UNIT 8: Ionic Equilibrium - Online Test

Ag₂C₂O₄ ↔ 2Ag⁺ + C₂O₄^2-

[ Ag⁺ ] = 2.24 × 10^-4 mol L^-1

[ C₂ O₄^2- ] = { 2.24×10^-4 }/2 = mol L^-1

 = 1.12×10^-4 mol L^-1

K_sp = [Ag⁺]²[C₂O₄^2-]

 = (2.24×10^-4 mol^-4 L^-1 )² (1.12×10^-4 mol L^-1)

= 5.619×10^-12mol³ L^-3

iii) 75 ml M/5 HCl + 25ml M/5 NaOH

No of moles of HCl = 0.2×75×10^-3 = 15 × 10^-3

No of moles of NaOH = 0.2 × 25 × 10^-3 = 5 × 10^-3

No of moles of HCl after mixing = 15 × 10^-3 - 5 × 10^-3

= 10 × 10^-3

concentration of HCl = No of moles of HCl / Vol in litre

= 10 ×10⁻³ / 100 ×10⁻³ = 0.1M

for (iii) solution, pH of 0.1M HCl = -log₁₀(0.1)

= 1.

BaSO₄ ↔ Ba²⁺ +SO₄^2-

K_sp =(s) (s)

K_sp =(s)²

= ( 2.42×10^-3 g L^-1 )²

= ( 2.42×10^-3 g L^-1 / 233g mol^-1 )

= ( 0.01038×10^-3 ) ²

= (1.038×10^-5 )²

= 1.077×10^-10

= 1.08×10^-10 mol² L^-2

Ca(OH)₂ ↔ Ca²⁺ + 2OH^-

Given that pH = 9

pOH = 14-9 = 5

[pOH = -log₁₀ [OH- ]]

∴[OH- ] = 10^{- pOH}

[OH^- ]=10^-5 M

K_sp =[Ca²⁺ ][OH^- ]²

= (10^-5 /2 )×(10^-5 )²

=0.5 ×10^-15

H₂O + H₂O ↔ H₃O⁺ + OH^-

acid 1 + base 1 ↔ acid 2 + base 2

HF + H₂O ↔ H₃O+ + F^-

acid 1 + base 1 ↔ acid 2 + base 2

Conjugate bases are OH^- and F^- respectively

Basic buffer is the solution which has weak base and its salt

 NH₄OH(200ml) + HCl(100ml) →NH₄Cl(Salt)  + H₂O + NH₄OH (100ml weak base)

BF₃ → elctron deficient → Lewis acid

PF₃ → electron rich → lewis base

CF₄ → neutral → neither lewis acid nor base

SiF₄^- → neutral → neither lewis acid nor base

BF₃ → elctron deficient → Lewis acid

PF₃ → electron rich → lewis base

CO → having lone pair of electron → lewis base

F^- → unshared pair of electron → lewis base

HCOONa Basic in nature. + H ⋅OH ↔ NaOH strong base  + H-COOH weak acid

C₆H₅NH₃Cl^- + H ⋅ OH ↔ H₃O⁺Acidic +C₆H₅ -NH₂ + Cl^-

KCN basic + H− OH ↔ KOH strong base + HCN weak acid

basic, acidic, basic is correct.

C₅H₅N + H-OH ↔ C₅H₅ NH + OH^-

(α²C) / (1-α ) =K_b

α²C =∼ K_b

α = √ [ K_b / C ] = √[ 1.7 ×10^-9  / 0.1 ]

= √1.7 × 10^-4

= √1.7 × 10⁻⁴ × 100

Percentage of dissociation  = 1.3 ×10^-2 = 0.013 %