QUIZ ON SOLUTIONS, SOLUBILITY, ACID-BASE THEORY.

*CATALYST’s  QUIZ*.

👉🏼Tuesday, 19th may 2020.

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INSTRUCTION.

POST YOUR ANSWERS INTO THE COMMENT BOX.

Show your working on question number 5.

QUESTION  1.

_____ is the conjugate acid of hydrogen oxide ?

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QUESTION  2.

_____ is the problem associated with Arrhenius’ theory of acids and bases ?

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QUESTION  3.

Whose theory of acid is exhibited by ethanoic acid in the reaction below?

CH3COOH + NaOH –> CH3COONa + H2O

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QUESTION 4.

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TOPIC 👉🏼 SOLUTIONS.

A solution in which Q > Ksp is known as a ____ solution ?

QUESTION  5.

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TOPIC 👉🏼 SOLUBILITY and Ksp

Calculate the solubility of Al2O3 at a given condition, given that it’s  solubility product constant (Ksp) is 8.83×10^–3

Good luck…

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THE CATALYST
May 20, 2020 2:17 pm

You would have gotten 5/5 but your answer to number 5 came late.

Unknown
May 20, 2020 11:14 am

Name is Ann
1. H3O
2. The limitation of Arrhenius definition of acid/base is that Water must be present to show if it's an acid or a base.
3. Bronsted-Lowry; because the acid donated a proton (H^+) and the base accepted a proton.
4. It is a super saturated solution
5. Al2+O3= 5moles
2[S]².3[S]³
2².S² .3³.S³
4.S² . 27.S³
4*27 .S²+³
Ksp=108(solubility)^5
8.83×10^-3=108(solubility)^5
Divide both side by 108
5th root of 0.000082 = 0.15

Nelson
May 20, 2020 10:44 am

1. H2O
2. Water as a necessity and Hydrogen ion or Hydroxyl ion must be produced
3. Brownsted_ Lowry theory because a proton or hydrogen ion was donated by ethanoic acid to the base which accepts the proton
4. Super saturated solution
5. Ksp = 2^2. 3^3 (S)^2+3
8.83×10^-3= 4×27 (S)^5
8.83×10^-3= 108 (S)^-5
8.83×10^-3/108= S^-5
8.176×10^-5= S^-5
S= 0.152

Unknown
May 20, 2020 10:36 am

My first answer is the correct one😪

Unknown
May 20, 2020 9:08 am

S=0.04522

Unknown
May 20, 2020 9:04 am

Invictus 101
3) Bronsted_Lowry thats my correction

Unknown
May 20, 2020 9:04 am

Name: Boniface

Unknown
May 20, 2020 9:02 am

1)H2O
2)the limitation is that it's acid theory cannot be explained using a reaction that does not occur in aqueous solution.
3)Bronsted Lowry
4) supersaturated solution
5) (2Al^3+) (3O^2-) = Ksp
Let S stand for solubility
(2S)^2. (3S)^3 = 8.83×10^-3
36S^5 = 8.83×10^-3/ 36
S^5 = 0.000245
S = 0.0783

Unknown
May 20, 2020 8:58 am

Final answer
1)H2O
2)the limitation is that it's acid theory cannot be explained using a reaction that does not occur in aqueous solution.
3)Bronsted Lowry
4) supersaturated solution
5) (2Al^3+) (3O^2-) = Ksp
Let S stand for solubility
(2S)^2. (3S)^3 = 8.83×10^-3
36S^5 = 8.83×10^-3/ 36
S^5 = 0.000245
S = 0.0783

Unknown
May 20, 2020 8:25 am

NAME: Ejiogu Chimdia

1. H3O^+ ____ Hydroxonium ion
2. The substance must be
dissolved in water
3. Bronsted-Lowry's theory of
acid
4. A super-saturated solution
5. Al2O3 —–> 2Al + 3O
8.83×10^-3 = [2s]^2 + [3s]^3
=2^2×s^2×3^3×s^3
=4×27×s^2+3
=108 ×s^5
8.83×10^-3 = 108s^5
s^5 = 8.83×10^-3 / 108
s^5 = 8.1759×10^-5
s = 5√8.1759×10^-5
s = 5th root of 8.1759×10^-5
s = 0.1522

Unknown
May 20, 2020 8:00 am

From xandra

Unknown
May 20, 2020 8:00 am

S=0.1522mol/dm³

Unknown
May 20, 2020 7:57 am

S=0.04522

Ricky_rizz
May 20, 2020 7:03 am

1. H30+
2. Water must be present and it must produce hydrogen ion if acid or hydroxyl ion if base
3. Lewis theory
4. Supersaturated
5. 8.83×10-³=6( solubility ) ^5

8.83×10-³/6=(Solubility)^5
0.00147=(solubilty)^5
5 root 0.00147=solubility
Solubility =0.187..

