38.6k views
1 vote
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)

User Zswqa
by
8.8k points

1 Answer

6 votes

Answer:

pH: 11

pOH: 3

Step-by-step 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.

User Shubham Azad
by
8.4k points

Related questions

2 answers
5 votes
136k views
asked Oct 4, 2024 103k views
Kwart asked Oct 4, 2024
by Kwart
8.7k points
1 answer
1 vote
103k views
asked Dec 19, 2024 149k views
Buc asked Dec 19, 2024
by Buc
8.0k points
1 answer
5 votes
149k views