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Ionic Equilibrium question

2010 · Shift 0 · Q4
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  5. /2010 · Shift 0 · Q4

Ionic Equilibrium question

2010 · Shift 0 · Q4

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
Solubility product of silver bromide is 5.0 ×\times× 10–13. The quantity of potassium bromide (molar mass taken as 120g of mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is :
  1. A
    1.2 ×\times× 10–10 g
  2. B
    1.2 ×\times× 10–9 g
  3. C
    6.2 ×\times× 10–5 g
  4. D
    5.0 ×\times× 10–8 g
View written solutionFree

Correct answer: B

  1. Condition for precipitation of AgBr\mathrm{AgBr}AgBr

    Precipitation starts when the ionic product becomes equal to the solubility product: [Ag+][Br−]=Ksp[\mathrm{Ag^+}][\mathrm{Br^-}] = K_{sp}[Ag+][Br−]=Ksp​

    Given: Ksp(AgBr)=5.0×10−13K_{sp}(\mathrm{AgBr}) = 5.0 \times 10^{-13}Ksp​(AgBr)=5.0×10−13

  2. Initial concentration of silver ion

    The solution is 111 L of 0.050.050.05 M AgNO3\mathrm{AgNO_3}AgNO3​.

    Since AgNO3\mathrm{AgNO_3}AgNO3​ is a strong electrolyte: [Ag+]=0.05 M[\mathrm{Ag^+}] = 0.05\,\text{M}[Ag+]=0.05M

  3. Minimum bromide ion concentration needed to start precipitation

    At the point precipitation just begins: [Br−]=Ksp[Ag+][\mathrm{Br^-}] = \frac{K_{sp}}{[\mathrm{Ag^+}]}[Br−]=[Ag+]Ksp​​

    Substituting values: [Br−]=5.0×10−130.05=1.0×10−11 M[\mathrm{Br^-}] = \frac{5.0 \times 10^{-13}}{0.05} = 1.0 \times 10^{-11}\,\text{M}[Br−]=0.055.0×10−13​=1.0×10−11M

  4. Moles of Br−\mathrm{Br^-}Br− needed in 1 L

    Since volume is 111 L, moles of Br−=1.0×10−11\text{moles of } \mathrm{Br^-} = 1.0 \times 10^{-11}moles of Br−=1.0×10−11

    KBr\mathrm{KBr}KBr dissociates completely: KBr→K++Br−\mathrm{KBr \rightarrow K^+ + Br^-}KBr→K++Br−

    Therefore, moles of KBr required=1.0×10−11\text{moles of KBr required} = 1.0 \times 10^{-11}moles of KBr required=1.0×10−11

  5. Mass of KBr required

    Molar mass of KBr=120 g mol−1\mathrm{KBr} = 120\,\text{g mol}^{-1}KBr=120g mol−1

    mass=moles×molar mass\text{mass} = \text{moles} \times \text{molar mass}mass=moles×molar mass =1.0×10−11×120= 1.0 \times 10^{-11} \times 120=1.0×10−11×120 =1.2×10−9 g= 1.2 \times 10^{-9}\,\text{g}=1.2×10−9g

  6. Match with options

    1.2×10−9 g1.2 \times 10^{-9}\,\text{g}1.2×10−9g corresponds to Option B.

Final Answer: B. 1.2×10−91.2 \times 10^{-9}1.2×10−9 g

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