Unknown
May 20, 2020 7:01 am

Success
1.H20^+
2.Acid must contain Hydrogen ion(H+) while base must contain Hydroxyl ion(OH-)
3.Bronsted_Lowry theory of acid and base
4.Super-saturated solution
5.Al2O3–>Al2 + O3
(2Al)² .(3O)³
Ksp=4(S)².27(S)³
Ksp=108(S)^5
108(8.83*10^-3)^5
5.79*10^-9

Unknown
May 20, 2020 6:42 am

Correction to number 5;
KSP= 108(S)^5
KSP/108=S^5
8.83×10^-³/108=S^5
8.17593×10^-5=S^5
5√8.17593= S
0.1522= S
Therefore, Solubility= 0.1522

Unknown
May 20, 2020 6:28 am

By Chidera

Unknown
May 20, 2020 6:28 am

1. H2O^+
2. It is limited to only water
3. Browsted_Lowry theory
4. Supersaturated solution
5. Al2O3 —-> 2Al + 3O
[2.S]^2 . [3.S]^3
2^2.S^2 . 3^3.S^3
4.S^2 . 27.S^3
108.S^2+3
108S^5

But S=8.83*10^-3

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108(8.83*10^-3)^5
5.8*10^-9

Unknown
May 20, 2020 6:22 am

The name's Miracle

Unknown
May 20, 2020 5:58 am

Orisakwe Emmanuel
Correction on number one

Conjugate of hydrogen oxide (H20) is

1)H3O+

Unknown
May 20, 2020 12:32 am

Sir please I used X instead of S in solving number 5…

I'm rewriting it again
5. Al2O3 – 2Al +3O
(2Al) ^². (3O) ^³
Ksp= 4(S) ². 27(S) ^³
Ksp= 108(S) ^5
8.83×10^-3 =108(S) ^5
Divide through by 108
S^5=8.18×10^-5
S=5√8.18×10^-5
S = 0.1522

Unknown
May 20, 2020 12:07 am

Steph onyeokoro
1 H3O^+.
2 water is a necessity… Acid must have hydrogen, bases must have hydroxyl ion
3 Bronsted_Lowry theory of acid and base
4 super saturated solution
5 Al2O3 – 2Al + 3O
(2Al) ^². (3O) ^³
Ksp= 4(x)^². 27(x) ^³
Ksp=108(x) ^5
8.83×10^-3=108(x) ^5
Divide through by 108
X^5=8.18×10^-5
X=5√8.18×10^-5
X= 0.1522

Godstime
May 19, 2020 11:33 pm

Godstime

1. H30^+

2. The presence of water as a necessity

3. Bronsted-lowry's theory

4. Super saturated solution

5. Ksp = 108(Solubility)^5
(Solubility)^5 = Ksp/108
Solubility = 5√Ksp/108
Solubility =
5√8.83×10^-3/108
Solubility = 0.15M

Unknown
May 19, 2020 10:32 pm

1.H2O
2. The recognition of water as a necessity.
3.bronsted_lowry theory
4. Supersaturated solution.
5. AL2O3
KSP=(2S)². (3S)³
4S². 27S³
108.S^5
Hence,
KSP=108.(solubility)^5
(Solubility)^5= KSP/108
=8.83×10^-3/108
=8.18×10^-5
Solubility= 5√8.18×10^-5
=1.522×10^-5

VICTOR C

Unknown
May 19, 2020 10:28 pm

1.H2O
2. The recognition of water as a necessity.
3.bronsted_lowry theory
4. Supersaturated solution.
5. AL2O3
KSP=(2S)². (3S)³
4S². 27S³
108.S^5
Hence,
KSP=108.(solubility)^5
(Solubility)^5= KSP/108
=8.83×10^-3/108
=8.18×10^-5
Solubility= 5√8.18×10^-5
=1.522×10^-5

Unknown
May 19, 2020 10:23 pm

Isaac M
1.H3O^+
2.Water as a necessity
3.Arrhenius
4.super saturated solution
5.Al2O3=2Al+3O
Ksp= (2S)^2 . (3S)^3
4S^2 . 27S^3
Ksp= 108S^5
8.83 •10^-3=108S^5
S^5=8.83•10^-3/108
S^5=8.1759•10^-5
S=5√8.1759•10^-5
S=0.152mol/dm^3

