PH of the solution containing equimolar amount of CH 3 COOH pK a -4-74 and CH 3 COO -at 50- C is-A- None of theseB- More than 4-74C- Less than 4-74D- Equal to 4-74

The correct option is D Equal to 4.74pH of an acidic buffer is given by pH=pK_a+log[salt][acid] pH=pK_a+log[CH_3COO^ ][CH_3COOH] since [CH_3COO^ ]=[CH_3COOH] p ...

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Byju's Answer

Standard XII

Chemistry

Acidic Buffer Action

pH of the sol...

Question

$p H$ of the solution containing equimolar amount of $C H_{3} C O O H$
$(p K_{a} = 4.74) and C H_{3} C O O^{-}$ at $50^{\circ} C$ is:

Less than 4.74

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More than 4.74

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Equal to 4.74

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None of these

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Solution

The correct option is C Equal to 4.74
$p H$ of an acidic buffer is given by
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$
$p H = p K_{a} + l o g \frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H]}$
since $[C H_{3} C O O^{-}] = [C H_{3} C O O H]$
$p H = p K_{a} = 4.74$

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Similar questions

p H

of the solution containing equimolar amount of

C H_{3} C O O H

(p K_{a} = 4.74) and C H_{3} C O O^{-}

50^{\circ} C

is:

p K_{a}

C H_{3} C O O H

is 4.74. The pH of 0.01 M

C H_{3} C O O N a

Q. To

1.0 L

solution containing

0.1

mol each of

N H_{3}

and

N H_{4} C l, 0.05

mol NaOH is added. The change in pH will be :

(p K_{a}

for

C H_{3} C O O H = 4.74

)

Q. The

p H

of the solution obtained by mixing

250 m l, 0.2 M C H_{3} C O O H

and

200 m l 0.1 M N a O H

is (Given :

p K_{a}

C H_{3} C O O H = 4.74, log 3 = 0.48)

p K_{a}

C H_{3} C O O H

is 4.74. The pH of 0.01 M

C H_{3} C O O N a

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pH of the solution containing equimolar amount of CH3COOH (pKa=4.74) and CH3COO− at 50∘C is:

The correct option is C Equal to 4.74pH of an acidic buffer is given by pH=pKa+log[salt][acid] pH=pKa+log[CH3COO−][CH3COOH] since [CH3COO−]=[CH3COOH] pH=pKa=4.74

$p H$ of the solution containing equimolar amount of $C H_{3} C O O H$
$(p K_{a} = 4.74) and C H_{3} C O O^{-}$ at $50^{\circ} C$ is:

The correct option is C Equal to 4.74
$p H$ of an acidic buffer is given by
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$
$p H = p K_{a} + l o g \frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H]}$
since $[C H_{3} C O O^{-}] = [C H_{3} C O O H]$
$p H = p K_{a} = 4.74$