Class 11th

For precipitation to take place, it is required that the calculated ionic product exceeds the K_sp value. So, following data on K_sp values should have been provided to answer the question.

KspforFeS=6.3*10⁻¹⁸,

MnS=2.5*10⁻¹³,

ZnS=1.6*10⁻²⁴

CdS=8.0*10⁻²⁷

Before mixing:

[S²⁻]=1.0*10⁻¹⁹M and [M²⁺]=0.04M 

volume =10mL volume =5mL 

After mixing:

[S²⁻]=? [M²⁺]=?  

volume = (10+5)=15mL volume =15mL 

[S²⁻]= (1.0*10⁻¹⁹*10)/ 15=6.67*10⁻²⁰M

[M²⁺]= (0.04*5) / 15=1.33*10⁻²M

Ionic product = [M²⁺] [S²⁻] 

= (1.33*10⁻²) (6.67*10⁻²⁰)

=8.87*10⁻²²

This ionic product exceeds the Ksp of ZnS and CdS. Therefore, precipitation will occur in CdCl₂ and ZnCl₂ solutions.

Since pH=3.19,  

[H3O⁺]=6.46*10⁻⁴M

C₆H₅COOH+H₂O↔C₆H₅COO⁻+H₃O 

K_a= [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH] 

[C₆H₅COOH] / [C₆H₅COO⁻]= [H₃O⁺] / Ka=6.46*10⁻⁴ / 6.46*10⁻⁵=10

Let the solubility of C₆H₅COOAg be xmol/L.

Then,

[Ag⁺]=x

[C₆H₅COOH]+ [C₆H₅COO⁻]=x 

10 [C₆H₅COO⁻]+ [C₆H₅COO⁻]=x 

[C₆H₅COO⁻]=x / 11 

K_sp = [Ag⁺] [C₆H₅COO⁻] 

= >2.5*10⁻¹3=x (x / 11)

= >x=1.66*10⁻⁶mol/L

Thus, the solubility of silver benzoate in a pH3.19 solution is 1.66*10⁻⁶mol/L.

Now, let the solubility of C₆H₅COOAg be x′mol/L.

Then, [Ag⁺]=x′Mand [C6H5COO⁻]=x′M

K_sp= [Ag⁺] [C6H5COO⁻] 

Ksp= (x′)²

x′= (Ksp)^1/2= (2.5*10⁻¹³)^1/2 = 5*10⁻⁷mol/L

∴x / x′= (1.66*10⁻⁶) / (5*10⁻⁷) =3.32

Hence, C₆H₅COOAg is approximately 3.317 times more soluble in a low pH solution.

1. Silver chromate:

Ag₂CrO₄→2Ag⁺+CrO₄²⁻

Then, [Ag⁺] = (2s), [CrO₄²⁻] = s

K_sp= [Ag⁺]² [CrO₄²⁻]

s = 0.65*10⁻⁴M

So, [Ag⁺] = 2s = 1.30 x 10⁻⁴M, [CrO₄²⁻] = 6.5 x 10^-5M

2. Barium Chromate:

BaCrO₄→Ba²⁺+CrO₄²⁻

[Ba²⁺] = [CrO₄²⁻] = s

Then, K_sp= [Ba²⁺] [CrO₄²⁻] = s x s

= > 1.2 x 10^-10M = s²

= > s = 1.09 x 10^-5M

3. Ferric Hydroxide:

Fe (OH)₃→Fe³⁺ + 3OH⁻

Then [Fe³⁺] = s, [OH⁻] = 3s

K_sp= [Fe³⁺] [OH⁻]³

Let s be the solubility of Fe (OH)₃

[Fe³⁺] = s = 1.38 x 10^-10M

[OH⁻] = 3s = 4.14 x 10^-10M

4. Lead Chloride:

PbCl₂→Pb²⁺+2Cl⁻

Then, [Pb²⁺] = s, [Cl⁻] = 2s

K_sp= [Pb²⁺] [Cl⁻]²

= >Ksp=s x (2s)² =4s³ 

⇒1.6*10⁻⁵=4s³

⇒0.4*10⁻⁵=s³ 

=>4*10⁻⁶=s³

[Pb²⁺] = s=1.58*10⁻²M

[Cl⁻] = 2s=3.16*10⁻²M

5. Mercurous Iodide:

Hg₂I₂→ 2Hg⁺+2I⁻

[Hg²⁺] = [I⁻] = 2s

K_sp= [Hg²⁺]² [I⁻]²

=>4.5 x 10^-29 = (2s)² (2s)² = 16s⁴

=> s = 4.09 x 10^-8 M

 Therefore, [Hg²⁺] = [I⁻] = 2s = 8.18 x 10^-8 M

(a) Total number of moles present in 10 mL of 0.2 M calcium hydroxide are  (10*0.2) / 1000? = 0.002 moles.
Total number of moles present in 25 mL of 0.1 M HCl are (25*2) / 1000? = 0.0025 moles.

Ca (OH)_2? + 2HCl → CaCl₂? + 2H₂? O
1 mole of calcium hydroxide reacts with 2 moles of HCl.
0.0025 moles of HCl will react with 0.00125 moles of calcium hydroxide.
Total number of moles of calcium hydroxide unreacted are 0.002−0.00125 = 0.00075 moles.
Total volume of the solution is 10 + 25 = 35 mL.
The molarity of the solution is  (0.00075*1000) / 35? = 0.0214M
[OH⁻] = 2 * 0.0214 = 0.0428M
pOH = −log0.0428 = 1.368
pH = 14 − pOH

 = 14 − 1.368 = 12.635

(b) Total number of moles present in 10 mL of 0.01 M H₂? SO₄?  is  (10*0.01) / 1000? = 0.0001 mol.
Total number of moles present in 10 mL of 0.01 M Ca (OH)₂? = (10*0.01) / 1000? = 0.0001 mole.
Ca (OH)₂? + H₂? SO₄? → CaSO₄? + 2H_2? O
0.0001 mole of Ca (OH)₂?  will react completely with 0.0001 mole of H₂? SO₄?
Hence, the resulting solution is neutral with pH 7.0.

(c) Total number of moles in 10 mL of 0.1 M H₂? SO₄? = (10*0.1) / 1000? = 0.001 mole.
Total number of moles present in 10 mL of 0.1 M KOH = (10*0.1? ) / 1000 = 0.001 mole.
2KOH + H₂? SO₄? → K₂? SO₄? + 2H₂? O
2 moles of KOH reacts with 1 mole of H₂? SO₄.
0.001 mole of KOH will react with 0.0005 mole of H₂? SO₄?
Number of moles of H₂? SO₄ left = 0.001−0.0005 = 0.005M
Volume of solution is 10+10 = 20mL
Molarity of the solution is  (0.0005*1000? ) / 20 = 0.025M.
[H⁺] = 2 * 2.5 * 10⁻² = 0.05M
pH = −log [H⁺] = −log0.05 = 1.3.