NEETChemistryIonic Equilibrium
Match the Column - I and Column - II ( array |l|l|l|l| & Column-I & & Column-II & (Species) & & (Conjugate acid) (a) & NH ₃ & (p) & H ₂ CO ₃ (b) & HCO ₃ ⁻ & (q) & H ₂ SO ₄ (c) & H ₂ O & (r) & NH ₄ ⁺ (d) & HSO ₄⁻ & (s) & H ₃ O ⁺ array ) (1) (a) ( ) (r); (b) ( ) (p); (c) ( ) (s); (d) ( ) (q) (2) (a) ( ) (p); (b) ( ) (r); (c) ( ) (q); (d) ( ) (s) (3) (a) ( ) (r); (b) ( ) (s); (c) ( ) (p); (d) ( ) (q) (4) (a) ( ) (q); (b
Options
- A( NH 3( r ) NH 4+ )
- B( HCO ₃-( p ) H ₂ CO 3 )
- C( H ₂ O ( s ) H 3 O + )
- D( HSO ₄ ) - (q) ( H ₂ SO ₄ )
Correct answer
A. ( NH 3( r ) NH 4+ )
Step-by-step solution
So, the correct answer is: (1) (a) ( ) (r); (b) ( ) (p); (c) ( ) (s); (d) ( ) (q) Explanation: - (a) Ammonia ( ( NH ₃ ) ) accepts a proton to form its conjugate acid, ammonium ( ( NH ₄ ) ). - (b) Bicarbonate ( ( HCO ₃ ) ) accepts a proton to form its conjugate acid, carbonic acid ( ( H ₂ CO ₃ ) ). - (c) Water ( ( H ₂ O ) ) accepts a proton to form its conjugate acid, hydronium ( ( H ₃ O ) ). - (d) Hydrogen sulfate ( ( HSO ₄ ) ) accepts a proton to form its conjugate acid, sulfuric acid ( ( H ₂ SO ₄ ) ).