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Question

If equilibrium constant of $${ CH }_{ 3 }COOH+{ H }_{ 2 }O\rightleftharpoons { CH }_{ 3 }{ COO }^{ - }+{ H }_{ 3 }{ O }^{ + }$$ is $$1.8\times { 10 }^{ -5 }$$, equilibrium constant for $${ CH }_{ 3 }COOH+{ OH }^{ - }\rightleftharpoons { CH }_{ 3 }{ COO }^{ - }+{ H }_{ 2 }O$$ is?


A
1.8×109
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B
1.8×109
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C
5.55×109
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D
5.55×1010
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Solution

The correct option is B $$1.8\times { 10 }^{ 9 }$$
$$CH_3 COOH+H_2O\equiv CH_3 COO^-+H_3O^+$$
$$K_1=\dfrac {[CH_3COO^-][H_3O^+]}{[CH_3COOH]}---(i)$$
$$CH_3COOH+OH^-\equiv CH_3COO^-+H_2O$$
$$K_2=\dfrac {[CH_3COO^-]}{[CH_3COOH][OH^-]}---(ii)$$
$$H_2O+H_2O\rightleftharpoons H_3O^+ +OH^-$$
$$K_w=[H_3O^+][OH^-]---(iii)$$
comparing all three quation we get
$$\dfrac {K_1}{K_2}=K_w$$
$$K_2=\dfrac {K_1}{K_w}=\dfrac {1.8\times O^{-5}}{1\times 10^{-19}}=1.8\times 10^{-9}$$

Chemistry

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