A 0.10 M HF solution is 8.4% ionized. Calculate the H^+ ion concentration. The "8.4% ionized" is throwing me. This is my line of thinking.
For every mole HF disolved yeilds one mole of H^+. Since there is 0.10 M HF and the solution is 8.4% ionized = 0.84 M. Is this correct?
A 0.10 M HF solution is 8.4% ionized. Calculate the H^+ ion concentration. The "8.4% ionized
No, your line of thinking is not correct. The term "8.4% ionized" means that only 8.4% of the HF molecules have dissociated into H+ ions and F- ions. In other words, only 8.4% of the HF has turned into H+ ions.
To calculate the H+ ion concentration, you need to take into account the percentage of ionization. In this case, 8.4% of the HF has dissociated, so the concentration of H+ ions will be 8.4% of the initial concentration of HF.
Given that the initial concentration of HF is 0.10 M, you can calculate the concentration of H+ ions as follows:
H+ concentration = 8.4% of 0.10 M
= (8.4/100) * 0.10 M
= 0.0084 * 0.10 M
= 0.00084 M
Therefore, the H+ ion concentration is 0.00084 M.
The term "8.4% ionized" means that only 8.4% of the HF molecules in the solution have dissociated into H+ ions and F- ions. To calculate the H+ ion concentration, you need to consider this percentage of ionization.
First, let's calculate the amount of HF that has ionized in the solution. If the HF solution is 0.10 M (molar), then the amount of HF that has ionized can be calculated as follows:
8.4% ionized = 0.084 (ratio of ionized HF to total HF)
Amount of ionized HF = 0.084 * 0.10 M = 0.0084 M
Now, since HF dissociates to produce one H+ ion for every one HF molecule, the concentration of H+ ions will also be 0.0084 M. Therefore, the H+ ion concentration in the solution is 0.0084 M.
So, you were not correct in assuming that the H+ ion concentration would be the same as the concentration of HF. The ionization of HF leads to the formation of H+ ions, and the degree of ionization needs to be taken into account.