A) 100 mL 0.1 M HCl+100mL 0.1 M KCl
B) 100 mL 0.1M \[{{H}_{2}}S{{O}_{4}}\]+100 mL 0.1M NaOH
C) 100 mL 0.1M\[{{\operatorname{CH}}_{3}}COOH\]+100mL 0.1M KOH
D) 50 mL 0.1M HCl+50mL 0.1 M \[C{{H}_{3}}COONa\]
Correct Answer: C
Solution :
| [a] \[[{{H}^{+}}]=\frac{100\times 0.1}{200}=0.05,pH<7\] |
| [b] Meq of \[{{\operatorname{H}}_{2}}S{{O}_{4}}=2\times 0.1\times 100=20\] |
| Meq of \[NaOH = 0.1 \times 100 = 10\] |
| Meq of excess \[{{\operatorname{H}}_{2}}S{{O}_{4}}=20-10=10;pH<7\] |
| [c] \[{{\operatorname{CH}}_{3}}COOH+KOH\xrightarrow{{}}C{{H}_{3}}COOK+{{H}_{2}}O\] |
| \[[Salt]=\frac{0.1\times 100}{200}=0.05\operatorname{M};pH>7\] due to hydrolysis of salt of weak acid with a strong base. |
| [d] \[C{{H}_{3}}COONa+ HCl \xrightarrow{{}} C{{H}_{3}}COOH +NaCl\] |
| \[[C{{H}_{3}}COOH]=\frac{0.1\times 50}{100}=0.05M,pH<7\] |
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