Consider the dissociation of the weak acid HX | vedclass

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(A) The dissociation of $HX$ is given by: $HX \rightleftharpoons H^{+} + X^{-}$.Initial concentration: $0.03 \ M$.Equilibrium concentration: $(0.03 -

Vedclass · Accepted Answer

$76$

Vedclass · Answer

(A) The dissociation of $HX$ is given by: $HX ightleftharpoons H^{+} + X^{-}$.Initial concentration: $0.03 \ M$.Equilibrium concentration: $(0.03 - x), x, x$.Since $K_{a}$ is very small,$0.03 - x \approx 0.03$.$K_{a} = \frac{x^2}{0.03} = 1.2 	imes 10^{-5}$.$x^2 = 3.6 	imes 10^{-7} = 36 	imes 10^{-8}$.$x = 6 	imes 10^{-4} \ M$.Total concentration of particles $C_{total} = (0.03 - x) + x + x = 0.03 + x = 0.03 + 0.0006 = 0.0306 \ M$.Osmotic pressure $\Pi = C_{total} 	imes R 	imes T$.$\Pi = 0.0306 	imes 0.083 	imes 300 = 0.76194 \ bar$.$\Pi = 76.194 	imes 10^{-2} \ bar$.Rounding to the nearest integer,we get $76 	imes 10^{-2} \ bar$.

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