PH of the solution containing equimolar amount of CH-3COOH -pK-a - 4-74- -and- CH-3COO- - at 50-C is-Less than 4-74More than 4-74None of theseEqual 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 ...

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 D 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:

Q. pH of 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

Q. 30.pH of a mixture which contain 0.1 M in CH3COOH and 0.05 M in (CH3COO)2Ba is [pKa of CH3COOH = 4.74] a) 4.74 b) 5.04 c) 4.44 d) 7.00

Q. A mixture contains 0.05 moles each of CH3COONa and CH3COOH. Its pH will be (pKa of CH3COOH = 4.74)

Q. pKb of CH3COO– ion is 9.26. pH of a solution when 0.01 M CH3COOH is neutralized 50% and at equivalent point using 0.01 M NaOH are respectively:a 4.63 8.22 b 4.74 8.22c 2.37 4.11 d 4.74 8.37

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

The correct option is D 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 D 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 $[C H_{3} C O O^{-}] = [C H_{3} C O O H]$
$p H = p K_{a} = 4.74$