Find the pH and pOH of a solution that is 1.04x10^-4 M HCl.
Do i have to use an ice table to solve this? or is there a short cut? please help
No. HCl is a strong acid (100% ionized), so the H^+ concentration is 1.04 x 10^-4 M.
To find the pH and pOH of a solution that is 1.04x10^-4 M HCl, you don't need to use an ice table or any calculations involving equilibrium.
Since HCl is a strong acid, it completely ionizes in water, meaning it dissociates into H^+ ions and Cl^- ions. In this case, the concentration of H^+ ions is equal to the concentration of HCl.
Therefore, the pH of the solution is equal to the negative logarithm (base 10) of the H^+ concentration:
pH = -log[H^+]
pH = -log(1.04x10^-4)
pH ≈ 3.98
Similarly, the pOH is equal to the negative logarithm (base 10) of the OH^- concentration, which can be calculated using the concept of Kw (the self-ionization constant of water).
Kw = [H^+][OH^-]
At a neutral pH, [H^+] = [OH^-], so we can assume the concentration of OH^- is the same as the concentration of H^+.
Therefore, pOH = -log[H^+]
pOH = -log(1.04x10^-4)
pOH ≈ 3.98
So, the pH and pOH of the given solution are both approximately equal to 3.98.
Since HCl is a strong acid, it fully dissociates in water, meaning that it completely ionizes into H+ ions and Cl- ions. So, the concentration of H+ ions is equal to the original concentration of HCl, which is 1.04 x 10^-4 M.
To find the pH of the solution, you can use the formula:
pH = -log[H+]
Substituting the concentration of H+ ions, we have:
pH = -log(1.04 x 10^-4)
pH = -(-3.983) (taking the negative logarithm value)
pH ≈ 3.983
To find the pOH of the solution, you can use the formula:
pOH = -log[OH-]
Since this is a strong acid solution, the concentration of OH- ions is negligible. Therefore, the pOH can be assumed to be 0.
So, the pH of the solution is approximately 3.983, and the pOH is 0.