Calculate the degree of hydrolysis ammonium acetate- Given K a -1-75 - 10-5 - K b -1-80 - 10-5-A- 2-3 - 10-6B- 5-63 - 10-3- 8-4 - 10-2D- 9-8 - 10-1

The correct option is B h = √(K_w/K_aK_b) = √(1 × 10^ 14/1.75 × 10^ 5 × 1.8 × 10^ 5) = 5.63 × 10^ 3 ...

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Salt of Weak Acid and Weak Base

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Question

Calculate the degree of hydrolysis ammonium acetate. Given $K_{a}$ = $1.75 \times 10^{- 5}$ ; $K_{b}$ = $1.80 \times 10^{- 5}$ .

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Solution

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(K_{h})

K_{a} =

1.8 \times

10^{- 5}

K_{b} =

2 \times

10^{- 5}

K_{w} = 10^{- 14}

0.15 M

K_{a}

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

1.8 \times 10^{- 5}

K_{b}

{N H}_{3}

1.8 \times 10^{- 9}

K_{a} = 1 \times 10^{- 5}

K_{b} = 1 \times 10^{10}

(h)

25^{\circ} C

K_{a}

C H_{3} C O O H

1.8 \times 10^{- 5}

K_{b}

4.6 \times 10^{- 10}

0.01 M

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

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