Answer:
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
