Difficult stuff

Question 1

Pale pink trans element ions

Answer 1

Mn^2+ and Co^2+

Question 2

Lilac trans element ions

Answer 2

Ti^2+ and V^2+

Question 3

Purple trans element ions

Answer 3

Ti^3+ and MnO^4+

Question 4

Green trans element ions

Answer 4

V^3+ and Cr^3+ ad Ni^2+ and MnO4^2-

Question 5

Pale green trans element ions

Answer 5

Fe^2+

Question 6

Pale blue trans element ions

Answer 6

Cu^2+

Question 7

Blue trans element ions

Answer 7

VO^2+ and Cr^2+

Question 8

Yellow trans element ions

Answer 8

VO2^+1 and Fe^3+ and CrO4^2-

Question 9

Orange trans element ions

Answer 9

Cr2O7^2-

Question 10

Precipitation rct of Cu^2+

Answer 10

Cu2+ (aq) + 2OH- –> Cu(OH)2 (s)

Blue solution –> Blue ppt

Question 11

Precipitation rct of Mn^2+

Answer 11

Mn^2+ (aq) + 2OH- –> Mn(OH)2 (s)

Pale pink solution –> Pale brown ppt

Question 12

Precipitation rct of Fe^2+

Answer 12

Fe^2+ (aq) + 2OH- –> Fe(OH)2 (s)

green solution –> green ppt

Question 13

Precipitation rct of Fe^3+

Answer 13

Fe^3+ (aq) + 3OH- –> Fe(OH)3 (s)

yellow solution –> brown ppt

Question 14

Reaction of Cu(H20)6^2+ (aq) with NH3

Answer 14

Cu(H20)6^2+ (aq) + 4NH3 (aq) –> Cu(NH3)4(H2O)2^2+ + 4H2O

Blue solution –> Dark blue solution

Question 15

Reaction of Co(H20)6^2+ (aq) with NH3

Answer 15

Co(H20)6^2+ (aq) + 6NH3(aq) –> Co(NH3)6^2+ + 6H2O

Question 16

Reaction of Cu(H2O)6^2+ with Cl-

Answer 16

Cu(H2O)6^2+ + 4Cl- —> CuCl4^2- + 6H2O

Blue solution –> Yellow/Green solution

Question 17

Reaction of Cu(H2O)6^2+ with Cl-

Answer 17

Co(H2O)6^2+ + 4Cl- –> CoCl4^2+ + 6H2O

Pink solution to blue solution

Question 18

Reaction of Cu and Cr with excess NH3

Answer 18

Cr(OH)3(H2O)3 (s) + 6NH3 (aq) –> Cr(NH3)6^3+ + 3H2O (l) + 3OH- (aq) Purple solution

Cu(OH)2(H2O)4 (s) + 4NH3 (aq) —> Cu(NH3)4(H2O)2 + 2H2O (l) + 2OH-(aq) Deep blue solution

Question 19

Precipitation rct of Cr3+

Answer 19

Cr(H20)6^3+ + 3OH- –> Cr(OH)3(H2O)3 (s) + 3H2O (l)

Green solution –> Grey-green ppt

Question 20

How is Fe2+ oxidised to Fe3+

Answer 20

MnO4^- + 8H+ + 5Fe2+ –> Mn^2+ + 4H2O + 5Fe^3+

Pale green –> Yellow

Question 21

How is Fe3+ reduced to Fe2+

Answer 21

2I- + 2Fe^3+ –> 2Fe^2+ + I2

Yellow –> Pale green

Question 22

How is Cr^3+ oxidised to CrO4^2-

Answer 22

3H2O2 + 2Cr(OH)6^3- –> 2OH- + 2CrO4 ^2- + 8H2O
Dark green to yellow
Warm H2O2 in alkaline conditions
Oxygen is reduced and Chromium is oxidised

Question 23

How is Cr2O7^2- reduced to Cr^3+

Answer 23

Cr2O7^2- + 14H+ +3Zn –> 2Cr^3+ + 7H2O + 3Zn^2+
Orange to Green
Acidified zinc
ZInc is oxidised and chromium is reduced

Question 24

How can trans elements be identified

Answer 24

- Add sodium hydroxide (aq) 
Cu^2+ = blue ppt
Fe^2+ = green ppt 
Fe^3+ = Yellow/orange ppt
Mn^2+ = pink/buff ppt
Cr^3+ =grey/green ppt

Question 25

How do you calculate a pH of a strong acid

Answer 25

• conc of H+ is same as conc of acid

Question 26

How to calculate pH of water

Answer 26

• Use Kw and square root as [H+] = [OH-]

Question 27

How to calculate pH of strong base

Answer 27

• Work out [H+] using kw expression (Kw = 1x10-14

Question 28

How to calculate pH of a weak acid

Answer 28

• use Ka = [H+] + [A-] / [HA]
• assume: [H+] = [A-] and [HA] (initial) = [HA] (eqm)
• simplify Ka to Ka = [H+] ^2/HA

Question 29

How to calculate pH of diluted strong acid

Answer 29

• [H+] = [H+] (old) x old volume/new volume

- pH = -log(H+)

Question 30

How to calculate pH of diluted strong alkali

Answer 30

• [OH-] = [OH-] (old) x old vol/new vol
• [H+] = Kw / [OH-]
• pH = -log(H+)

Question 31

How to calculate the pH of buffer solution

Answer 31

Ka = [H+] [A-] / [HA]
Assume the initial conc of the acid remained constant as the amount that has dissociated is small
Rearrange to [H+] = Ka x [HA] / [A-]

Question 32

How to calculate change in ph of buffer on addition of small amount of acid/alkali

Answer 32

• Work out moles of acid and salt in initial buffer solution
• Work out moles of acid and salt after addition of acid/alkali by taking away/adding conc of acid/alkali to initial moles of acid and salt
• Sub in values into Ka expression