Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Ionic Equilibrium question

2007 · Shift 0 · Q5
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Ionic Equilibrium
  5. /2007 · Shift 0 · Q5

Ionic Equilibrium question

2007 · Shift 0 · Q5

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
In a sautrated solution of the sparingly soluble strong electrolyte AgIO3AgIO_3AgIO3​ (Molecular mass = 283) the equilibrium which sets in is AgIO3AgIO_3AgIO3​(s) ⇋\leftrightharpoons⇋ Ag+Ag^+Ag+(aq) +IO3−IO_3^-IO3−​ If the solubility product constant Ksp of AgIO3AgIO_3AgIO3​ at a given temperature is 1.0 ×\times× 10−8, what is the mass of AgIO3AgIO_3AgIO3​ contained in 100 ml of its saturated solution?
  1. A
    28.3 × 10−2 g
  2. B
    2.83 × 10−3 g
  3. C
    1.0 × 10−7 g
  4. D
    1.0 × 10−4 g
View written solutionFree

Correct answer: B

  1. Write the dissolution equilibrium
AgIO3(s)⇌Ag+(aq)+IO3−(aq)AgIO_3(s) \rightleftharpoons Ag^+(aq) + IO_3^-(aq)AgIO3​(s)⇌Ag+(aq)+IO3−​(aq)

If the solubility of AgIO3AgIO_3AgIO3​ is sss mol L−1^{-1}−1, then at equilibrium:

[Ag+]=s,[IO3−]=s[Ag^+] = s, \qquad [IO_3^-] = s[Ag+]=s,[IO3−​]=s
  1. Use the solubility product expression
Ksp=[Ag+][IO3−]=s⋅s=s2K_{sp} = [Ag^+][IO_3^-] = s \cdot s = s^2Ksp​=[Ag+][IO3−​]=s⋅s=s2

Given:

Ksp=1.0×10−8K_{sp} = 1.0 \times 10^{-8}Ksp​=1.0×10−8

So,

s2=1.0×10−8s^2 = 1.0 \times 10^{-8}s2=1.0×10−8 s=1.0×10−8=1.0×10−4 mol L−1s = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4} \text{ mol L}^{-1}s=1.0×10−8​=1.0×10−4 mol L−1
  1. Find moles in 100 mL saturated solution
100 mL=0.100 L100 \text{ mL} = 0.100 \text{ L}100 mL=0.100 L

Moles of AgIO3AgIO_3AgIO3​ dissolved in 0.1000.1000.100 L:

moles=s×V=1.0×10−4×0.100=1.0×10−5\text{moles} = s \times V = 1.0 \times 10^{-4} \times 0.100 = 1.0 \times 10^{-5}moles=s×V=1.0×10−4×0.100=1.0×10−5
  1. Convert moles into mass

Molecular mass of AgIO3=283AgIO_3 = 283AgIO3​=283

mass=moles×molar mass\text{mass} = \text{moles} \times \text{molar mass}mass=moles×molar mass mass=1.0×10−5×283=2.83×10−3 g\text{mass} = 1.0 \times 10^{-5} \times 283 = 2.83 \times 10^{-3} \text{ g}mass=1.0×10−5×283=2.83×10−3 g
  1. Match with the options
2.83×10−3 g2.83 \times 10^{-3} \text{ g}2.83×10−3 g

This corresponds to Option B.

PreviousNext

More from Ionic Equilibrium

  • The first and second dissociation constants of an acid H2​A are 1.0 × 10−5 and 5.0 × 10−10 respectively. The overall dissociation constant of the acid will be :2007 · MCQ
  • Hydrogen ion concentration in mol / L in a solution of pH = 5.4 will be :2005 · MCQ
  • The solubility product of a salt having general formula MX2​, in water is: 4 × 10-12 . The concentration of M2+ ions in the aqueous solution of the salt is :2005 · MCQ
  • What is the conjugate base of OH−?2005 · MCQ
  • The conjugate base of H2​PO4−​ is :2004 · MCQ
  • The molar solubility (in ol L-1) of a sparingly soluble salt MX4​ is "s". The corresponding solubility product is Ksp. 's' is given in term of Ksp by the relation :2004 · MCQ
  • Which one of the following statements is not true?2003 · MCQ
  • When rain is accompanied by a thunderstorm, the collected rain water will have a pH value :2003 · MCQ