'One of the keys to acid/base chemistry is that the equilbria are always coupled to an equilibrium of water auto ionization. Thus, there are always at least two equilibria to think about: the equilibrium of the acid (or base) with the water and the equilibrium of water itself with H3O+ and OH-. It is important to remember that in any given solution, the concentration of any compound is the same in all the equilibrium expressions. Thus the idea that they are "coupled" together.
Finally, we like to classify solutions based upon the concentration of H3O+ and OH-. Because these are generally small numbers (much less than 1), it is convenient to look at the log of the concentrations. Because the log is a negative number, we look at the negative log of the concentration. Thus the pH scale.
A video with an introduction to aqueous equilibriaWater is always in equilibrium with H3O+ and OH- ions. The following reaction is what we refer to as the autoionization of water.
\[\rm{2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)}\]
You can see in this reaction that one water molecule is essentially transferring a proton to another water molecule. Thus water is acting both as an acid and as a base. We call such a molecule amphiprotic (both proton donor and proton acceptor).
The extent of this reaction is very, very small since the standard free energy of the water is significantly lower than the standard free energy of the ions. Nonetheless, it is important since these ion species are so reactive.
The equilibrium constant expression for this reaction is
\[\rm{K_w = [H_3O^+][OH^-]}\]
where we have K with the subscript w for "water". The water does not appear in the equilibrium expression since it is a liquid and its activity is taken as one. Kw is also called the "ion product for water".
Kw is a very small number. At room temperature it is very close to being exactly 10-14.
Note: This is a point of great confusion since this makes for very nice round numbers for concentrations at room temperature. It is important to note that: one, it is by chance that it ends up as such a nice round number in the units we have chosen since Kw is temperature dependent (like any equilibrium constant). Thus it has a different value (that is not such a nice round number) at different temperatures.
This means the product of the H3O+ and OH- concentrations is a constant (much like solubility). So if one of the concentrations increases, the other must decrease as the equilibrium adjusts. For example, imagine that we increase the H3O+ concentration by the addition of a strong acid. This will push the equilibrium back to the reactant side to try to "reduce" the H3O+ concentration. This will mean that the OH- concentration will be reduced.
At room temperature Kw is 1x10-14. This allows us to solve for the concentrations of H3O+ and OH- at room temperature.
Given the reaction
\[\rm{2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)}\]
we know that in pure water the concentrations of H3O+ and OH- must be equal. Since for each mole of H3O+ (proton donor) formed, we get one mole of OH- (proton acceptor). Since these concentrations are equal we can give them some variable name. We'll call them both x. Putting this in the equilibrium expressions yields
\[\rm{K_w = [H_3O^+][OH^-] = (x)(x) = x^2 \hskip20pt x = \sqrt{K_w}}\]
\[\rm{x = [H_3O^+]=[OH^-] = \sqrt{K_w} = \sqrt{1x10^-{14}} = 1x10^{-7} M}\]
Thus at room temperature where Kw = 1x10-14 the concentrations of both H3O+ and OH- are 1x10-7 M. (Note: The units are M just like they are for any aqueous equilibrium because we have taken 1 M to be the standard state for aqueous concentrations).
This is at room temperature. What about other temperatures? Since the dissociation of water is endothermic, as we increase the temperature the equilibrium constant will increase. Thus, at higher temperature the concentrations of both the hydronium ion and hydroxide ion will increase. They will no longer be 1x10-7 M; they will be slightly larger. As the concentration is no longer 1x10-7, the pH will no longer be 7. However, the solution will still be neutral! This is because [H3O+] = [OH-]. They are simply no longer equal to 1x10-7M since Kw is no longer equal to 10-14.
Solutions can be classified in comparison to pure water in terms of the concentrations of H3O+ and OH-.
In pure water
\[\rm{[H_3O^+] = [OH^-] \hskip 24pt Neutral}\]
When these concentrations are equal, the solution is neutral. Note, at room temperature this would be when [H3O+] = [OH-] = 1x10-7 M. At other temperatures in a neutral solution the concentrations of hydroxide and hydronium still be equal to each but they will have a different numerical value.
If the concentration of [H3O+] > [OH-], the solution is acidic.
\[{\rm{[H_3O^+] > [OH^-] \hskip 24pt Acidic}}\]
If the concentration of [H3O+] < [OH-], the solution is basic (or alkaline).
\[\rm{[H_3O^+] < [OH^-] \hskip 24pt Basic}\]
The equilibrium expression for the autoionization of water can be used to relate the strength of an acid to its conjugate base (and vice versa).
