Question

If pka=-log ka=4 and ka=cx2 and c=0.01m then vant hoff factor for weak monobasic acid

Answer 1

Answer:

It is given that,

pka = - log ka = 4

⇒ log ka = -4  

⇒ ka = 10⁻⁴ …… (i)

Also given, ka = cx² ….. (ii)

Now, equating eq. (i) & (ii), we get

cx² = 10⁻⁴

or, 0.01 * x² = 10⁻⁴ ….. [since c = 0.01 M given]

or, x² = 10⁻⁴ / 0.01 = 0.01

or, x = √0.01 = 0.1

                         HA         ↔      H⁺  +       A⁻

Initial moles:    0.01 M

At equilibrium: (0.01 - x)            x             x

Thus,  

The Vant Hoff factor, “i” for weak monobasic acid is  

= [(0.01 – x) + x + x] / 0.01  

= (0.01 + 0.1)/0.01  

= 0.11/ 0.01  

= 1.1