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Standard XII

Chemistry

Salt of Weak Acid and Strong Base

0.1 M formic ...

Question

$0.1 M$ formic acid solution is titrated against $0.1 M N a O H$ solution. What would be the difference in $p H$ between $\frac{1}{5}$ and $\frac{4}{5}$ stages of neutralization of acid?

2 log 3 / 4

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2 log 1 / 5

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log 1 / 3

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2 log 4

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Solution

The correct option is D $2 log 4$
As we know,
$p H = p K_{a} + l o g s a l t / a c i d$
At first stage:
$p H_{1} = p K_{a} + l o g 2 / 8 = p K_{a} - l o g 4$
At last stage:
$p H_{2} = p K_{a} + l o g 8 / 2 = p K_{a} + l o g 4$
Now,
$p H_{2} - p H_{1} = l o g 4 - (- l o g 4) = 2 l o g 4.$

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0.1 M formic acid solution is titrated against 0.1 M NaOH solution. What would be the difference in pH between 15 and 45 stages of neutralization of acid?

The correct option is D 2log4As we know,pH=pKa+logsalt/acidAt first stage:pH1=pKa+log2/8=pKa−log4At last stage:pH2=pKa+log8/2=pKa+log4 Now,pH2−pH1=log4−(−log4)=2log4.

$0.1 M$ formic acid solution is titrated against $0.1 M N a O H$ solution. What would be the difference in $p H$ between $\frac{1}{5}$ and $\frac{4}{5}$ stages of neutralization of acid?

The correct option is D $2 log 4$
As we know,
$p H = p K_{a} + l o g s a l t / a c i d$
At first stage:
$p H_{1} = p K_{a} + l o g 2 / 8 = p K_{a} - l o g 4$
At last stage:
$p H_{2} = p K_{a} + l o g 8 / 2 = p K_{a} + l o g 4$
Now,
$p H_{2} - p H_{1} = l o g 4 - (- l o g 4) = 2 l o g 4.$