- hydrolysis of 0-1 M CH3COONH4 when Ka-Kb-1-8-10-5 is -

H=√(K_wK_a× K_b)=√(10^ 141.8×10^ 5×1.8×10^ 5) % h=5.55×10^ 3×100=0.55% ...

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Introduction and Degree of Hydrolysis

% hydrolysis ...

Question

$%$ hydrolysis of $0.1 M C H_{3} C O O N H_{4}$ when $K_{a} = K_{b} = 1.8 \times 10^{- 5}$ is -

0.55

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7.63

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0.55 \times 10^{- 2}

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7.63 \times 10^{- 3}

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Solution

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Similar questions

%

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

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

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

K_{a} (C H_{3} C O O H) = K_{b} (N H_{4} O H) = 1.8 \times 10^{- 5}

%

0.1 M C H_{3} C O O N H_{4}

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

p H

0.1 M

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

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

{N H}_{4} O H

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

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

(K_{h})

K_{a} =

1.8 \times

10^{- 5}

K_{b} =

2 \times

10^{- 5}

K_{w} = 10^{- 14}

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