To find the resulting pH of the final mixture, we need to determine the concentration of hydroxide ions (OH-) in the solution.
First, let's find the number of moles of potassium hydroxide (KOH) in 4.62 grams. We can do this by using the formula:
Number of moles = Mass (g) / Molar mass (g/mol)
The molar mass of KOH is calculated as follows:
Molar mass (KOH) = (atomic mass of K) + (atomic mass of O) + (atomic mass of H)
= (39.10 g/mol) + (16.00 g/mol) + (1.01 g/mol)
= 56.11 g/mol
Substituting the values into the formula:
Number of moles = 4.62 g / 56.11 g/mol
≈ 0.0824 mol
Since we mixed the potassium hydroxide in 1.00 L of water, the concentration of KOH in the solution is:
Concentration (KOH) = Number of moles / Volume (L)
= 0.0824 mol / 1.00 L
= 0.0824 M
Next, we need to determine the change in concentration of hydroxide ions after adding the 250.0 mL of 0.100 M HCl solution.
Using the equation:
Molarity (M) = Moles / Volume (L)
We can calculate the number of moles of HCl in the 250.0 mL solution:
Moles (HCl) = Molarity (HCl) * Volume (L)
= 0.100 mol/L * 0.250 L
= 0.0250 mol
Since HCl is a strong acid, it completely dissociates in water, forming one mole of H+ ions for every mole of HCl. Therefore, the change in concentration of hydroxide ions will be equal to the change in concentration of H+ ions.
In this case, the change in concentration of H+ ions is:
Change in concentration (H+) = - Moles (HCl)
= - 0.0250 mol
Now, let's calculate the concentration of hydroxide ions in the final solution:
Concentration (OH-) = [Initial concentration (KOH) - Change in concentration (H+)]
= [0.0824 M - (-0.0250 mol)]
= 0.0824 M + 0.0250 M
= 0.1074 M
Finally, to find the pH, we can use the formula:
pOH = -log (Concentration of hydroxide ions, [OH-])
pOH = -log (0.1074 M)
pOH ≈ 0.97
Since pH + pOH = 14 (at 25 degrees Celsius), we can calculate the pH as follows:
pH = 14 - pOH
= 14 - 0.97
≈ 13.03
Therefore, the resulting pH of the final mixture is approximately 13.03.