# Acidity and pH

#### pivoxa15

In the textbook it stated "In aqeous solutions, the concentration of hydronium ions is always related to the concentration of hyroxide ions. Therefore we can still measure or calculate the pH of solutions of bases like NaOH, even though they are not acidic."

The book also stated that pH is a measure of the concentration of hydonium in the solution. 10^-1M indicates its very acidic. 10^-14 indicates its very basic.

OH from NaOH will react with the hyrdonium in the solution only. Just say at the start, in pure water there are 10^-7M of hydronium and hyroxide due to self ionisation. In the end there will be less hyronium left in the solution and how much left will be an indication of the acidity of NaOH? It will depend on how much NaOH is added wouldn't it? The more its added the less hyrodium so the more basic NaOH seems. There will still be 10^-7M of hydroxide ions. But how is the concentration of hydronium related to hyrdoxide ions as the book claims?

#### Gokul43201

Staff Emeritus
Gold Member
OH from NaOH will react with the hyrdonium in the solution only. Just say at the start, in pure water there are 10^-7M of hydronium and hyroxide due to self ionisation. In the end there will be less hyronium left in the solution and how much left will be an indication of the acidity of NaOH? It will depend on how much NaOH is added wouldn't it?
Yes, the pH depends on the concentration of the solution. As you add more NaOH to the water, you are increasing the concentration, and consequently increasing the pH.

The more its added the less hyrodium so the more basic NaOH seems.
It's the solution that is becoming more basic because its concentration is increasing. And strictly speaking, the terms 'acidic' and 'basic' refer to the strength of the acid or base (i.e., its dissociation constant), and not to the pH.

There will still be 10^-7M of hydroxide ions.
Not true. As you add more NaOH, you make more OH- ions. When NaOH(s) is dissolved in water, most of it dissociates into Na+ and OH- ions. If you chuck 1 mole (40g) of NaOH(s) into a liter of water, you will have nearly a 1M concentration of OH- ions.

But how is the concentration of hydronium related to hyrdoxide ions as the book claims?
As you put more OH- ions into the solution, they consume more and more H+ ions, making H2O, thus reducing the resultaing concentration of H+. The fact that the autodissociation of water, $2H_2O \leftrightarrow H_3O^+ + OH^-$ has the equilibrium constant $K_w = [H_3O^+] \cdot [OH^-] \approx 10^{-14}$ tells you that in any aqueous solution in equilibrium, the product of [H3O+] and [OH-] is a constant. If one is high, the other must be low to keep their product (or the sum of their logarithms) a constant.

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#### pivoxa15

Not true. As you add more NaOH, you make more OH- ions. When NaOH(s) is dissolved in water, most of it dissociates into Na+ and OH- ions. If you chuck 1 mole (40g) of NaOH(s) into a liter of water, you will have nearly a 1M concentration of OH- ions.
How did you calculate 1M concentration of OH- ions. These ions inclusde NaOH ions as well as self ionised water?

The amount of hydronium in the solution will be neglible compared to the amount of OH-(mostly coming from NaOH). So all the hydronium will have reactd with OH-, with pretty much none left. The pH is a measure of the concentration of hydronium in the solution so in this case it would be close to 14?

As you put more OH- ions into the solution, they consume more and more H+ ions, making H2O, thus reducing the resultaing concentration of H+. The fact that the autodissociation of water, $2H_2O \leftrightarrow H_3O^+ + OH^-$ has the equilibrium constant $K_w = [H_3O^+] \cdot [OH^-] \approx 10^{-14}$ tells you that in any aqueous solution in equilibrium, the product of [H3O+] and [OH-] is a constant. If one is high, the other must be low to keep their product (or the sum of their logarithms) a constant.
Is this constant value of 10^-14 including hyrdoxide or hydronium ions from other compounds like NaOH? So its not just self ionisation constant? Why is it a constant?

#### Gokul43201

Staff Emeritus
Gold Member
How did you calculate 1M concentration of OH- ions. These ions inclusde NaOH ions as well as self ionised water?
Yes, but I neglected the ions coming from water (because that's small in comparison) and I assumed that all the NaOH dissociated into ions in solution. In reality, I think it's about 50 to 60% for NaOH.

The amount of hydronium in the solution will be neglible compared to the amount of OH-(mostly coming from NaOH). So all the hydronium will have reactd with OH-, with pretty much none left.
You have to remember, that in equilibrium chemistry, you can never use the word "all". The fact that H+, OH- and H2O have their own dynamic equilibrium means that none of these 3 concentrations can ever go to zero.

The pH is a measure of the concentration of hydronium in the solution so in this case it would be close to 14?
That's probably true, though not obviously true (unless you do the calculation). There is nothing that actually limits the pH to values between 0 and 14.

Is this constant value of 10^-14 including hyrdoxide or hydronium ions from other compounds like NaOH?
If I understand your question correctly, then yes, it is the product of [total H+] and [total OH-] that is a constant.
So its not just self ionisation constant?
It is the self-ionization constant, because any H= and OH- that gets added to the solution must participate in the self-ionization equilibrium.

Why is it a constant?
You are asking why the equilibrium constant for a reaction is a constant (independent of concentrations of various things put into the reaction vessel). That argument comes from thermodynamics (and is the basis of le Chatelier's principle) - the best I can do is refer you to a text like Atkins (Physical chemistry).

While it might appear that we are unfairly applying the autodissociation constant to a case where something else is added to the water, we must recognize that as far as the three species H+, OH- and H2O are concerned, all other species in the solution are irrelevant to their mutual equilibrium condition, and additionally that they have no way of knowing "where they originally came from" (nor would they consider that relevant).

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