NEETChemistryIonic Equilibrium
Match the Column - I and Column - II ( array |l|l|l|l| & Column-I & & Column-II (a) & Al ( OH )₃ & (p) & S = ( K _ sp 6912 )^ 1 / 7 (b) & BaCrO ₄ & (q) & S = ( K _ sp 4 )^ 1 / 3 (c) & Zr ₃ ( PO ₄ )₄ & (r) & S = ( K _ sp 27 )^ 1 / 4 (d) & Hg ₂ ~L ₂ & (s) & S = ( K _ s )^ 1 / 2 array )
Options
- A(a) ( ) (r); (b) ( ) (s); (c) ( ) (p); (d) ( ) (q)
- B(a) ( ) (s); (b) ( ) (p); (c) ( ) (r); (d) ( ) (q)
- C(a) ( ) (p); (b) ( ) (q); (c) ( ) (s); (d) ( ) (r)
- D(a) ( ) (q); (b) ( ) (r); (c) ( ) (p); (d) ( ) (s)
Correct answer
B. (a) ( ) (s); (b) ( ) (p); (c) ( ) (r); (d) ( ) (q)
Step-by-step solution
Here's the matched column: (a) ( Al ( OH )₃( p ) 1 / 3 Ksp [ Al 3+][ OH -] 3= 1 27 ) (b) ( BaCrO ₄( q ) 1 / 2 Ksp [ Ba 2+][ CrO 42-]= 1 4 ) (c) ( Zr ₃ ( PO ₄ )₄( r ) 1 / 4 Ksp [ Zr ₄+][ PO ₄³⁻]₄= 1 27 ) (d) ( Hg ₂ I ₂( ~s ) Ksp [ Hg ^2₂+][ I -] 2=1 ) So, the correct answer is: (2) (a) ( ) (p); (b) ( ) (q); (c) ( ) (r); (d) ( ) (s) Explanation: - (a) Aluminum hydroxide: ( Al ( OH )₃ Al ₃++3 OH -, Ksp =[ Al ₃+][ OH -] 3 ) - (b) Barium chromate: ( BaCrO ₄ Ba ²⁺+ CrO ₄²⁻, Ksp =[ Ba ₂+][ CrO ₄²⁻] ) - (c) Zirconium phosp