Consider the dissociation of the weak acid HX as given below HX(aq) H+(aq) + X– (aq), Ka = 1.2 × 10–5 [Ka : dissociation constant] The osmotic pressure of 0.03 M aqueous

Question

Prashant Kotian · Accepted Answer

Correct answer: .. HX  H+ + X– Ka = 1.2 × 10–5 0.03M 0.03 – x x x Ka = 1.2 × 10–5 = x 0.03 x  0.03 – x  0.03 (Ka is very small) x 0.03 = 1.2 × 10–5 x = 6 × 10–4 Final solution : 0.03 – x + x + x = 0.03 + x = 0.03 + 6 × 10–4  = (0.03 + (6 × 10–4)) × 0.083 × 300 = 76.19 × 10–2  76 × 10–2

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