If 1 mM solution of ethylamine produces pH = | vedclass

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(A) The dissociation reaction of ethylamine is: $C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^-$Given concentration $C = 1 \ mM = 10^{-3} \

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$7$

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(A) The dissociation reaction of ethylamine is: $C_2H_5NH_2 + H_2O ightleftharpoons C_2H_5NH_3^+ + OH^-$Given concentration $C = 1 \ mM = 10^{-3} \ M$.Given $pH = 9$,so $pOH = 14 - 9 = 5$.Thus,$[OH^-] = 10^{-pOH} = 10^{-5} \ M$.Since $[OH^-] = [C_2H_5NH_3^+] = 10^{-5} \ M$ and $[C_2H_5NH_2] \approx C = 10^{-3} \ M$ (as degree of ionization is neglected),$K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]} = \frac{10^{-5} 	imes 10^{-5}}{10^{-3}} = 10^{-7}$.Comparing $10^{-7}$ with $10^{-x}$,we get $x = 7$.

If $1 \ mM$ solution of ethylamine produces $pH = 9$,then the ionization constant $(K_b)$ of ethylamine is $10^{-x}$. The value of $x$ is . . . . . . (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]

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