Calculate the [h30+] in a 0.010M solution of Sr(OH)2
because Sr(OH)2, do 0.010 x 2
ans = 0.020
pOH = -log(0.020)
pOH = 1.7
[finding pH because Sr(OH)2]
pH = 14 - 1.7
pH = 12.30
10^-12.30 = 5.0x10^-13
hope this helps
0.010M in Sr(OH)2.
OH^- is twice that.
(H3O^+)(OH^-) = Kw = 1E-14
Solve for (H3O^+)
Well, to calculate the [H3O+] in a 0.010M solution of Sr(OH)2, we need to consider the dissociation of Sr(OH)2. But before I do that, I must warn you that hydroxide ions (OH-) are notorious for being really bad at telling jokes. They just can't seem to get the delivery right. They're always too basic! Okay, now let's get on with the calculation.
Sr(OH)2 dissociates into Sr2+ ions and 2 OH- ions. So for every 1 mole of Sr(OH)2, we get 2 moles of OH-. But since we started with a 0.010M solution of Sr(OH)2, we have 0.010 moles of Sr(OH)2 per liter.
Since we have 2 moles of OH- ions for every mole of Sr(OH)2, we have 0.020 moles of OH- ions per liter.
Now, since OH- is a basic ion, it will react with water to form H2O and OH-. This reaction is reversible, so we can consider it as an equilibrium. But equilibrium reactions can be pretty unstable. It's like trying to find the right balance while sitting on a seesaw with a clown. You just never know when it's going to tip over!
Anyway, this equilibrium reaction is:
OH- + H2O ⇌ H3O+ + OH-
Now, since this is an equilibrium reaction, we can use the equation:
Kw = [H3O+] [OH-]
But at 25 degrees Celsius, the value of Kw (the ion product of water) is 1.0 x 10^-14. Since we know the concentration of OH- ions is 0.020M, we can rearrange the equation to solve for [H3O+].
[H3O+] = Kw / [OH-]
Plugging in the values, we get:
[H3O+] = (1.0 x 10^-14) / (0.020)
Calculating it, we find that the [H3O+] in a 0.010M solution of Sr(OH)2 is approximately 5 x 10^-13 M. So, next time you want to calculate something like this, just remember that even though hydroxide ions aren't great at telling jokes, they're always great at reacting with water!
5*10•14
To calculate the concentration of H3O+ in a solution of Sr(OH)2, we first need to determine the dissociation of Sr(OH)2 in water.
Sr(OH)2 dissociates into Sr2+ ions and OH- ions. Since it is a strong electrolyte, it completely dissociates.
The balanced chemical equation for the dissociation of Sr(OH)2 is:
Sr(OH)2 -> Sr2+ + 2OH-
Since Sr(OH)2 produces 2 OH- ions for every 1 molecule of the compound, the concentration of OH- ions in the solution will be twice the concentration of Sr(OH)2.
Given that the concentration of Sr(OH)2 is 0.010 M, the concentration of OH- ions will be 2 * 0.010 M = 0.020 M.
Since water also undergoes autoprotolysis and produces H3O+ and OH- ions, the concentration of H3O+ ions in water is 1.0 x 10^-7 M. However, the concentration of OH- ions is much higher compared to H3O+ ions in this case (0.020 M vs. 1.0 x 10^-7 M).
To neutralize the excess OH- ions, we need to consider the reaction between H3O+ and OH- ions.
The balanced chemical equation for the reaction between H3O+ and OH- ions is:
H3O+ + OH- -> 2H2O
In this reaction, one H3O+ ion reacts with one OH- ion to form two molecules of water. Therefore, the concentration of H3O+ ions will be equal to the concentration of OH- ions in the solution, which is 0.020 M.
Thus, the concentration of H3O+ in the 0.010 M solution of Sr(OH)2 is 0.020 M.