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.