Question

For a weak monobasic acid, if pK, = 4, then at a
concentration of 0.01 M of the acid solution, the
van't Hoff factor is​

Answer 1

$\huge\sf{\underline{\pink{Answer:-}}}$

Let a weak monobasic acid is HCN

Hence the degree of dissociation of HCN is:

HCN→H + CN − 1001−αα a(1−α)aαaα

We know that

$- ph = - log_{10}( {H}^{ + } )$

${H}^{ + } = {10}^{ - ph}$

Given : a = 0.01

Again we know that the Hoff factor,i =

$\frac{1 + a}{1}$

${H}^{ + } = a \alpha$

${10}^{ - p \: H} = 0.01 \alpha$

${10}^{ - 4} = 0.01 \alpha$

$\alpha = \frac{ {10}^{ - 4} }{0.01} = 0.01$

Hence vant Hoff factor,i

$\frac{1 + \alpha }{1}$

= 1 + 0.01

= 1.01

hope this will help you..

-3idiots29

Answer 2

Answer:

If at a particular temperature then van't Hoff factor, i for 0.01 M monobasic acid is,

Solutions

A1.01

B1.02

C11.0

D1.20

Solution:

Answer (c) 11.0

Explanation:

Hope its help you