Question

is 0.2 m solution of monoprotic
acid is 40% ionized the ionization constant of the acid​

Answer 1

Ionization constant for monoprotic acid is 0.032 or 3.2*10^-2

We know that for a monoprotic acid x=(k/c)^1/2

Where x is degree of dissociation or ionized value of acid

k is the ionization constant and c is concentration of an acid

In the given question we have x=40% =40/100=0.4

c or concentration is 0.2 m

So x=(k/c)^1/2

Hence (x)^2=k/c so x^2=0.4*0.4=0.16

so x^2*c=k substituting values we have:--

0.16*0.2=k so k=0.032 or 3.2*10^-2

Hence ionization constant for acid is 3.2*10^-2