A solution is prepared by mixing 4.62 grams of potassium hydroxide in 1.00 L of water. After that, 250.0 mL of a 0.100 M solution of HCl is added to the solution. What is the resulting pH of the final mixture?

mols KOH = grams/molar mass = ?

mols HCl = M x L
Which is in excess. How much is the excess. If H in excess, (H^+) = mols/L solution. If OH in excess, ((OH^-) = mols/L. Then pH = -log(H^+) if H^+ is in excess. If OH is in excess it is pOH = -log(OH^-), the pH + pOH = pKw = 14. You will know pKw and pOH, solve for pH.

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.