What is the difference in pH for 13 and 23 stages of neutralisation of 0-1 M CH3COOH with 0-1 M NaOH-

The correct option is B 2 log 2 0.1CH_3COOH + 0.1NaOH→ CH_3COONa + H_2O At 13 neutralisation, pH = pK_a + log(salt)(acid) pH_1 = pKa ...

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Standard XII

Chemistry

Solubility Product

What is the d...

Question

What is the difference in $p H$ for $\frac{1}{3}$ and $\frac{2}{3}$ stages of neutralisation of $0.1 M$ $C H_{3} C O O H$ with $0.1 M N a O H$ ?

- 2 log 3

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2 log \frac{1}{4}

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2 log \frac{2}{3}

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- 2 log 2

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Solution

The correct option is B $- 2 log 2$
$C H_{3} C O O H 0.1 + N a O H 0.1 \to C H_{3} C O O N a + H_{2} O$
At $\frac{1}{3}$ neutralisation,
$p H = p K_{a} + l o g \frac{(s a l t)}{(a c i d)}$
$p H_{1} = p K a + l o g \frac{1 / 3}{2 / 3}$ ....(1)
At $\frac{2}{3}$ neutralization,
$p H_{2} = p K a + l o g \frac{2 / 3}{1 / 3}$ .....(2)
Hence, the difference is as follows:
$p H_{1} - p H_{2} = l o g \frac{1}{2} - l o g 2$ $= - 2 l o g 2$

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What is the difference in pH for 13 and 23 stages of neutralisation of 0.1 M CH3COOH with 0.1 M NaOH?

The correct option is B −2 log 2CH3COOH0.1+NaOH0.1→CH3COONa+H2OAt 13 neutralisation,pH=pKa+log(salt)(acid)pH1=pKa+log1/32/3 ....(1)At 23 neutralization,pH2=pKa+log2/31/3 .....(2)Hence, the difference is as follows:pH1−pH2=log12−log2 =−2log2

What is the difference in $p H$ for $\frac{1}{3}$ and $\frac{2}{3}$ stages of neutralisation of $0.1 M$ $C H_{3} C O O H$ with $0.1 M N a O H$ ?