A) \[C{{H}_{3}}COONa<{{C}_{6}}{{H}_{5}}COON{{H}_{4}}\]\[<N{{H}_{4}}Cl<KN{{O}_{3}}\]
B) \[KN{{O}_{3}}<N{{H}_{4}}C<C{{H}_{3}}COONa\]\[<{{C}_{6}}{{H}_{5}}COONa\]
C) \[N{{H}_{4}}Cl<KN{{O}_{3}}<C{{H}_{3}}COONa\]\[<{{C}_{6}}{{H}_{5}}COON{{H}_{4}}\]
D) \[N{{H}_{4}}Cl<KN{{O}_{3}}<{{C}_{6}}{{H}_{5}}COON{{H}_{4}}\]\[<C{{H}_{3}}COONa\]
Correct Answer: D
Solution :
[d] (i) \[KN{{O}_{3}}\] is a salt of strong acid and strong base, hence its aqueous solution is neutral; pH = 7 |
(ii) \[C{{H}_{3}}COONa\] is a salt of weak acid and strong base/hence, its aqueous solution is basic ; pH > 7 |
(iii) \[N{{H}_{4}}Cl\] is a salt of strong acid and weak base, hence its aqueous solution is acidic; pH < 7 |
(iv) \[{{C}_{6}}{{H}_{5}}COON{{H}_{4}}\] is a salt of weak acid, \[{{C}_{6}}{{H}_{5}}COOH\] and weak base, \[N{{H}_{4}}OH\]. But \[N{{H}_{4}}OH\] is slightly stronger than \[{{C}_{6}}{{H}_{5}}COOH\]. Hence, pH is slightly > 7. |
Therefore, increasing order of pH of the given salts is |
\[<C{{H}_{3}}COONa\] |
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