How does Bronsted-Lowry definition of base apply to NaOH exactly?

[zenterix]

Homework Statement

$\mathrm{NaOH(s)}$ is an ionic solid that dissolves in water.

Relevant Equations

A solution of this compound thus contains $\mathrm{Na^+}$ and $\mathrm{OH^-}$.

$\mathrm{OH^-}$ participates in the autoprotolysis of water.

$$\mathrm{2H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)}$$

Now, the definition of a Bronsted-Lowry base is a compound that is a proton acceptor.

$\mathrm{OH^-}$ accepts a proton and is thus a Bronsted-Lowry base.

So, why do we say that $\mathrm{NaOH}$ is a Bronsted-Lowry base?

 

[zenterix]

I'd like to show my doubts with an example.

I think that much of what is done in the calculations below involve small approximations that cause conceptual confusion.

Suppose we have an aqueous solution of $\mathrm{9.78\times 10^{-8}M}$ of $\mathrm{NaOH}$ and we would like to know the pH.

A naive calculation would be

$$\mathrm{pOH=-\log{(9.78\times 10^{-8})}}\tag{1}$$

$$\mathrm{pH=14-pOH=6.99}\tag{2}$$

This is incorrect because it neglects the effects of $\mathrm{NaOH}$ on the equilibrium autoprotolysis of water.

The calculation did partially take into account autoprotolysis: it used the value $10^{-14}$ for $K_w$.

But here I already have one doubt.

We have $\mathrm{K_w=[H_3O^+][OH^-]}=$ from which we can obtain $\mathrm{pH+pOH=pK_w}$ which equals 14 at $\mathrm{25^\circ C}$.

The book I am reading says, based on these equations that "the pH and pOH of a solution have complementary values: if one increases, the other decreases such that their sum remains constant".

This statement is approximately true, right? After all, we don't have pure water anymore, we have $\mathrm{Na^+}$ as well.

Isn't $\mathrm{pK_w=14}$ true for only pure water?

In addition, for example, how can pure water ever have a pH of 2 and a pOH of 12?

I mean, if you were to add more hydronium to the pure water, thus shifting the autoprotolysis reaction towards more water (at the expense of hydroxide), and thus decreasing pH, would this still be pure water?

I guess not.

"Pure water" represents one single equilibrium (at a specific temperature), in which $\mathrm{pH=pOH=7}$. The way I understand it is, this is what would occur if you started just with water molecules.

A dilute solution is almost pure water (at $\mathrm{25^\circ C}$). Solutes are additions to the pure water. If we approximate the solution as just pure water, these solutes are disturbances to the chemical equilibrium but the equilibrium constant, which is a function of temperature, stays the same.

If we add hydroxide ions for example, the reaction shifts towards reactants (ie, water molecules). We end up with less hydronium than before and so the pH increases.

Anyway, moving on, here is a more accurate calculation

$$\mathrm{K_w=[H_3O^+][OH^-]=10^{-14}}\tag{3}$$

$$\mathrm{[OH^-]=[H_3O^+]+[Na^+]}\tag{4}$$

$$\mathrm{[K^+]=[NaOH]_{init}}\tag{5}$$

(5) tells us that cations all come from the initial molecules of $\mathrm{NaOH}$ (this is the material balance equation).

(4) tells us that the number of hydroxide ions is the same as the number of hydronium ions plus sodium ions, for charge balance.

(3) tells us that in water the equilibrium constant for autoprotolysis is $\mathrm{10^{-14}}$.

We can already tell, without doing calculations, that we'll have a smaller concentration of hydronium than hydroxide and so pH will be basic.

In fact, it comes out to 7.20.

Finally, suppose that $\mathrm{NaOH}$ were not a strong base.

Already this last statement is confusing because as far as I can tell $\mathrm{OH^-}$ is actually the base.

If the base were ammonia, $\mathrm{NH_3}$ then being a weak base would mean that protonation would be incomplete: not all the atoms of ammonia in solution would become protonated.

For $\mathrm{NaOH}$ however, an ionic compound, we get hydroxide ions directly.

So, is it even possible to have a weak base from an ionic compound? Does it mean simply that an ionic compound is not fully dissolved in water?

In this hypothetical situation we would have to compute (not sure about this, because dissociation of $\mathrm{NaOH}$ does not seem to be the correct reaction to compute the basicity constant $K_b$).

$$\mathrm{K_b=\frac{[Na^+][OH^-]}{[KOH]}}$$

and we would replace equation (5) with

$$\mathrm{[Na^+]+[NaOH]=[NaOH]_{init}}$$

The smaller $K_b$ would be, the less $\mathrm{[OH^-]}$ ions would come from $\mathrm{NaOH}$ and the less the autoprotolysis of water equilibrium would be shifted towards reactants. Thus, the rise in pH would be less than in the case of full protonation.

In the limits, if $K_b=0$ then we get a pH of 7, and if $K_b=\infty$ then this means that $\mathrm{[NaOH]=0}$ and we get the initial result of ph=7.20.

 

[Borek]

So, why do we say that $\mathrm{NaOH}$ is a Bronsted-Lowry base?

Do we?

Yes, colloquially I would say NaOH is a base, but I would never call it a Bronsted-Lowry base. It is OH^- that is the BL base.

 

[zenterix]

Do we?

Yes, colloquially I would say NaOH is a base, but I would never call it a Bronsted-Lowry base. It is OH^- that is the BL base.

I mean, when people say something is a base, are they not usually using the BL definition of a base?

 

[Borek]

We have $\mathrm{K_w=[H_3O^+][OH^-]}=$ from which we can obtain $\mathrm{pH+pOH=pK_w}$ which equals 14 at $\mathrm{25^\circ C}$.

The book I am reading says, based on these equations that "the pH and pOH of a solution have complementary values: if one increases, the other decreases such that their sum remains constant".

This statement is approximately true, right? After all, we don't have pure water anymore, we have $\mathrm{Na^+}$ as well.

Isn't $\mathrm{pK_w=14}$ true for only pure water?

Original book statement doesn't say anything about "pure water", you just added it to complicate things.

Yes, to some extent presence of other ions changes the equilibrium constant (not to mention the fact pKw depends on the temperature). But 14 is a reasonable approximation.

 

[Borek]

I mean, when people say something is a base, are they not usually using the BL definition of a base?

No.