what is the [h3o+] in a 9.93 x 10^-6 M Ba(OH)2 solution?
How many moles of OH are there?
then [OH ][H3O]=1E-14 solve for the cncentration of the hydroniim ion
thats all that's given
To determine the concentration of hydronium ions ([H3O+]) in a solution, we need to consider the dissociation of water and the reaction of the base (Ba(OH)2) with water.
First, let's write the balanced chemical equation for the dissociation of water:
H2O ⇌ H+ + OH-
Water molecules can donate a proton (H+) to form hydronium ions (H3O+), and hydroxide ions (OH-) are formed as a result.
Now, let's write the equation to show the reaction of the base (Ba(OH)2) with water:
Ba(OH)2(s) → Ba2+(aq) + 2OH-(aq)
For every one mole of Ba(OH)2 that dissolves in water, it forms one mole of Ba2+ ions and two moles of OH- ions. Since Ba(OH)2 completely dissociates in water, the concentration of OH- ions is twice the concentration of Ba(OH)2.
Given the concentration of Ba(OH)2 as 9.93 x 10^-6 M, the concentration of OH- ions is twice that value:
[OH-] = 2 * (9.93 x 10^-6 M) = 1.986 x 10^-5 M
Since we know that [H3O+] * [OH-] = Kw (the ion-product constant for water at a given temperature), and at 25 °C Kw = 1.0 x 10^-14, we can calculate the concentration of [H3O+]:
[H3O+] = Kw / [OH-]
[H3O+] = (1.0 x 10^-14) / (1.986 x 10^-5)
[H3O+] ≈ 5.04 x 10^-10 M
Therefore, the concentration of hydronium ions ([H3O+]) in the solution is approximately 5.04 x 10^-10 M.