(C3H8) from the following data; 1), C3H8+5O2->3CO2+4H2O; -222.0kJ/mol 2), C+O2->CO2;-395.5kJ/mol 3),H2+O->H2O; 285.8kJ/mol
Josephine
let eventually of formation of propane = X X + (-222)=3×(-395.5)+4×(-286) rearrange to find X
Paul
wat is electrolysis?
Mgbachi Reply
it is the chemical decomposition of a substance when electric current is passed through it either in molten form or aqueous solution
Nuru
list the side effect of chemical industries
Chelsea Reply
how do you ionise an atom
Rabeka Reply
many ways ,but one of them is when the atom becomes heated to a certain temperature the surface electron becomes too energetic and leaves the atom because the attraction between the nucleus and the electron becomes overpowered by the energetic eletron
sunday
also hitting of two atoms can cause transfer of surface electrons
sunday
and when this transfers occur the atom becomes ionised
sunday
who is doing Cape chemistry tomorrow?
caramel Reply
What is hybridization
edmondnti Reply
the mix between different breeds of species in one
Jared
it is the blending of orbitals.
stanley
the mixing of orbital
caramel
are covalent bonds influenced by factors such as temperature and pressure?
patrick Reply
what is catalyst used for mirror test
Sanjay Reply
Practice Key Terms 5

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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