PH of 0.1M solution of acid HX is 3.The value of Ka of this acid is equal to?

1.01 ×10^-5

Explanation:

HX +H2O ===>X- + H³O+

pH=3

ph=-log(H30+)

[H3O+] = 10^-3

[H3O+] = 1 × 10^-3 M

now notice that every mole of HX that ionises produce 1mole of X- , the conjugate base of the acid and 1mole of hydroxonium cations , therefore

[X-] =[H3O+] = 1× 10^-3M

the initial concentration of the acid will decrease because some of the molecules ionise to produce X- and H3O+

so the new concentration of HX will be

[HX]= [HX]initial - [ H3O+]

= 0.1 - 1.0×10^-3

[HX] = 0.099M

also recall that ka = [x-][H30+] /[HX]

1×10^-3×1×10^-3 divided by 0.099

ka = 1.01 ×10^-5

lol for the pointsss