Take a generic acid, HA in water. It will dissociate and form its conjugate base A- and the hydronium ion.
\[\rm{HA(aq) + H_2O(l) \rightleftharpoons A^-(aq) + H_3O^+(aq)}\]
The equilibrium constant for this reaction is
\[{\rm K_a = {{[H_3O^+][A^-]} \over {[HA]}}}\]
Conversely the conjugate base of HA is A-. When A- is placed in water it will accept a proton from water (protonate) and form HA and the hydroxide ion
\[\rm{A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^-(aq)}\]
The equilibrium constant for this reaction is
\[{\rm K_b = {{[OH^-][HA]} \over {[A^-]}}}\]
If we add these two reactions together we can see that the overall reaction for the two combined reactions is simply the auto-ionization of water.
\[\rm{2H_2O(l) \rightleftharpoons OH^-(aq) + H_3O^+(aq)}\]
The equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions. Thus the equilibrium constant for the two reactions added together is Kb x Ka. And because the net reaction is the auto-ionization of water with the equilibrium constant Kw, we know that
\[{\rm K_w = {K_a \times K_b}}\]
We can use this relationship to convert between the Ka for an acid HA and the Kb for its conjugate base A-. Or we can convert between the Kb for a base B and the Ka for its conjugate acid BH+.
Because the product of Ka and Kb is a constant, there is a natural relationship between the strengths of the acids and bases that form a conjugate pair. The acid strength depends on the magnitude of Ka. The larger the Ka the stronger the acid. The strength of the base depends similarly on the Kb. The larger the Kb the stronger the base.
For conjugated acid/base partners both Ka and Kb can't be large. Because their product is a constant, the larger the Ka of the acid the smaller the Kb of its conjugate base (and vice versa). Thus the stronger the acid, the weaker its conjugate base. Or the stronger the base, the weaker its conjugate acid.
For example, the Ka for acetic acid, CH3COOH, is 1.8x10-5. From this we can find the Kb of its conjugate base partner the acetate ion, CH3COO-. Kb = Kw/Ka = 1x10-14/1.8x10-5 = 5.6x10-10. This can be compared to the acid/base strength of the conjugate pair for hydrofluoric acid, HF/F-. The Ka for HF, is 7.2x10-4. Using this and Kw we find the Kb for F- is 1.4x10-11. HF is a stronger acid than CH3COOH (it has a larger Ka). However, its conjugate base F- is a weaker base than acetic acid's conjugate base, the acetate ion, CH3COO- since Kb for F-1 is smaller than Kb for CH3COO-. For all pairs, the stronger the acid the weaker the base; the stronger the base, the weaker the acid.
This relationship is typically used for weak acids and bases. However, the same is true for strong acids/bases. Because strong acids are assumed to be "infinitely strong (have essentially infinite Ka's), their conjugate bases have Kb that are zero. or "infinitely weak". This makes the conjugate partner for strong acids and bases infinitely weak. They are so weak that we do not even consider them to be acids/bases. Instead these are the ions we refer to as spectator ions.
Which of the following of these acids has the strongest conjugate base? (touch choices to toggle feedback on/off)
Logarithms are a useful mathematical tool that are extremely helpful when dealing with extremely large or extremely small numbers. There are a number of useful relationships to remember
log(xy) = log(x) + log(y)
log(x/y) = log(x) - log(y)
log(x/y) = -log(y/x)
The pH scale is a scale that expresses the hydronium ion concentration, [H3O+], in an aqueous solution using log base 10.
When you are dealing with very large or very small numbers it is convenient to use logarithms. Note: We will be using the logarithm base 10, log10, which is denoted as "log" (this is not the log "base 2", log2; or the log base "e", loge, that is the natural logarithm noted as ln).
Pure neutral water at room temperature has a hydronium ion concentration, [H3O+] = 1x10-7M.
The pH is defined as:
\[pH = -log([H_3O^+])\]
This means for pure water at room temperature
\[pH = -log([H_3O^+])= -log(1x10^{-7}) = -(-7) = 7\]
Thus the pH for neutral water at room temperature is 7. (Again it is important to realize there is nothing "magic" about 7. It just so happens that at room temperature, Kw is nearly exactly a perfect factor of 10 so the math works out in nice round numbers.). Why take the logarithm? Convenience. Rather than having to say "[H3O+] = 7.2x10-5M" you can simply say "pH = 4.14" (-log(7.2x10-5)=-(-4.14)=4.14). Also, in the days before calculators the use of logarithms simplified many calculations. The irony being that now the use of logarithms makes the mathematics more difficult for many students.
You need to remember how to perform basic logarithm math. In particular, you need to be able to take the inverse log. Thus to convert from pH to [H3O+]
\[\rm{[H_3O^+]} = 10^{-pH}\]
pH can vary widely. However, there are some natural limits to the value since [H3O+] has some limits. For example, it is not possible in any aqueous solution for [H3O+] = 200 M. This is simply absurd since one liter of water has only 55 moles of water. It is not possible to have 200 moles of H3O+ in solution that has only 55 moles of water. However, the hydronium ion concentration can easily be larger than 1 M. The pH of these solutions is negative (as the log of numbers larger than 1 is a positive number and pH = -log([H3O+]). On the dilute end of the spectrum, it is not possible to obtain hydronium ion concentrations much less than 1x10-14 M as such extremely low values imply extremely large values of [OH-] as the two are linked by Kw.
Practically speaking [H3O+] range from 1x10-14 M up to 1 M ( but they can be slightly smaller or larger than this). That means that typical pH values range from 14 to 0. Below is a pH scale that shows both the pH and the [H3O+] concentration relative to pH = 7. It is important to note that since pH is a log scale, changes by one unit are really changes in concentration of 10x.
