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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)

There is a reaction between an acid and a base when a compound containing hydrated aluminium ions are added to water.

(1.) How can this be shown experimentally?

(2.) Write the formula of the reacting acid.

(3.) Write the formula of the reacting base.

Predict whether 0.01 M solutions of the following substances will be acidic, neutral or basic:

(1.) rubidium iodide

(2.) iron(II) bromide

(3.) ammonium dihydrogenphosphate

(4.) lithium acetate

(5.) potassium hydrogenphosphate

(6.) ammonium chloride

(7.) copper(II) nitrate

(8.) magnesium nitrate

(9.) barium hydroxide

What mass of ammonium chloride must be dissolved in water (200 cm³) to give a solution whose pH is 4.75?

What is the pH of a water solution made by dissolving sodium ethoxide (34 g) and making the volume up to 1.00 litre?

Give the pH of aqueous solutions prepared from equimolar amounts of the following pairs of substances:

(1.) H₃PO₄ / NaH₂PO₄

Calculate the -

(1.) pH of 0.10 M acetic acid

(2.) pH of a solution 1.0 M with regard to both acetic acid and sodium acetate

(3.) pH of a solution 0.10 M with regard to both acetic acid and sodium acetate

Calculate the ratio [CH₃COOH] : [CH₃CO₂^-] in acetic acid - acetate solutions with the following pH values:

(1.) 2.00

(2.) 3.00

(3.) 5.00

(4.) 7.00

(5.) 9.00

(1.) An acid - base reaction with water will involve a change in [H₃O⁺], so it can be shown experimentally by noting a pH change upon addition of the aluminium salt (such as Al(NO₃)₃, where the anion does not hydrolyse in water.)

(2.) The reacting acid is [Al(H₂O)₆]³⁺ (ie this is the proton donor)

(3.) The reacting base is water, H₂O (ie water accepts a proton to become H₃O⁺)

(1.) PO₄^3- cannot act as an acid as it has no H atoms in its structure. Its conjugate acid is a weak acid, therefore PO₄^3- is a weak base and its water solution is basic.

PO₄^3- + H₂O OH^- + HPO₄^2-

(2.) HPO₄^2- is amphiprotic (can act as either acid or base). In water solution:

(3.) H₂PO₄^- is amphiprotic (H₃PO₄ is it's conjugate acid, HPO₄^- it's conjugate base)

(1.) RbI. Neither Rb⁺ nor I^- enter into acid-base reactions with water, so the solution is neutral. Rb⁺ is not acidic due to it's low charge density. I^- is not a base in water as it's conjugate acid HI is a strong acid, so the reaction

I^- + H₂O HI + OH^- cannot occur.

(2.) FeBr₂. Br^- does not hydrolyse water, so it has no effect on the pH. Fe²⁺ exists in solution as the hydrated species Fe(H₂O)₆²⁺. The positively charged Fe ion can draw enough electron density from the bound water H-O bonds to release a proton:

(3.) NH₄H₂PO₄. In this case both NH₄⁺ and H₂PO₄^- could enter into acid-base reactions with water and contribute to the pH of the solution, so K_A (or pK_A) values need to be considered. NH₄⁺ is a weak acid, and the two possible reactions are:

(4.) CH₃CO₂Li. Dissociation in water gives lithium and acetate ions. Li⁺ does not behave as an acid in water, so it has no effect on the pH. CH₃CO₂^- is the conjugate base of acetic acid, CH₃COOH, a weak acid, so the following equilibrium is set up:

(5.) Basic: HPO₄^2- + H₂O H₂PO₄^- + OH^-, the K⁺ cation has no effect on pH.

(6.) Acidic: NH₄⁺ + H₂O NH₃ + H₃O⁺, the Cl^- anion has no effect on pH.

(7.) Acidic: [Cu(H₂O)₄]²⁺ + H₂O [Cu(H₂O)₃OH]⁺ + H₃O⁺, the NO₃^- anion has no effect on pH.

(8.) Acidic: [Mg(H₂O)₆]²⁺ + H₂O [Mg(H₂O)₅OH]⁺ + H₃O⁺, the NO₃^- anion has no effect on pH.

(9.) Basic: Barium hydroxide is soluble, releasing Ba²⁺ and OH^-ions in solution.

Ba(OH)₂ + H₂O → Ba²⁺ + 2OH^-, the Ba²⁺ is one of the large 2+ charged metal cations which do not react with water and so have no effect on pH.

