Sodium hydroxide (NaOH) is a strong base, which means it should dissociate (more or less) fully. This means there's a 1:1 ratio of Na to OH⁻, making concentration calculations easy.
First, use the molar mass to calculate how many moles of OH⁻ is in 192 mg of NaOH. (The molar mass of NaOH is approximately 40.0 g/mol.)
192 g NaOH × 1 mol / 40.0 g NaOH = 4.80 mol NaOH (and OH⁻)
Then, find the concentration by dividing that by by the volume of liquid. We must first convert it to liters:
150. mL × 1 L / 1000 mL = 0.150 L
4.80 mol OH⁻ / 0.150 L = 32.0 mol/L OH⁻
Then, we apply −㏒ to the concentration.
−㏒32.0 = −1.505.
We are not done. This is pOH, and we want pH. There's a few ways to find pH, but the easiest one (given that the solution's at STP) is to subtract the answer from 14 (pH + pOH = 14).
14 − (−1.505) = 15.505 (note that with pH, only the figures after the decimal point count as significant figures).
Another way is to take the concentration of OH- and find the concentration of H₃O⁺. We take Kw (1.0 x 10⁻¹⁴) and divide it by 32.0, since [OH⁻] x [H₃O⁺] = Kw.
1.0 x 10⁻¹⁴ / 32.0 M OH⁻ = 3.13 x 10⁻¹⁶ M H₃O⁺
Then apply −㏒ to the concentration:
−㏒3.13 x 10⁻¹⁶ = 15.505.
The answer is pH = 15.505.
