Calculate the hydrogen ion concentration, [H+], in 0.00988 M Ba(OH)2. Careful!
Ba(OH)2 ionizes 100%; therefore,
..........Ba(OH)2 ==> Ba^2+ + 2OH^-
I.........0.00988.....0.........0
C.......-0.00988....0.00988..2*0.00988
E.........0.........0.00988..0.0198
.........H2O --> H^+ + OH^-
I........liquid...0....0.0198
C........liquid...x.....x
E........liquid...x...0.0198+x
I assume this is standard Kw.
Kw = (H^+)(OH^-)
Substitute the E line into Kw and solve for x = (H^+)
You can make the assumption that 0.0198+x = 0.0198
C.......
To calculate the hydrogen ion concentration in a solution of Ba(OH)2, we first need to determine the dissociation reaction of Ba(OH)2.
The formula for barium hydroxide is Ba(OH)2. When it dissolves in water, it dissociates into Ba2+ ions and OH- ions. Since we are interested in the concentration of hydrogen ions, we need to consider that the concentration of hydroxide ions (OH-) is twice the concentration of barium hydroxide (Ba(OH)2).
The balanced dissociation reaction is:
Ba(OH)2(s) → Ba2+(aq) + 2 OH-(aq)
Given the concentration of Ba(OH)2 as 0.00988 M, we know that the concentration of OH- ions will be twice that value.
Concentration of OH- ions = 2 * 0.00988 M = 0.01976 M
Now, we can use the fact that in water, the concentration of H+ ions multiplied by the concentration of OH- ions is equal to 1x10^-14 M^2 (known as the ion product of water).
[H+][OH-] = 1x10^-14
Substituting the given values:
[H+](0.01976 M) = 1x10^-14
[H+] = (1x10^-14) / (0.01976 M)
[H+] ≈ 5.06x10^-14 M
Therefore, the hydrogen ion concentration in 0.00988 M Ba(OH)2 is approximately 5.06x10^-14 M.
To calculate the hydrogen ion concentration, [H+], in a solution of Ba(OH)2, we need to consider the dissociation of the compound.
Ba(OH)2 dissociates into Ba2+ ions and OH- ions in water, according to the following balanced chemical equation:
Ba(OH)2 -> Ba2+ + 2OH-
Since we are interested in the concentration of hydrogen ions, we need to consider the concentration of hydroxide ions and apply the concept of water autoionization.
Water undergoes autoionization in small amounts, represented by the following chemical equation:
H2O <=> H+ + OH-
In pure water, the concentration of H+ and OH- ions is equal and is referred to as the dissociation constant, Kw. For water at room temperature, Kw is approximately 1.0 x 10^-14.
From the balanced equation for the dissociation of Ba(OH)2, we can see that for every Ba(OH)2 molecule that dissociates, two OH- ions are formed. Therefore, the concentration of OH- ions is twice the concentration of Ba(OH)2.
Given that the concentration of Ba(OH)2 is 0.00988 M, the concentration of OH- ions is 2 * 0.00988 M = 0.01976 M.
Now, because of the autoionization of water, the concentration of OH- ions will react with H+ ions in a 1:1 ratio. Therefore, the concentration of H+ ions in the solution is also 0.01976 M.
So, the hydrogen ion concentration, [H+], in 0.00988 M Ba(OH)2 is 0.01976 M.