Question

48g of potassium hydroxide (NaoH) dissolved in 133 ML of solution. Calculate the PH of the solution.
(Na=23, O=16, H=1)

Answer 1

First I’ll assume you mean sodium hydroxide (NaOH) since you listed Na = 23, O = 16, H = 1.

1. Molar mass NaOH = 23 + 16 + 1 = 40 g·mol⁻¹
2. Moles NaOH = 48 g / 40 g·mol⁻¹ = 1.20 mol
3. Volume = 133 mL = 0.133 L, so [OH⁻] = 1.20 mol / 0.133 L = 9.02 M
4. pOH = −log10[OH⁻] = −log10(9.02) = −0.955
5. pH = 14.00 − pOH = 14.00 − (−0.955) = 14.96

So pH ≈ 14.96 (using Kw = 1.0×10⁻¹⁴ at 25 °C and ideal behavior).

If you actually meant potassium hydroxide (KOH, molar mass ≈ 56 g·mol⁻¹), the result would be pH ≈ 14.81 (same method, [OH⁻] ≈ 6.45 M → pH ≈ 14.81).