Calculate the pOH and pH of a solution which contains 0.001 M NaOH. Assume 100% ionization. (Need an in-depth explanation with formulas please)

pH: 11

pOH: 3

Explanation:

NaOH is a strong base which means that it dissociates completely in water. It will break apart into Na⁺ and OH⁻ ions.

NaOH ⇒ Na⁺ + OH⁻

Because NaOH dissociates completely into its respective ions in water, the moles of NaOH is equal to the moles of hydroxide ions. So, [OH⁻] = 0.001 M.

Now to find the pOH, use the formula pOH = -log[OH⁻].

pOH = -log[OH⁻]

= -log(0.001)

= 3

The pOH of the solution is 3.

To find the pH, subtract the pOH from 14 since pH + pOH = 14.

14 - 3 = 11

The pH of the solution is 11.

Hope that helps.

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