A) \[C{{H}_{3}}COOK\]
B) \[N{{a}_{2}}C{{O}_{3}}\]
C) \[N{{H}_{4}}Cl\]
D) \[NaN{{O}_{3}}\]
Correct Answer: B
Solution :
\[pH=\log \frac{1}{[{{H}^{+}}]}\]\[pH\]is inversely proportional to hydrogen ion concentration. As concentration of\[{{H}^{+}}\]decreases \[pH\] increases and vice versa, (a) Ammonium chloride \[(N{{H}_{4}}Cl)\]is a salt of weak base and strong acid. So, its aqueous solution will be acidic as\[N{{H}_{4}}Cl+{{H}_{2}}O\xrightarrow{{}}\underset{weak\,\,base}{\mathop{N{{H}_{4}}OH}}\,+\underset{strong\,\,acid}{\mathop{HCl}}\,\] So, \[pH\] of \[N{{H}_{4}}Cl\] is less than 7. (b) \[NaN{{O}_{3}}\] (Sodium nitrate) is the salt of strong acid and strong base. So, its aqueous solution is neutral as \[NaN{{O}_{3}}+{{H}_{2}}O\xrightarrow{{}}\underset{strong\,\,base}{\mathop{NaOH}}\,+\underset{strong\,\,acid}{\mathop{HN{{O}_{3}}}}\,\] So, \[pH\] of \[NaN{{O}_{3}}\] is 7. (c) \[C{{H}_{3}}COOK\] (Potassium acetate) is a salt of strong base and weak acid. Its aqueous solution will be basic and pH value will be\[>7\approx 8.8\] \[C{{H}_{3}}COOK+{{H}_{2}}O\xrightarrow{{}}\underset{weak\,\,acid}{\mathop{C{{H}_{3}}COOH}}\,+\underset{strong\,\,base}{\mathop{KOH}}\,\] (d)\[N{{a}_{2}}C{{O}_{3}}\](Sodium carbonate) is a salt of strong base and weak acid. Its aqueous solution is also basic and its \[pH\] value will be more than 10. \[N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O\xrightarrow{{}}\underset{strong\,\,base}{\mathop{2NaOH}}\,+\underset{weak\,\,acid}{\mathop{{{H}_{2}}C{{O}_{3}}}}\,\]
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