Question

The pH of 0.1M monobasic acid solution os 2.Calculate degree of dissociation and dissociation constant.

Answer 1

Osmotic pressure (π)=iCRT

Where C=Molarity of solution

α=i−1m−1

0.1=i−12−1

i = 1.1

∴π=iCRT

∴=1.1×0.1RT=0.11RT

Answer 2

Answer:

Explanation:

ph of 0.1 m is 2

0.1 alpha = h plus = 10^-ph= 10^-2

alpha=10^-1=degree of dissociation

alpha= root ka/c

(alpha)^2 * c = dissociation constant

10^-2*10^-1=ka

10^-3=ka