What is the [h3o + ] in a 0.050 m solution of ba(oh)2?

The H3O+ in a 0.050M  solution of Ba(OH)2  is calculated as below

write  the equation for the dissociation of Ba(OH)2

Ba(OH)2  =  Ba^2+  +2OH^-

calculate the OH-  concentration

by use of mole ratio between Ba(OH)2  to OH^- which is 1:2 the concentration of OH  =  0.050 x2  = 0.1 M

by  use of  the  formula ( H3O+)(OH-) =  1 x10 ^-14

by  making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-

substitute  for OH-

H3O+ =  (1 x10^-14 )/0.1

=  1  x10^-3  M

Answer is: concentration of hydronium ions are 10⁻¹³ M.
Chemical dissociation of barium hydroxide in water: 
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.050 M.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
c(OH⁻) = 0.10 M = 10⁻¹ M.
c(OH⁻) · c(H₃O⁺) = 1·10⁻¹⁴ M².
c(H₃O⁺) = 10⁻¹⁴ M² ÷ 10⁻¹ M.
c(H₃O⁺) = 10⁻¹³ M.