Acids and Bases1 - Questions

Link to Data Page

Your feedback on these self-help problems is appreciated. Click here to send an e-mail.

For on-line help on this topic, please see the following Chemcal modules:

Acids and Bases (strong acids and bases; conjugate acid/base pairs; pH;self-dissociation of water)

Weak Acids and Bases (weak acids; weak bases; Ka and Kb)

Calculations with Weak Acids and Bases (calculations involving weak acids and bases and buffers)

Acid-base Titrations (titration curves for acid/base titrations; indicators)

 

Shortcut to Questions

Q: 1 2 3 4 5 6 7 8 9 10 11

1

Provide the conjugate base for the following acids:

(1.) H3O+

(2.) NH4+

(3.) HCl

(4.) HCO3-

(5.) [Fe(H2O)6]3+

(6.) C6H5OH

(7.) HPO42-

Provide the conjugate acid for the following bases:

(8.) SO42-

(9.) O2-

(10.) [Al(H2O)5OH]2+

(11.) N(CH3)3

(12.) N3-

(13.) H2PO4-

(14.) NH3

2

(1.) Write an equation for the self-ionization of water

(2.) Give an algebraic expression for Kw

(3.) Define pKw and state its value at 298 K to 2 significant figures.

3

For the following solutions, determine: [H+], [OH-], pH, pOH

(1.) HBr (1.0 x 10-3 M)

(2.) Ba(OH)2 (1.0 x 10-5 M)

4

(1.) Write the equation for the equilibrium which is established when propanoic acid (CH3CH2COOH) is added to water.

(2.) Write the algebraic expression for KA of propanoic acid.

(3.) A water solution of propanoic acid had, at equilibrium, [CH3CH2COOH] = 0.10 M and [H3O+] = 1.2 x 10-3 M. Calculate KA for propanoic acid.

5

(1.) Write the equation for the equilibrium which is established when a quantity of ammonia gas is dissolved in water, and

(2.) the algebraic expression for KB of ammonia.

(3.) A sample of "ammonia water" was 1.0 M in NH3 and 4.2 x 10-3 M in OH-. Calculate KB for ammonia.

6

For the weak acid HA:

(1.) Write the algebraic expression for KA

(2.) Write the algebraic expression for KB of the conjugate base of HA

(3.) Show that KB = KW / KA

(4.) Calculate KB for CN- given that dissolution of hydrogen cyanide gas in water gave, at equilibrium, [HCN] = 0.010 M and [H3O+] = 2.5 x 10-6 M.

7

(1.) Using acid dissociation constant tables, list the following in order of increasing acid strength:

H2S, HCN, HCO3-, H2PO4-, HCOOH

(2.) Using KA tables for their conjugate acids, list the following in order of increasing base strength:

CN-, N2H4, NH3, HS-, H2PO4-, HSO4-

8

Write an equation for the self-ionisation of

(1.) sulfuric acid

(2.) liquid ammonia

9

(1.) Write an equation illustrating hydration

(2.) Write an equation illustrating hydrolysis

10

(1.) Why is Na+ in water solution hydrated to an indefinite rather than a definite degree?

(2.) Why is the hydrated Al3+ an acid whereas the hydrated Na+ is not?

11

For the acid/base reaction, in water solution,

HA + B- A- + HB

(1.) Find the expression for the concentration equilibrium constant KC,

(2.) Find the condition which must be satisfied for the products to predominate at equilibrium, and

(3.) Show that KC = KA(HA) / KA(HB) and that the condition for the products to predominate at equilibrium is given by KA(HA) > KA(HB)

Acids and Bases 1 (Answers)  

1

A conjugate acid - base pair differ by H+. For example, when H3O+ reacts as an acid and loses a proton, the conjugate base (H2O) is formed: H3O+ H2O + H+. Hence H2O is the conjugate base of H3O+ (and, therefore, H3O+ is the conjugate acid of H2O).

(1.) H2O

(2.) NH3

(3.) Cl-

(4.) CO32-

(5.) [Fe(H2O)5OH]2+

(6.) C6H5O-

(7.) PO43-

Conjugate acids are the species formed when a conjugate base accepts a proton.

(8.) HSO4-

(9.) OH-

(10.) [Al(H2O)6]3+

(11.) (CH3)3NH+

(12.) HN3

(13.) H3PO4

(14.) NH4+

2

(1.) 2H2O H3O+ + OH-

(2.) Kw = [H3O+] [OH-]

(3.) pKw = -log Kw = 14.00 at 298 K

Note the conversion of significant figures between K and pK - the number of significant figures in K becomes the number of decimal places in the pK value (and vice versa), since pK is a logarithmic scale. So in the example above, 2 significant figures of accuracy in the K data results in 2 decimal places of accuracy in the resultant pK value.

3

(1.) HBr is a strong acid, so we assume complete ionisation in solution:

HBr + H2O ® H3O+ + Br-.

Note the use of a single-headed arrow to represent complete reaction to the product side. This implies that the equilibrium concentration of H3O+ equals the initial concentration of HBr, ie [H3O+] = 1.0 x 10-3 M.

pH = -log [H3O+] = 3.00

Kw = [H3O+] [OH-].