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The concentration of the hydroxide ion, [OH-] can similarly be expressed on a logarithmic scale. The pOH is defined as the negative log of the hydroxide ion concentration
\[\rm{pOH = -log([OH^-])}\]
Since the [H3O+] and [OH-] concentrations are related by the Kw (the ion product of water) equilibrium, we can find a shortcut for converting between these two using pH and pOH
\[\rm{K_w = [H_3O^+][OH^-]}\]
Taking the negative logarithm of both sides of this equation yields
\[{\rm -log(K_w) = -log([H_3O^+]) -log([OH^-])}\]
Assuming we are working at room temperature where the ion product of water is 1x10-14 then gives
\[-log(1x10^{-14}) = -(-14) = 14 = pH + pOH\]
Thus at room temperature, where Kw is 1x10-14, we can convert between pH an pOH since their sum must equal fourteen (just like the product of the concentrations must equal 1x10-14).
It is extremely straightforward to find the pH of a solution of an acid in water. This is because it is not any different than a standard equilibrium problem.
For a strong acid the problem is extremely straight forward. The dissociation of the acid is assumed to be complete such that the concentration of the hydronium ion concentration is.
\[ {\rm[H_3O^+] \approx (C_{HA})}\]
Once we have the hydronium ion concentration we simply can relate it to the pH.
\[ {\rm pH = -log[H_3O^+]} \]
When will this fail? In very dilute solutions. Why? We are assuming that all of the hydronium ion concentration is coming from the acid. However, we know that in neutral water (at room temperature) the concentration of hydronium ion is 10-7 M. This is not much. But if we make some ridiculously low concentration acid solution (10-8 M HCl) then we cannot assume the initial concentration is zero. How can it be solved? Simply set up an equilibrium problem where we assume the initial concentration is 10-7 M. It requires some more algebra, but it isn't different than any other equilibrium problem.
The pH of a weak acid solution is also straight forward to solve as this is also simply an equilibrium problem. Writing out the typical problem for a generic weak acid of concentration CHA in some concentration in pure water we can find the concentration of all the species at equilibrium if we know Ka for the acid.
\[\rm{HA(aq) + H_2O(l) \rightleftharpoons A^-(aq) + H_3O^+(aq)}\]
The equilibrium constant for this reaction is
\[{\rm K_a = {{[H_3O^+][A^-]} \over {[HA]}}}\]
Previously it was shown that this results in the following algebraic expression where "x" is the concentration of the acid that dissociates. Again the assumption is made that the concentration of hydronium ion in the neutral water was zero when in fact it was 10-7
.\[{\rm K_a = {{[H_3O^+][A^-]} \over {[HA]}}}= {{(x)(x)} \over (C_{HA}-x)}\]
We can again make the assumption that the change in concentration of the acid is negligible. This gives
\[{\rm K_a }= {{(x)(x)} \over (C_{HA}-x)} \approx {{(x)(x)} \over (C_{HA})}= {x^2 \over C_{HA}}\]
Solving this we get
\[ x = \sqrt{{\rm (K_a)(C_{HA})}}\]
Remembering that we defined "x" as the change in concentration of the weak acid we know that this is also the concentration of the hydronium ion at equilibrium. This yields a very useful approximate formula, that the concentration of hydronium ion at equilibrium in a solution that contains only a weak acid in water is given by
\[ {\rm[H_3O^+] = \sqrt{(K_a)(C_{HA})}}\]
The hydronium ion concentration can then be converted into pH.
The same expression is useful for the a weak acid that is a protonated base (such as NH4+). We can denote this compound as BH+. It is identical in all respects to other weak acids, but we can identify it as it has a positive charge.
A video that shows examples of calculating the pH of a strong acid solution and a weak acid solutionThe pH of a solution of strong base in water can be found by first finding the concentration of the hydroxide ion and then converting to pH. For a strong base with concentration CB we have.
\[ {\rm[OH^-] \approx (C_{B})}\]
note: we have to keep in mind that some strong bases produce two moles of hydroxide for each mole of compound.
The [OH-] can be converted to pOH and then pH. Alternatively we can use Kw to convert [OH-] to [H3O+] and then to pH. Either way the result is the same.
For a weak base we have a very similar result to that for a weak acid. Fro a solution with concentration CB and Kb
\[{\rm K_b = {{[OH^-][BH^+]} \over {[B]}}}\]
Using an identical derivation, we arrive at the approximate result with CB is the concentration of the weak base.
\[ {\rm[OH^-] = \sqrt{(K_b)(C_B)}}\]
[OH-] can then be converted into pOH and then pH.
When do these approximations fail? Again the assumptions in all cases is that the initial concentration of the ions in the water are zero. This breaks down for dilute solutions. Also for the weak base, the assumption is that is reacts only very slightly. Thus the assumption gets better the weaker the base (with the caveat that when it gets to be weaker than water it is very poor again).
What is the pH of a 0.1M NaOH solution? (mouse over choices to get answer)
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