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; K_a and K_b)

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

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

Provide the conjugate base for the following acids:

(1.) H₃O⁺

(2.) NH₄⁺

(3.) HCl

(4.) HCO₃^-

(5.) [Fe(H₂O)₆]³⁺

(6.) C₆H₅OH

(7.) HPO₄^2-

(8.) SO₄^2-

(9.) O^2-

(10.) [Al(H₂O)₅OH]²⁺

(11.) N(CH₃)₃

(12.) N₃^-

(13.) H₂PO₄^-

(14.) NH₃

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

(2.) Ba(OH)₂ (1.0 x 10^-5 M)

For the weak acid HA:

(1.) Write the algebraic expression for K_A

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

(3.) Show that K_B = K_W / K_A

(4.) Calculate K_B 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.

(1.) sulfuric acid

(2.) liquid ammonia

(1.) Write an equation illustrating hydration

(2.) Write an equation illustrating hydrolysis

(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?

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

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

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

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.) CO₃^2-

(5.) [Fe(H₂O)₅OH]²⁺

(6.) C₆H₅O^-

(7.) PO₄^3-

(8.) HSO₄^-

(9.) OH^-

(10.) [Al(H₂O)₆]³⁺

(11.) (CH₃)₃NH⁺

(12.) HN₃

(13.) H₃PO₄

(14.) NH₄⁺

(1.) 2H₂O H₃O⁺ + OH^-

(2.) K_w = [H₃O⁺] [OH^-]

(3.) pK_w = -log K_w = 14.00 at 298 K

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

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

(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:

(2.) K_A = [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.)).

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

(2.) K_B = [NH₄⁺] [OH^-] / [NH₃]

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

(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

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

(1.) 2H₂SO₄ HSO₄^- + H₃SO₄⁺

(2.) 2NH₃(l) NH₂^- + NH₄⁺

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

(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:

(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.

HA + B^- A^- + HB

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

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

(3.) K_A(HA) = [A^-] [H₃O⁺] / [HA]