Equilibria

Question 1

Ionic product of water

Answer 1

Kw = [H+] [OH−] = 1.00 x 10^−14
(as [H+] = [OH−] so, kw = [H+]^2 and thus [H+] conc. can be found)

Question 2

Formula of pH & [H+]

Answer 2

pH = − log10 [H+]
[H+] = 10^ −pH

Question 3

Why can pH values of alkalis and pOH values of acids be calculated?

Answer 3

Even though they are alkalis, they have small conc. of H+ ions and this conc. is used to calculate pH and vice versa.

Question 4

Ionisation of strong monobasic acids i.e. HCl

Answer 4

They are completely ionised and the [H+] conc. is approximately same as the acid.

(Note: the concentration of H+ ions we get from the ionisation of water molecules is very small
compared with the concentration of H+ ions we get from the acid, so it’s considered to be negligible.)

Question 5

Effect of Dilution on pH of an acid

Answer 5

Diluting the acid ten times reduces the value of the H+ ion concentration to one-tenth and increases the pH by a value of one.

Question 6

Calculating the pH of strong bases

Answer 6

To calculate the pH of a solution of strong base we need to know:
♦ the concentration of OH− ions in solution.
♦ the equilibrium expression for the ionisation of water ➡ Kw = [H+] [OH−]
♦ the value of Kw for water
♦ As Kw = [H+] [OH−]
♦ [H+] = Kw / [OH−] and thus using [H+] conc. pH can be found.
………………………………….2nd way………………………………….

♦ Find –log10 [OH−] (here –log10 [OH−] = –log10 (0.0500) = 1.3)
♦ Subtract this value from 14 (in this
example 14 – 1.3 = 12.7)
♦ This works because –log10 [H+] – log10 [OH−] = 14.

Question 7

Ka & pKa formulae

Answer 7

Ka = [H+] [A−] / [HA] or [H+]^2 / [HA]
pKa = − log10 Ka

Question 8

What do high and low values of Ka indicate?

Answer 8

➊ A high value for Ka (for example, 40 mol dm−3) indicates that the position of equilibrium lies to the right. The acid is almost completely ionised.
➋ A low value for Ka (for example, 1.0 × 10−4 mol dm−3) indicates that the position of equilibrium lies to the left. The acid is only slightly ionised and exists mainly as HA molecules and comparatively few H+ and A− ions.

Question 9

What do the values of pKa indicate?

Answer 9

The less positive the value of pKa, the more strongly acidic is the acid.

Question 10

Assumptions involved while calculating Ka

Answer 10

1. The concentration of hydrogen ions produced by the ionisation of the water molecules present in the solution is ignored as the ionic product of water (1.00× 10−14 mol2 dm−6) is negligible compared with the values for most weak acids.
2. The ionisation of the weak acid is so small that the concentration of undissociated HA molecules present at equilibrium is approximately the same as that of the original acid.

Question 11

Buffer solution

Answer 11

A solution that minimises changes in pH when moderate amounts of acid or bases are added. A buffer solution is either a weak acid and its conjugate base or a weak base and its conjugate acid, and there are reserve supplies of the acid/base and its conjugate base / conjugate acid. The pH of a buffer solution depends on the ratio of the concentration of i.e. the acid and the concentration of its conjugate base. If these do not change very much, the pH changes very little.

Question 12

Acid / Conjugate base & Base / Conjugate acid examples

Answer 12

Textbook page no. 448, 449 (must read)

Question 13

[H+] and pH formulae for buffer solutions

Answer 13

[H+] = Ka × [acid] / [salt]
pH = pKa + log10 [salt] / [acid]

(Note: Just remember the formula i.e.
Ka = [H+][CH3COO−] / [CH3COOH] )

Question 14

Buffer solution examples

Answer 14

Textbook page no. 451
………………………………………………………………………………………
CO2(aq) + H2O(aq) ⇌ H+(aq) + HCO3−(aq)
» The equilibrium between carbon dioxide and
hydrogencarbonate is the most important buffering system in the blood.

Question 15

Solubility

Answer 15

Solubility is generally quoted as the number of grams or number of moles of compound needed to saturate 100 g or 1 kg of water at a given temperature.

Question 16

Solubilty product, Ksp

Answer 16

The product of the concentrationsof each ion ina saturated solution of a sparingly soluble salt at 298K, raised to the power of their relative concentrations. The idea of solubility product only applies to ionic compounds that are slightly soluble.
(Note: For any solid, the concentration of the solid phase remains constant and can be combined with the value of Kc thus giving Ksp.)
.

Question 17

Value of Ksp indicates

Answer 17

The smaller the value of Ksp
the lower is the solubility of the salt.

Question 18

Soluble and Insoluble salts

Answer 18

Soluble salts include salts of Group I elements, all nitrates and ammonium salts and many sulfates. Halides are generally soluble except for lead(II) halides and silver halides.

Question 19

Common Ion Effect

Answer 19

Common ion effect is the reduction in the solubility of a dissolved salt achieved by adding a solution of a compound that has an ion in common with the dissolved salt. This often results in precipitation.

Question 20

Values of ‘Solubility Product’ and ‘Solubility’ indicate

Answer 20

(Solubility product)
more positive ➡ ppt will form,
more negative ➡ ppt will not form
(Solubility)
more positive ➡ more soluble
more negative ➡ less soluble

Question 21

Dynamic equilibrium

Answer 21

i.e. Ammonia molecules are moving from the aqueous layer to the organic layer at the same rate as they are moving from the organic layer to the aqueous layer:
NH3(aq) ⇌ NH3(organic solvent)
……………………………………………………………………………………….
(The solute X should be in the same physical state and molecular state in both solvents)

Question 22

Solubility depends on

Answer 22

Solubility depends on the strength of the intermolecular bonds between the solute and the solvent. This in turn depends on the polarity of the molecules of solute and solvent.
For example,
➊ Ammonia is a polar molecule that forms hydrogen bonds with water. So ammonia is very soluble in water. Ammonia is less soluble in organic solvents such as trichloromethane (CHCl3) because the bonding between ammonia and the solvent is permanent dipole–permanent dipole, which is weaker than hydrogen bonding.
➋ Iodine dissolves in organic solvents such as cyclohexane. Both are non-polar molecules. Iodine forms instantaneous dipole–induced dipole bonds with cyclohexane.

Question 23

Partition coefficient and Paper Chromatography

Answer 23

(Textbook page no. 458, 459.)
The solutes in the mixture being separated are partitioned to different extents between the solvents in the mobile and stationary phases. The greater the relative solubility in the mobile phase, the faster the rate of movement as the mobile phase passes over the stationary phase.