Calculate the [H3O+] of each of the following H2SO4 solutions.
For a 0.50 M soln
How do I reach the answer? Do I have to use a ice table?
Not for the first H^+ since it is 100% ionized. The second H^+ is not 100% ionized (k2 is about 10^-2) and you need an ICE table for that. But it's a little more complicated than that.
As dilute as that is, assume all dissociation has taken place (you get two protons for each H2SO4 molecule)
So the concentration of the H ions is 2*.5=1M
and the answer.
Now for more concentrated solutinos, you would have to use Keq and an ice table. I would do that for anything over 8M solutions.
Remember: STrong acids (HCl, HNO3, H2SO4) "completely" dissociate in water solutions. IT is not exactly so, but it is very close.
how do i reach the answer then? what are my k values if i need to use a ice table
To calculate the [H3O+] (hydronium ion concentration) of a solution of H2SO4, you can use the ionization constant of the acid, known as Ka.
Yes, using an ICE (Initial Change Equilibrium) table is one approach to calculate the [H3O+]. Here's how you can do it:
Step 1: Write the balanced chemical equation for the ionization of H2SO4 in water:
H2SO4 + H2O → H3O+ + HSO4-
Step 2: Write the expression for the ionization constant, Ka:
Ka = [H3O+][HSO4-] / [H2SO4]
Step 3: Since H2SO4 is a strong acid, it completely ionizes in water. Therefore, the concentration of hydronium ions, [H3O+], is equal to the concentration of H2SO4, which is given as 0.50 M.
Step 4: Substitute the values into the expression for Ka:
Ka = [H3O+]^2 / 0.50
Step 5: Rearrange the equation to solve for [H3O+]:
[H3O+]^2 = Ka * 0.50
Step 6: Take the square root of both sides to get [H3O+]:
[H3O+] = √(Ka * 0.50)
Step 7: Look up the ionization constant Ka for H2SO4, which is 1.0 x 10^-2.
Step 8: Substitute the value of Ka into the equation:
[H3O+] = √(1.0 x 10^-2 * 0.50)
Step 9: Calculate the value using a calculator:
[H3O+] ≈ 0.071 M
Therefore, the hydronium ion concentration, [H3O+], for a 0.50 M solution of H2SO4 is approximately 0.071 M.