Ka for CH3COOH is 1-8- 10-5 and Kb for NH4OH is 1-8 - 10-5- The pH of ammonium acetate will be-

The correct option is D 7.0 pH=7+1/2(pK_a pK_b) Ka=1.8× 10^ 5 pKa= log (1.8× 10^ 5)=5 log1.8 =4.744 pK_b= log(1.8× 10^ 5)=4.744 ...

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Byju's Answer

Standard IX

Chemistry

pH of a Solution

Ka for CH3C...

Question

$K_{a}$ for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$ and $K_{b}$ for $N H_{4} O H$ is $1.8 \times 10^{- 5}$ . The pH of ammonium acetate will be:

7.005

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4.75

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7.0

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between

6

and

7

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Solution

The correct option is D $7.0$
$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b})$

$K a = 1.8 \times 10^{- 5}$

$p K a = - log (1.8 \times 10^{- 5}) = 5 - l o g 1.8$
$= 4.744$

$p K_{b} = - l o g (1.8 \times 10^{- 5}) = 4.744$

$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) = 7$

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Similar questions

Q. What will be the

p H

0.1 M

{C H}_{3} C O O {N H}_{4}

? Dissociation constants of

{C H}_{3} C O O H

and

{N H}_{4} O H

are

K_{a} = 1.8 \times 10^{- 5}

and

K_{b} = 1.8 \times 10^{- 5}

respectively.

Q. The degree of hydrolysis of

0.15 M

solution of ammonium acetate is:

[

K_{a}

for

{C H}_{3} C O O H

1.8 \times 10^{- 5}

and

K_{b}

for

{N H}_{3}

1.8 \times 10^{- 9}

]

Q. What is the

H^{+}

ion concentration in ammonium acetate solution?

(K_{a} = 1.8 \times 10^{- 5}, K_{b} = 1.8 \times 10^{- 5}, K_{w} = 1.0 \times 10^{- 14})

Calculate the e.m.f. of cell in Volts.

\begin{matrix} P t_{H_{2}} 1 a t m \end{matrix} ∣ ∣ ∣ \begin{matrix} C H_{3} C O O H 0.1 M \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} N H_{4} O H 0.01 M \end{matrix} ∣ ∣ ∣ \begin{matrix} P t_{H_{2}} 1 a t m \end{matrix}

K_{a}

for

C H_{3} C O O H = 1.8 \times 10^{- 5}

;

K_{b}

for

N H_{4} O H = 1.8 \times 10^{- 5}

If answer is

x

, then write as nearest integer to

10 | x |

Q. Hydrolysis constant

(K_{h})

of ammonium acetate is :
(Given that

K_{a} =

1.8 \times

10^{- 5}

K_{b} =

2 \times

10^{- 5}

K_{w} = 10^{- 14}

)

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Ka for CH3COOH is 1.8×10−5 and Kb for NH4OH is 1.8×10−5. The pH of ammonium acetate will be:

The correct option is D 7.0pH=7+12(pKa−pKb)Ka=1.8×10−5pKa=−log(1.8×10−5)=5−log1.8=4.744pKb=−log(1.8×10−5)=4.744pH=7+12(pKa−pKb)=7

$K_{a}$ for $C H_{3} C O O H$ is $1.8 \times 10^{- 5}$ and $K_{b}$ for $N H_{4} O H$ is $1.8 \times 10^{- 5}$ . The pH of ammonium acetate will be:

The correct option is D $7.0$
$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b})$

$K a = 1.8 \times 10^{- 5}$

$p K a = - log (1.8 \times 10^{- 5}) = 5 - l o g 1.8$
$= 4.744$

$p K_{b} = - l o g (1.8 \times 10^{- 5}) = 4.744$

$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) = 7$