(1.) F^- + H₂O HF + OH^- (pH > 7)

(2.) [Fe(H₂O)₆]²⁺ + H₂O [Fe(H₂O)₅OH]⁺ + H₃O⁺ (pH < 7)

(3.) [Fe(H₂O)₆]³⁺ + H₂O [Fe(H₂O)₅OH]²⁺ + H₃O⁺ (pH < 7)

(4.) S^2- + H₂O HS^- + OH^- (pH > 7)

(5.) Neither K⁺ nor Cl^- react with water (pH = 7)

(6.) Neither Ba²⁺ nor I^- react with water (pH = 7)

(7.) [Al(H₂O)₆]³⁺ + H₂O [Al(H₂O)₅OH]²⁺ + H₃O⁺ (pH < 7)

(1.) The hydride of nitrogen is ammonia, NH₃. Ammonia dissolves in water and causes hydrolysis to form a weakly basic solution:

NH₃ + H₂O NH₄⁺ + OH^-.

(2.) The hydride of bromine is hydrogen bromide, HBr. Hydrogen bromide dissolves in water and completely ionises to form an acidic solution:

HBr + H₂O H₃O⁺ + Br^-

(3.) The hydride of sulfur is hydrogen sulfide, H₂S. Hydrogen sulfide dissolves in water and dissociates to a small extent to form an acidic solution. H₂S is a diprotic acid, the first dissociation forms HS^-, the second forms S^2- (to a much smaller degree)

(1.) [NH₄⁺]_init = 0.175 mol L^-1

(2.) From K_A(NH₄⁺) x K_B(NH₃) = K_W = 1.0 x 10^-14

CH₃CH₂O^- + H₂O → OH^- + CH₃CH₂OH

CH₃CH₂O^- is a stronger base than OH^-, so reaction goes to completion.

There are 0.500 mol CH₃CH₂ONa in 1 L

∴ [OH^-] = 0.500 mol L^-1 (from above equation)

Calculating [H₃O⁺] for pH:

[H₃O⁺] = 1.0 x 10^-14 / 0.500 = 2.0 x 10^-14 mol L^-1

∴ pH = -log (2.0 x 10^-14) = 13.70

K_A = [H₃O⁺] [A^-] / [HA]

Now in the case of a buffer, [H₃O⁺] does not equal [A^-] because of the presence of additional conjugate base. Rather, [H₃O⁺] = K_A [HA] / [A^-], showing that the [H₃O⁺] is determined by (i) the Ka of the weak acid and (ii) the ratio of weak acid to weak base in the buffer. If the acid / conjugate base pair are present in equimolar quantities (ie [HA] = [A^-]), they cancel out, leaving [H₃O⁺] = K_A. It then follows that pH = pK_A.

(1.) The conjugate acid / base pair in this system is H₃PO₄ / H₂PO₄^-. They are present in equimolar quantities, so from the above working pH = pK_A of H₃PO₄ = 2.13, and so pH of the buffer = 2.13. Note here that the pK_A value used here must be that of the weak acid in the acid / base pair (ie H₃PO₄). Do not confuse with pK_A of H₂PO₄^-!

(2.) The conjugate acid / base pair in this system is H₂PO₄^- / HPO₄^2-. They are present in equimolar quantities, so from the above working pH = pK_A of H₂PO₄^- = 7.20, and so pH of the buffer = 7.20.

(3.) The conjugate acid / base pair in this system is HPO₄^2- / PO₄^3-. They are present in equimolar quantities, so from the above working pH = pK_A of HPO₄^2- = 12.36, and so pH of the buffer = 12.36.

(1.) CH₃COOH + H₂O CH₃CO₂^- + H₃O⁺

(2.) K_A = [H₃O⁺] [A^-] / [HA]

(3.) As for part (2.), the acid and it's conjugate base are present in equimolar amounts, so pH = pK_A (CH₃COOH) = 4.76

The acid / base ratio is found from K_A:

K_A = [H₃O⁺] [A^-] / [HA]

∴ [HA] / [A^-] = [H⁺] / K_A

The [H⁺] can be obtained from the quoted pH values:

[H⁺] = 10^-pH

Substituting into the previous equation we obtain

[HA] / [A^-] = [H⁺] / K_A

∴ [CH₃COOH] / [CH₃COO^-] = [H⁺] / K_A

From the data source, K_A for CH₃COOH = 10^-4.76

A worked example is provided below:

(1.) pH = 2

(2.) pH = 3.00, [CH₃COOH] / [CH₃COO^-] = 5.8 x 10¹ : 1

(3.) pH = 5.00, [CH₃COOH] / [CH₃COO^-] = 5.8 x 10^-1 : 1

(4.) pH = 7.00, [CH₃COOH] / [CH₃COO^-] = 5.8 x 10^-3 : 1

(5.) pH = 9.00, [CH₃COOH] / [CH₃COO^-] = 5.8 x 10^-5 : 1