Acids & Bases: Maths

Question 1

How to calculate/define pH

Answer 1

-log[H+]

Question 2

How to calculate [H+] from pH

Answer 2

1 x 10^-pH

Question 3

Finish the reaction and determine which is an acid and a base

HCl (g) + H2O (l) ->

Answer 3

HCl (g) + H2O (l) -> H3O+ (aq) + Cl- (aq)

HCl = Acid
H2O = Base

Question 4

Water partial dissociation reaction

Answer 4

H2O (l) ⇌ H+ (aq) + OH- (aq)

Question 5

What is Kw derived from

Answer 5

H2O being a constant value as it hardly ionises in aq solution

Question 6

Kw equation

Answer 6

Kw = [H+][OH-]

Question 7

Value of Kw at 25C or 298K

Answer 7

1x10^-14

Question 8

Calculate the pH of 2.20moldm-3 HNO3

Answer 8

[acid] = [H+]

[HNO3] = [H]

pH = - log(2.20) = -0.34

Question 9

What does it mean that moles of acid = moles of H

Answer 9

[Acid] = [H+]

Question 10

What to for diprotic acids when calculating pH or H+ concentratioj

Answer 10

pH = Times calculated [H+] by 2

[H+] = Divide calculated [acid] by 2

Question 11

As [H+] and [OH-] are the same mole values, how can the Kw equation be rewritten?

Answer 11

Kw = [H+]^2

Question 12

Effect of increasing temperature (above 298K) on Kw and pH of water

Answer 12

[H+] increases and Kw increases (above 1x10^-14)

pH decreases below 7

Question 13

Effect of decreasing temperature (below 298K) on Kw and pH of water

Answer 13

[H+] decreases and Kw decreases (below 1x10^-14)

pH increases above 7

Question 14

What is the pH of 0.65moldm-3 NaOH at standard conditions?

Answer 14

[NaOH] = [OH-] = 0.65

kW = [H+][OH-]

[H+] = kW / [OH-]

1x10^-14 / 0.65 = 1.54x10^-14

pH = -log(1.54x10^-14)

= 13.81

Question 15

How could you work out pH from pOH?

Answer 15

pOH = -log[OH-]

14 = pH + pOH

Question 16

pH changes during dilution:

What is the pH change if 20cm3 0.1moldm-3 HCl has 30cm3 water added?

Answer 16

[H+] = 0.1

pH = -log(0.1) = 1.00

Adding water does not change HCl moles

moles = 0.1 x 20/1000 = 0.002

New volume = 20 + 30 = 50cm3 = 0.05dm3

HCl new moles = 0.002 / 0.05 = 0.04

pH = -log(0.04) = 1.40

Question 17

What is Ka? (Association constant)

Answer 17

Dissociation constant for a weak acid

Larger Ka = stronger acid

Question 18

What 2 equations can be formed from

HA + H2O ⇌ H3O+ + A-

Answer 18

Ka = [H+][A-] / [HA]

Ka = [H+]^2 / [HA]

Question 19

Route to find the pH of weak acids

Answer 19

Ka = [H+]^2 / [HA]

[H+]^2 = Ka x [HA]

[H+] = √Ka x √[HA]

pH = -log[H+]

Question 20

2 equations to calculate Ka into pKa

Answer 20

pKa = -log(Ka)

Ka = 10^-pKa

Question 21

pKa and Ka value of a strong and weak acid

Answer 21

Strong acid = Larger Ka = Smaller pKa

Weak acid = Smaller Ka = Larger pKa

Question 22

Equation process of of reacting an acid and alkali of different volumes: Acid in excess

Answer 22

[H+] = Excess H+ moles / new volume

pH = -log[H+]

Question 23

Equation process of of reacting an acid and alkali of different volumes: Base in excess

Answer 23

[H+] = Excess OH- moles / new volume

[H+] = Kw / [OH-]

pH = -log[H+]

Question 24

Equation process of of reacting an acid and alkali of different volumes: Weak acid in excess

Answer 24

[H+] = Ka x ([HA] / [A-])

[H+] = Ka x (excess HA mol / A- mol)

pH = -log[H+]