Given the Ka for acetic acid is 1-77- 10-5 and Kb for ammonium hydroxide is 1-82- 10-5-Calculate the pH of an aqueous solution of ammonium acetate CH3COONH4 at 25- C-

PK_a= log(K_a) pK_a= log(1.77× 10^ 5)=(5 0.25) pK_a=4.75 pK_b= log(K_b) pK_b= log(1.82× 10^ 5)=(5 0.26) pK_b=4.74 For a salt of weak acid and weak base nbsp ...

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Byju's Answer

Standard XI

Chemistry

Salt of Weak Acid and Weak Base

Given the Ka ...

Question

Given the $K_{a}$ for acetic acid is $1.77 \times 10^{- 5}$ and $K_{b}$ for ammonium hydroxide is $1.82 \times 10^{- 5}$ .
Calculate the pH of an aqueous solution of ammonium acetate $(C H_{3} C O O N H_{4})$ at $25^{\circ} C$ .

8.115

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7.01

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7.65

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8.05

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Solution

The correct option is B 7.01
$p K_{a} = - l o g (K_{a}) p K_{a} = - l o g (1.77 \times 10^{- 5}) = (5 - 0.25) p K_{a} = 4.75$

$p K_{b} = - l o g (K_{b}) p K_{b} = - l o g (1.82 \times 10^{- 5}) = (5 - 0.26) p K_{b} = 4.74$

For a salt of weak acid and weak base pH is given as:
$p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) p H = 7 + \frac{1}{2} (4.75 - 4.74) p H = 7 + \frac{0.01}{2} p H = 7.005$

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Similar questions

Q. Given the

K_{a}

for acetic acid is

1.77 \times 10^{- 5}

and

K_{b}

for ammonium hydroxide is

1.82 \times 10^{- 5}

.
Calculate the pH of an aqueous solution of ammonium acetate

(C H_{3} C O O N H_{4})

25^{\circ} C

Q. The dissociation constant of acetic acid is

1.8 \times 10^{- 5}

and that of ammonium hydroxide is also

1.8 \times 10^{- 5}

25^{o} C

. Hence, an aqueous 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_{4} O H

1.8 \times 10^{- 5}

. The pH of ammonium acetate will be:

Q. Calculate the pH at the equivalence point when a solution of

0.1 M

acetic acid is titrated with a solution of

0.1 M

sodium hydroxide.

K_{a}

for acetic acid

= 1.9 \times 10^{- 5}

p k_{a}

of acetic acid and

p K_{b}

of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.

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Given the Ka for acetic acid is 1.77×10−5 and Kb for ammonium hydroxide is 1.82×10−5. Calculate the pH of an aqueous solution of ammonium acetate (CH3COONH4) at 25∘C.

The correct option is B 7.01pKa=−log(Ka)pKa=−log(1.77×10−5)=(5−0.25)pKa=4.75 pKb=−log(Kb)pKb=−log(1.82×10−5)=(5−0.26)pKb=4.74 For a salt of weak acid and weak base pH is given as: pH=7+12(pKa−pKb)pH=7+12(4.75−4.74)pH=7+0.012pH=7.005

Given the $K_{a}$ for acetic acid is $1.77 \times 10^{- 5}$ and $K_{b}$ for ammonium hydroxide is $1.82 \times 10^{- 5}$ .
Calculate the pH of an aqueous solution of ammonium acetate $(C H_{3} C O O N H_{4})$ at $25^{\circ} C$ .