The ionization constant of a monobasic acid H | vedclass

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(A) Given: $m = 0.025 \ m$,$\Delta T_f = 0 - (-0.060) = 0.060 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.Using the formula $\Delta T_f = i 	imes K_f 	ime

Vedclass · Accepted Answer

$3 	imes 10^{-3}$

Vedclass · Answer

(A) Given: $m = 0.025 \ m$,$\Delta T_f = 0 - (-0.060) = 0.060 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.Using the formula $\Delta T_f = i 	imes K_f 	imes m$:$0.060 = i 	imes 1.86 	imes 0.025$$i = \frac{0.060}{0.0465} \approx 1.29$.For a monobasic acid $HA ightleftharpoons H^+ + A^-$,the van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.$1 + \alpha = 1.29 \implies \alpha = 0.29$.The ionization constant $K_a = \frac{c \alpha^2}{1 - \alpha} = \frac{0.025 	imes (0.29)^2}{1 - 0.29} = \frac{0.025 	imes 0.0841}{0.71} \approx 0.00296 \approx 3 	imes 10^{-3}$.

The ionization constant of a monobasic acid $HA$ is to be determined. If a $0.025 \ m$ aqueous solution of the acid freezes at $-0.060 \ ^{\circ}C$,calculate the ionization constant $K_a$. (Assume molality = molarity and $K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$).

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