Calculate pH, pOH, [H+], [OH-], and Ka for a 0.026 M solution of HI (a strong acid). Please help me :(

strong acid; then it is completely dissociated.

[H]=.026
pH= -log (.026)=1.58
pOH= 14-pH= you do it.
[OH]=1E-14/.026=3.85E-13

thank you! :D

I'd be happy to help you with your question!

To calculate the pH, pOH, [H+], [OH-], and Ka for a 0.026 M solution of HI (a strong acid), we can follow these steps:

Step 1: Determine the concentration of hydrogen ions [H+]:
Since HI is a strong acid, it completely dissociates in water to yield hydrogen ions (H+) and iodide ions (I-). Therefore, the concentration of [H+] in the solution is equal to the concentration of HI, which is 0.026 M.

[H+] = 0.026 M

Step 2: Calculate the pOH:
The pOH can be calculated using the formula:

pOH = -log[OH-]

Since HI is a strong acid, it does not produce hydroxide ions (OH-) in significant amounts. Therefore, the concentration of [OH-] in the solution is assumed to be negligible.

So, the pOH in this case is 0.

pOH = 0

Step 3: Calculate the pH:
The pH can be calculated from the pOH using the relationship:

pH + pOH = 14

So, pH = 14 - pOH

Substituting the value of pOH we obtained in Step 2:

pH = 14 - 0
pH = 14

pH = 14

Step 4: Calculate [OH-]:
Since HI is a strong acid, the concentration of hydroxide ions (OH-) is assumed to be negligible. Therefore, [OH-] in this case is approximately 0.

[OH-] = 0

Step 5: Calculate Ka:
For a strong acid like HI, Ka is not applicable since it completely dissociates in water. Ka is only applicable for weak acids.

Therefore, for a 0.026 M solution of HI, the results are as follows:

pH = 14
pOH = 0
[H+] = 0.026 M
[OH-] = 0
Ka = N/A (since HI is a strong acid)

I hope this helps! Let me know if you have any more questions.

Of course, I can help you with that! To calculate the pH, pOH, [H+], [OH-], and Ka for a 0.026 M solution of HI (hydroiodic acid), we need to understand a few concepts and formulas.

First, let's start with the definition of pH and pOH. The pH is a measure of the acidity or basicity of a solution, while the pOH is a measure of the alkalinity (opposite of acidity). They are defined as the negative logarithms (base 10) of the concentrations of H+ and OH-, respectively. The formulas are as follows:

pH = -log[H+]
pOH = -log[OH-]

Next, let's consider the dissociation of HI in water. Since HI is a strong acid, it completely dissociates into H+ and I- ions. The balanced chemical equation is:

HI(aq) -> H+(aq) + I-(aq)

Now, the concentration of H+ ions in the solution is the same as the initial concentration of HI, which is 0.026 M. Therefore, [H+] = 0.026 M.

On the other hand, the concentration of OH- ions can be determined using the fact that the concentration of H+ ions in water is always equal to the concentration of OH- ions. This is known as the water autoprotolysis equilibrium:

H2O(l) ⇌ H+(aq) + OH-(aq)

Since water is neutral, the concentration of H+ and OH- ions is the same, so [OH-] = [H+] = 0.026 M.

Now, let's calculate the pH and pOH using the formulas mentioned earlier:

pH = -log(0.026) ≈ 1.59
pOH = -log(0.026) ≈ 1.59

Notice how the pH and pOH are the same, which is expected for a neutral solution.

Lastly, let's calculate the Ka, which is the acid dissociation constant. The Ka is a measure of the strength of an acid and is defined as the ratio of the concentration of products (H+ and I-) to the concentration of the undissociated acid (HI). In this case, since HI is a strong acid, it completely dissociates, so the concentration of HI is the same as the concentration of H+.

Therefore, the Ka for this system is equal to the concentration of H+ squared divided by the concentration of HI:

Ka = [H+]^2 / [HI] = (0.026)^2 / 0.026 = 0.026 M

So, to summarize:
- pH = pOH ≈ 1.59
- [H+] = [OH-] = 0.026 M
- Ka = 0.026 M

I hope this helps! If you have any further questions, feel free to ask.