Calculate the pOH and the pH of a 5.0 x 10-2 M solution of NaOH.

pOH = 1.3, pH = 12.7

Explanation:

Since NaOH is a strong base, it will completely ionize; further, since it completely ionizes, our hydroxide concentration (a product of the ionization) will be the same as the given concentration of NaOH.

NaOH -> Na⁺ + OH⁻, [OH⁻] = 5.0 x 10^-2 M

pOH is the negative log of the hydroxide concentration, so plug our hydroxide concentration in:

pOH = -log[OH⁻] = -log[5.0 x 10^-2 M] = 1.3

Since pH + pOH = 14, we can plug in pOH and solve for pH:

pH + 1.3 = 14

pH = 14 - 1.3 = 12.7

Thus, our pOH = 1.3 and pH = 12.7.

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