a 0.120 ml portion of a 50.0% (w/w) KOH solution having a density of 1.50*10^3 g/l is diluted to 250 ml. What is the pH of the resulting solution?

50% w/w KOH means 50 g KOH/100 g solution.

Convert 50 g KOH to mols. mol = grams/molar mass.
Convert 100 g soln (mass) to volume using the density; I would convert that density you have in the problem to g/mL first.
Then M = mols/L soln for the original solution = ?. After dilution it is
M = ?M x original volume/250 (I wonder if that is really 0.120 mL).
Calculate OH^- and pOH, then convert to pH from that.

To find the pH of the resulting solution, we need to calculate the concentration of KOH in the diluted solution.

First, let's determine the mass of KOH in the initial 0.120 mL portion of the 50.0% KOH solution:

Mass of KOH = Volume of solution (mL) × Density (g/mL) × Concentration (%w/w)
Mass of KOH = 0.120 mL × 1.50*10^3 g/L × 50.0% = 9.0 g

Next, we need to determine the final concentration of KOH after dilution:

Concentration (C1V1) = Concentration (C2V2)
(50.0%)(0.120 mL) = C2(250 mL)
C2 = (50.0%)(0.120 mL)/(250 mL)
C2 = 0.030%

Now, we can convert the concentration to moles per liter:

Concentration (mol/L) = Concentration (%) × (10^-2 g/1 mol) / Molar Mass (g/mol)
Concentration (mol/L) = 0.030% × (10^-2 g/1 mol) / 56.11 g/mol
Concentration (mol/L) = 0.030 × 10^-2 / 56.11
Concentration (mol/L) ≈ 5.35 × 10^-6 mol/L

Now that we have the concentration of KOH in the diluted solution, we can calculate the pOH:

pOH = -log10[OH-] = -log10(5.35 × 10^-6) ≈ 5.27

Finally, we can find the pH using the relation:

pH + pOH = 14
pH = 14 - pOH ≈ 14 - 5.27 ≈ 8.73

Therefore, the pH of the resulting solution is approximately 8.73.