Consider the dissociation of the weak acid HX | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $76 	imes 10^{-2} \ bar$.

Similar Questions

Which of the following aqueous solutions has the highest boiling point,under identical conditions having a concentration of $0.03 \ m$?

Which of the following $0.1 \ M$ solutions will exhibit the minimum freezing point?

For a $0.2 \ m$ aqueous solution of a weak acid $HX$,the degree of ionization is $0.3$. If $K_f$ for water is $1.85 \ K \ kg \ mol^{-1}$,the freezing point of the solution will be close to .......... $^oC$.

The freezing point of a $2\%$ aqueous solution of $[Ag(CH_3NH_2)_2]Cl$ is:

$pH$ of a $0.1 \ M$ monobasic acid is found to be $2$. Hence its osmotic pressure at a given temperature $T \ K$ is (in $RT$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module