Tycoons wurld
May 19, 2020 10:22 pm

1) H30^+
2)(i) Presence of Water as a necessity .
(ii) Substances must exhibit the possession of H^+ to qualify as an acid and OH^- to qualify as a base.
3)Bronsted-Lowry Theory.
4) A Supersaturated Solution.
5)Al2O3=Al2+O3
Ksp=(2S)^2.(3S) ^3
Ksp=2^2.S^2.3^3.S^3
Ksp= 4.S^2.27.S^3
Ksp= 4×27S^5
Ksp=108S^5
Finding Solubility, S
After solving Critically
S=5√8.83×10^-3÷108
S= 0.152 =1.52×10^-1 mol/dm³

Unknown
May 19, 2020 10:16 pm

1)H2O
2)Acids produce H3O+ when dissolved in water, while base produces OH- when dissolved in water(Water is a necessity)
3)Arrehenius Theory
4)Super-Saturated solution
5)Al2O3—->Al3+ +2O^2-
Ksp=[Al3+] [2O^2-]
8.83×10^-3=[X^3+] [2X^2-]
8.83×10^-3=[2X^1]
8.83×10^-3= X
—————
2
X=4.415×10^-3 Mol/dm^3

Unknown
May 19, 2020 10:10 pm

Promise prosper
5. Solubility = 5 root of 8.83^-3÷ by108 = 0.152mol/dm^3

Unknown
May 19, 2020 9:53 pm

Josh_Solution

1. H3O^+

2. Water is a necessity. Mainly there must be a proton which is (a hydrogen ion) that is being donated or accepted. If there is no hydrogen ion in the reaction, an Arrhenius acid or base can not be determined.

3. Arrhenius Theory

4. Supersaturated Solution

5. Al2O3
Ksp= (2S)² × (3S)³
Ksp= 4.S² × 27.S³
Ksp= 108 (S)^5
Ksp= 108 (8.83×10-³)^5
Ksp= 108 (5.37×10^-11)
Ksp= 5.80×10^-9

Unknown
May 19, 2020 9:44 pm

Promise prosper
5. (8.83×10^-3 ×108)^5
= 0.7887

Unknown
May 19, 2020 9:41 pm

Promise prosper
1. H2O
2. Necessity of water
3. Bronsted lowry
4. Super saturated solution
5.0.7887mol/dm^3

Unknown
May 19, 2020 9:37 pm

Ksp = 8.83 * 10^-3
8.83 *10^-3
—————
108
S=5√ 8.17 *10^-5
S = 0.045M

Unknown
May 19, 2020 9:32 pm

Danny

1 H30+
2 water is a necessity and it's important in order to find an acid or a base

3 bronsted_lowry theory
4 super saturated solution

5 Al2CO3
Al2CO3 = ksp
(2S)².(3S)³ = ksp
4S².27S³ = ksp
108.S^5 =ksp
108.S^5 = 8.83 ×10^-13
S^5 = 8.83 × 10^-13
——————-
108
S^5 =8.17 ×10^-15
S = 5√8.17 ×10^-15
S = 4.51 ×10^-7

Unknown
May 19, 2020 9:32 pm

FCBOY

1. H3O^+

2. It must produce H^+ to qualify as acid and OH^- as a base

Ii) water must be present

3. Arrhenius

4. Super saturated solution

5.
S = 5√ksp/108
S = 5√8.83×10^-3/108
S = 0.15223mol/dm^3

Unknown
May 19, 2020 9:30 pm

From Benitha Benson
1)H20
2)water as a necessity
3) Arrhenius Theory
4)super saturated solution
5)(2s)^2×(3s)^3
4×s^2×27×s^3
4×27×s^5
108×s^5
Ksp=108(s)^5
(Solubility)^5=8.83×10^-3/108
(Solubility)^5=8.18×10^-5
Solubility=5√8.18×10^-5
Solubility=0.045mol/dm^3
Dis is my final & correct ans..d two previous ones are wrong..

Unknown
May 19, 2020 9:21 pm

RITA

1. H2O

2. Water must be present before a substance can be called either acid or base ( water necessity).

3. Bronsted – Lowry theory of acid and base

4. supersaturated

5. Al2O3 dissociate to give 2Al + 3O2
Total number of moles=5
(2S)² x (3S)³
2² x S² x 3³ x S³
4 x S² x 27 x S³
4x27x S²xS³
108xS²+³
108x S ^5
S means solubility
108(solubility)^5
108( 8.83×10^-3)^5
S=5.8×10^-9

Unknown
May 19, 2020 9:11 pm

H30+ for question 2

Danny by name

Ebube
May 19, 2020 8:52 pm

Ebube
1) H3O
2) Water is a necessity ie water must be involved
3) Arrhenius theory
4) super saturated solution
5. Al2O3___ 2Al + 3O
Ksp= [2Al]^2 . [3O]^3
=4[Al]^2 .27[O]^3
= 4[S]^2 .27[S]^3
= 108[S]^5
= 108[8.83*10^-3]^5
= 5.797*10^-9(mol/dm^3)^2.

