What should be the pH of a 0.1M NaOH solution?
- mrjeffy321Lv 71 decade agoFavorite Answer
pH = -log [H+]
Where [H+] is the concentration of the H+ ions present in the solution.
pOH = -log [OH-]
Where [OH-] is the concentration of the OH- ions present in the solution.
pH + pOH = 14
So, pH = 14 – pOH = 14 – (-log [OH-])
We are given the concentration of an NaOH solution. NaOH is a strong electrolyte and will disassociate completely into Na+ and OH- ions in solution. These OH- ions from the NaOH will contribute to raise the pH (and lower the pOH) of the solution.
For every 1 mole of NaOH dissolved in solution, 1 mole of OH- ions are releases (one to one ratio). For if you have a .1 Molar NaOH solution, you have a .1 Molar OH- solution.
Since the concentration of the NaOH solution given in the problem is MUCH greater than the natural OH- concentration in the water due to water’s tendency to disassociate with itself (H2O --> H+ + OH-), we can neglect the water’s contribution to the [OH-] and focus solely on the NaOH’s contribution to the [OH-].
The pH of pure water is 7, meaning that it disassociates into H+ and OH and the equilibrium lies where [H+] = [OH-] = 10^-7 Molar = .0000001 Molar.
.1 Molar >> .0000001 Molar, so we are justified in making this estimation, the [OH-] due to the NaOH is 1 million times greater than due to the water.
[NaOH] = [OH-] = .1 Molar
pOH = -log [OH-] = -log [NaOH] = -log (.1) = 1
pOH = 1
pH = 14 – pOH = 14 – 1 = 13
pH = 13
So the pH of a .1 Molar solution of Sodium Hydroxide is 13.
- 1 decade ago
OH is a strong base
[OH] = .1
pOH = -log[OH]
pOH = -log[.1] = 1
pH = 14- pOH
pH = 13
- Gervald FLv 71 decade ago