Therefore [OH-] = Kw / [H3O+]

= 1.0 x 10-14 M2 / 1.0 x 10-3 M = 1.0 x 10-11 M.

pOH = -log [OH-] = 11.00

(2.) Ba(OH)2 is soluble, releasing the strong base OH- into solution.

Ba(OH)2 ® Ba2+ + 2OH-

[OH-] = 2 x 1.0 x 10-5 M = 2.0 x 10-5 M

pOH = -log [OH-] = 4.70

[H3O+] = Kw / [OH-]

= 1.0 x 10-14 M2 / 2.0 x 10-5 M = 5.0 x 10-10 M

pH = -log [H3O+] = 9.30

4

(1.) Propanoic acid, CH3CH2COOH, is a weak acid, so the dissociation is a reversible process at equilibrium (unlike strong acids which are considered completely ionised in water). Note that the acidic carboxylic hydrogen is lost:

CH3CH2COOH + H2O CH3CH2CO2- + H3O+

(2.) KA = [CH3CH2CO2-] [H3O+] / [CH3CH2COOH]

(3.) If [H3O+] = 1.2 x 10-3 M then [CH3CH2CO2-] must

also equal 1.2 x 10-3 M (from part (2.)).

KA = [CH3CH2CO2-] [H3O+] / [CH3CH2COOH]

= (1.2 x 10-3 M)2 / 0.10 M = 1.4 x 10-5 M.

5

(1.) Ammonia is a weak base, so the hydrolysis is an equilibrium process.

NH3 + H2O NH4+ + OH-

(2.) KB = [NH4+] [OH-] / [NH3]

(3.) Given [OH-] = 4.2 x 10-3 M, [NH4+] must also equal 4.2 x 10-3 M.

\ KB = (4.2 x 10-3 M)2 / 1.0 M = 1.8 x 10-5 M.

6

(1.) KA = [H3O+] [A-] / [HA]

(2.) KB = [OH-] [HA] / [A-]

(3.) KW / KA = [H3O+] [OH-] [HA] / [H3O+] [A-]

= [OH-] [HA] / [A-] = KB.

(4.) HCN + H2O H3O+ + CN-

Given [H3O+] = 2.5 x 10-6 M, [CN-] must also equal 2.5 x 10-6 M.

KA of HCN = (2.5 x 10-6 M)2 / 0.01 M = 6.25 x 10-10 M.

\ KB of CN- = KW / KA(HCN)

= 1.0 x 10-14 M2 / 6.25 x 10-10 M = 1.6 x 10-5 M.

7

(1.) In order of increasing acid strength:

HCO3-, HCN, H2PO4-, H2S, HCOOH

(2.) In order of increasing base strength:

HSO4-, HS-, H2PO4-, N2H4, CN-, NH3

8

Self ionisation involves the one species acting as both proton donor and proton acceptor:

(1.) 2H2SO4 HSO4- + H3SO4+

(2.) 2NH3(l) NH2- + NH4+

9

(1.) Hydration is the addition of H2O to a molecule to form a new species, for example:

· Cu2+ + 4H2O ® Cu(H2O)42+

· CaO + H2O ® Ca(OH)2

(2.) Hydrolysis is the splitting of water (hence the name hydro-lysis). Note that where hydration involves the addition of one or more H2O molecules, hydrolysis always involves the cleavage of a water O-H bond during reaction, often leading to the formation of H+ or OH-, for example:

· Hydrolysis of a hydrated cation:

M(H2O)xn+ + H2O H3O+ + M(H2O)x -1(OH)(n -1)+

eg. Fe(H2O)63+ + H2O H3O+ + Fe(H2O)5(OH)2+

· Hydrolysis of a hydrated anion:

X- + H2O HX + OH-

eg. CH3CO2- + H2O CH3COOH + OH-

· Hydrolysis of an organic molecule:

eg. CH3COCl + H2O ® CH3COOH + HCl

10

(1.) The charge density of the Na+ ion is too small to hold H2O molecules sufficiently strongly to keep them in a definite arrangement, as occurs for example with Fe3+ in [Fe(OH2)6]3+.

(2.) The high charge density of the aluminium ion has a strong attraction for the O atom of the hydrating water molecules, and breaks some of their O-H bonds, thereby releasing H+ to the solvent. Na+ has too small a charge density to attract electrons from the O-H bonds of surrounding H2O molecules sufficiently strongly to cause O-H bonds to break.

11

HA + B- A- + HB

(1.) KC = [A-] [HB] / [B-] [HA]

(2.) KC must be greater than 1 for products to predominate.

(3.) KA(HA) = [A-] [H3O+] / [HA]

KA(HB) = [B-] [H3O+] / [HB]

Cancelling common terms, KA(HA) / KA(HB) = [A-] [HB] / [B-] [HA] = KC.

Therefore for products to predominate, the numerator of the expression for KC must be greater than the denominator, ie. KA(HA) > KA(HB).