Unknown
May 19, 2020 8:52 pm

Francis_Best
(1) H3O+
(2) Water as a necessity i.e aqueous.
(3) Arrhenius theory of acids and base.
(4) Supersaturated solution
(5) Al2O3 → Al^2+ + O3^2-

Ksp= 2Al² +3O³
Ksp=4Al +27O²+³
Ksp= 4x 27 (Al2O3) ^5
Ksp=108(solubility) ^5

8.83×10-³=108(solubility) ^5
Solubility^5=8.83 x10-³/108
Solubility = 5√ 8.2 x10^-5
Solubility =0.0453

Unknown
May 19, 2020 8:43 pm

Invictus 101
1) H3O
2) Water is a necessity ie water must be involved
3) Arrhenius theory
4) super saturated solution
5) Al2O3
Al2 + O3
(2S)^2 . (3S)^3
4S^2 × 27S^3
(4×27) S^2+3
108 S^5
Ksp = 108S^5
Recall ksp = 8.83×10^-3
= 8.83×10^-3 = 108 S^5
S^5 = 8.83×10^-3÷108
S^5 = 8.1759×10^-5
S= 5√8.1759×10^-5
S= 0.1522
S= 0.152

Unknown
May 19, 2020 8:43 pm

Amira
1.H3O+
2.Necessity of water to produce H+ for acids and OH- for bases
3.BRONSTED-LOWRY
4.supersaturated solution
5.(2Al)^2. (3O)^3
(2S)^2. (3S)^3
4S^2. 27S^3=108.S^5
Ksp=108(solubility)^5
8.83×10^-3=108(solubility)^5
S^5=8.83×10^-3/108
S^5=8.18×10^-5
S=5√8.18×10^-5
Solubility=4.52×10^-2

Unknown
May 19, 2020 8:42 pm

I missed another 4 seriously is not good to rush is 0.004415 or 4.415×10-³

Mmesoma u
May 19, 2020 8:40 pm

1. H3O^+
2. He uses water as a necessity
3. Arrhenius theory of acid
4. Supersaturated solution
5. Al2O3___ 2Al + 3O
Ksp= [2Al]^2 . [3O]^3
=4[Al]^2 .27[O]^3
= 4[S]^2 .27[S]^3
= 108[S]^5
= 108[8.83*10^-3]^5
= 5.797*10^-9(mol/dm^3)^2

Unknown
May 19, 2020 8:40 pm

Oh am sorry 0.00415

Unknown
May 19, 2020 8:37 pm

Sir pls pls I 4got to input my No. 5
Al2O3> Al³+ +O2²
Ksp= [Al³+] . [2O]²-
8.83×10-³=Al³+ .2O²-
8.83×10-³=S³+.2S²-
8.83×10-³=2S¹
Solubility=8.83×10-³/2
=0.4415Mol/l ans

Uchenna Stephen
May 19, 2020 8:33 pm

0.152Mol/dm^-3

Unknown
May 19, 2020 8:31 pm

Benson Benitha
(2s)^2×(1s)^1
2^2×s^2×I^1×s^1
4×s^2×1×s^1
4s^3
Ksp=4(s)^3
(S)^3=Ksp/4
(S)^3=8.83×10^-3/4
(S)^3=2.2075×10^-3
S=3√2.2075×10^-3
S=0.13mol/dm^3..for no. 5
No.4=supersaturated solution
No.3=Arrhenius Theory
No.2=water as necessity
No.1=H20..
The previous one was wrong..dis is d right one

Anonymous Heymeka
May 19, 2020 8:30 pm

1. H30
2. Water as necessity
3. Bronsted_Lowry
4. Super Saturated (Q>ksp)
5. Al203(bauxite)
(2s)². (3s)³
4s² . 27s³
Ksp= 108s^5
To find 'S' …Ksp= 8.83×10^-3

Therefore
8.83×10^-3= 108s^5
Making 'S' the subject of formula

S= 5th root of 8.83×10^-3/108

S= 0.1522moldm^-3

Unknown
May 19, 2020 8:29 pm

Mikky
1H30^+
2.Water as a necessity
3. Bronsted Lowry
4 Unsaturated
5. Ksp=108.S^5
8.83*10^-3= 108.S^5
S^5= 8.83*10^-3/108
S=5√8.83*10^-3/108
S=0.152mol/dm^3

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