The pH of an aqueous solution at 25- C having 0-1 M NaOH and 0-3 M to acetic acid pKa - 4-76 would be nearly

PH = pK_a log[acid]/[salt] 0.1 M NaOH will react with acid to form 0.1 M CH_3COONa and therefore 0.3 nbsp;CH_3COOH will be reduced to 0.2 M. pH = 4.76 – ...

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Byju's Answer

Standard XII

Chemistry

Salt of Weak Acid and Strong Base

The pH of an ...

Question

The pH of an aqueous solution at $25^{\circ} C$ having 0.1 M NaOH and 0.3 M to acetic acid ( $p K_{a}$ = 4.76) would be nearly

4.25

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4.45

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4.75

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5.05

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Solution

The correct option is B
4.45

pH = $p K_{a} - l o g \frac{[a c i d]}{[s a l t]}$

0.1 M NaOH will react with acid to form 0.1 M $C H_{3} C O O N a$ and therefore 0.3 $C H_{3} C O O H$ will be reduced to 0.2 M.

pH = 4.76 – log $\frac{0.2}{0.1}$ = 4.45

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Similar questions

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The pH of an aqueous solution at 25∘C having 0.1 M NaOH and 0.3 M to acetic acid (pKa = 4.76) would be nearly

The correct option is B 4.45 pH = pKa−log[acid][salt] 0.1 M NaOH will react with acid to form 0.1 M CH3COONa and therefore 0.3 CH3­COOH will be reduced to 0.2 M. pH = 4.76 – log0.20.1 = 4.45

The pH of an aqueous solution at $25^{\circ} C$ having 0.1 M NaOH and 0.3 M to acetic acid ( $p K_{a}$ = 4.76) would be nearly

The correct option is B
4.45

pH = $p K_{a} - l o g \frac{[a c i d]}{[s a l t]}$

0.1 M NaOH will react with acid to form 0.1 M $C H_{3} C O O N a$ and therefore 0.3 $C H_{3} C O O H$ will be reduced to 0.2 M.

pH = 4.76 – log $\frac{0.2}{0.1}$ = 4.45