Calculate the Ph and POH of an aqueous soltuion that is 0.500 M in HCl(aq) and 0.085 M in HBr (aq) at 25 degrees C.
HCl is a strong acid; i.e., it ionizes 100%.
Same for HBr.
So the combined M in the solution is 0.5+0.085.
pH = -log(H^+), then
pH + pOH = pKw = 14.
To calculate the pH and pOH of the given aqueous solution, we need to use the concept of dissociation of acids in water.
1. Dissociation of HCl:
HCl(aq) → H+(aq) + Cl-(aq)
2. Dissociation of HBr:
HBr(aq) → H+(aq) + Br-(aq)
Since both HCl and HBr are strong acids, they will ionize 100% in water. This means all the HCl and HBr will dissociate into their respective ions.
To find the pH and pOH, we need to calculate the concentration of H+ ions in the solution. Since HCl is fully dissociated, the concentration of H+ ions from HCl is 0.500 M. Similarly, the concentration of H+ ions from HBr is 0.085 M.
Step 1: Calculate the total concentration of H+ ions:
[H+] = [H+](from HCl) + [H+](from HBr)
[H+] = 0.500 M + 0.085 M
[H+] = 0.585 M
Step 2: Calculate the pOH:
pOH = -log[OH-]
In a neutral solution, [H+] = [OH-], so we can assume that [OH-] is also 0.585 M.
pOH = -log(0.585)
pOH ≈ 0.233
Step 3: Calculate the pH:
pH = 14 - pOH
pH = 14 - 0.233
pH ≈ 13.767
Therefore, the pH of the given solution is approximately 13.767, and the pOH is approximately 0.233.