Question

What is the hydronium ion concentration of 0.25 m ha solution (ka = 4 10-8)?

Answer 1

Answer:-

Solution:- The given acid is HA. It's a monoprotic acid and the ionization equation is written as:

$HA(aq)+H_2O(l)\leftrightarrow H_3O^+(aq)+A^-(aq)$

Initial concentration of HA is given as 0.25M. If the change in concentration is x then the equilibrium concentration for each of the product is x and for HA it would be (0.25 - x).

Equilibrium expression for the above equation is written as:

$Ka=\frac{[H_3O^+][A^-]}{[HA]}$

Let's plug in the values in the equilibrium expression and solve it for x, where x is also the equilibrium concentration of hydronium ion:

$4*10^-^8=\frac{x^2}{(0.25-x)}$

From Ka value, the acid is very weak and so the (0.25 - x) could be taken as 0.25.

$4*10^-^8=\frac{x^2}{0.25}$

On keeping similar terms on same side:

$x^2=(4*10^-^8*0.25)$

$x^2=1*10^-^8$

On taking square root to both sides:

x = $1*10^-^4$

So, the hydonium ion concentration is $1*10^-^4$ M.

Answer 2

Shout cut

Using formula

HA=✓ka×c

= ✓4×10^-8 × 0.25

=✓10^-8

So root se bahar nikalne pe

10^-4 and .