Equilibria, pH and pOH

Acid-Base Equilibria

According to the Bronsted-Lowry theory of acids and bases, an acid is a substance which when dissolved in water generates solvated protons, whereas a base is a substance which when dissolved in water sequesters solvated protons. When an acid releases its proton, what remains is said to be the conjugate base, and similarly, when a base accepts a proton, what results is said to be the conjugate acid. An acid is said to be strong if it dissociates to near completion, whereas one which only partially dissociates is a weak acid. In general, if a strong acid dissociates, it produces a weak conjugate base, and when a weak acid dissociates, its conjugate base is said to be strong.

Representing the equilibrium as a generalised acid, HA, we have

`HA(aq)+H_2O(l) earr H_3O^+(aq)+A^(-) (aq)`

where HA is the generalised acid, and A- is the conjugate base

and for a base, B,

`B(aq)+H_2OearrOH^(-)+BH^+(aq)`

where B is the generalised base, and BH+ is the conjugate acid

As with all equilibria, we can express their equilibrium concentrations through their equilibrium expression. For a generalised equilibrium,

`alphaA+betaBearrgammaC+deltaD`

We have the equilibrium expression,

`K=([C]^gamma[D]^delta)/([A]^alpha[B]^beta)`

where K is a constant, characteristic of the species, temperature and pressure. Note, pure substances, solids and liquids are defined to have unit concentrations

Thus for an acid we have the acid dissociation constant, Ka,

`K_a=([H_3O^(+)(aq)][A^(-)(aq)])/([HA(aq)][H_2O(l)])`

...but, water is a pure liquid and so it is defined to have a unit concentration, therefore

`K_a=([H_3O^(+)(aq)][A^(-)(aq)])/([HA(aq)])`

and for a base we have the base dissociation constant, Kb,

`K_b=([OH^(-)(aq)][BH^(+)(aq)])/([B(aq)][H_2O(l)])`

again, the concentration of water is one, thus

`K_b=([OH^(-)(aq)][BH^(+)(aq)])/([B(aq)])`

Auto-Dissociation of Water

Also called the 'dissociation-product', 'self-dissociation', 'self-ionisation' and other various permutations!

For a weak acid (or base) the equilibrium expression takes on this form. Consider, water is both a weak acid and weak base - it dissociates, partially, to give a mixture of H3O+ and OH-,

`H_2O + H_2O earr H_3O^+ + OH^(-)`

Thus the equilibrium expression for the auto-dissociation of water is

`K_w=([OH^(-)][H_3O^(+)])/[H_2O]^2`

However, since water is a pure substance, it has unit concentration, thus,

`K_w=[OH^(-)][H_3O^(+)]`

Water only dissociates very slightly. At equilibrium, at STP, the dissociation lays far toward H2O - away from H+ and OH-, infact, water dissociates only to the extent that the concentration of solvated protons is 1x10-7moldm-3. Also consider, that since one mole of water dissociates to produce one mole of protons, and one mole of hydroxide, then, for pure water,

`[H_3O^(+)]=[OH^(-)]=1xx10^(-7)moldm^(-3)`

Thus we can express the auto-dissociation of water equilibrium constant as,

`K_w=[H^(+)]^2=[OH^(-)]^2=1xx10^(-14)mol^2dm^(-6)`

It is common to represent equilibrium constants, which can span many orders of magnitude (for example, 1x10-30 to 1x1030), using logarithms. This contracts the scale to one more manageable. Whilst `lgK_w` may suffice, we actually use use the notation, `pK`, with `pK=-lgK` (where p means cologarithm). According to this formulation, for water, the equilbrium constant can be expressed as,

`pK_w=lg([H^(+)])+lg([OH^(-)])=14`

...which brings us onto the definition of pH. Introduced in 1909 by Sørensen, the pH is simply defined as the negative base-10 logarithm (ie., the cologarithm) of the proton concentration,

`pH=-lg([H^+])`

Similarly defined, pOH, is the negative base-10 log of the hydroxide concentration,

`pOH=-lg([OH^(-)])`

Hence,

`pK_w=pH+pOH`

pH of Strong Acids

For a strong acid, for which dissociation is close to 100%, then according to the equilibrium, HA reacts with water to produce a stoiciometric quantity of solvated protons. Provided the amount of strong acid added is signficantly greater than the concentration of protons in pure water ie., `[HA]>1xx10^(-7)moldm^(-3)` then in which case,

`[H^+]_"equilibrium"=[HA]_"initial"`

Thus we can express the pH in terms of the concentration of acid,

`pH=-lg([H^+])=-lg([HA]_"initial")`

Example I:

Calculate the pH of a solution containing 0.5moldm-3 of the strong acid, HI.

`pH=-lg([HI]_"initial")=-lg(0.5)=0.3`

Example II:

Calculate the pH of a solution containing 0.1moldm-3 of the strong acid, HCl.

`pH=-lg([HCl]_"initial")=-lg(0.1)=1`

pH of Strong Bases

A strong base is defined as one which sequesters a near stoichiometric quantity of protons from water when dissolved. Ie.,

`B_("strong")(aq)+H_2O(l)earrBH^(+)(aq)+OH^(-)(aq)`

Consequently, at equilibrium,

`[OH^(-)]_"equilibrium"=[B]_"initial"`

Through the equilibrium Kw, we can express the proton concentration in terms of the hydroxide concentration,

`K_w/([OH^(-)])=[H^(+)]`

Hence,

`[H^(+)]=K_w/[B]_"initial"`

We can, alternatively, express the relationship in a different form, using the pK notation, ie.,

`pK_w=pH+pOH rarr pK_w-pOH=pH`

and so,

`pH=pK_w-pOH=14+lg([B]_"initial")`

Example III:

Calculate the pH of an aqueous solution containing 0.05moldm-3 of the strong base, KOH.

`[H^(+)]=K_w/[B]_"initial"=(1xx10^-14mol^2dm^(-6))/(0.05moldm^(-3))=2xx10^(-13)moldm^(-3)`, and then

`pH=-lg([H^(+)])=-lg(2xx10^(-13))=12.7`

Example IV:

Calculate the pH of an aqueous solution in which 0.15moldm-3 of the strong base, NaOH has been dissolved.

`pH=pK_w-pOH=14+lg([B]_"initial")=14+lg(0.15)=13.2`

pH of Weak Acids

The calculation of weaks acids is slightly more difficult. We begin with the equation for the dissociation of weak acid, HA,

`HA(aq)earrH^(+)(aq)+A^(-)(aq)`

Then we state that, at equilibrium, the equilibrium expression always holds true and the ratio is constant, ie.,

`K_a=([H^(+)(aq)][A^(-)(aq)])/([HA(aq)])`

The calculate the equilibrium concentration of H3O+ we can use the ICE method.

Example V:

Estimate the pH of a solution of 0.5M acetic acid (pKa=1.8x10-5moldm-3). To start, we state the equilibrium expression,

`K_a=([H^(+)(aq)][AcO^(-)(aq)])/([AcOH(aq)])`

We use the ICE method to establish the equilibrium concentrations in terms of a reaction variable, x,

  `AcOH` `stackrel(K_a) earr H^(+)` `+AcO^(-)`
I `A_0` 0 0
C `-x` `+x` `+x`
E `A_0-x` `x` `x`

At equilibrium we have,

`K_a=([H^(+)][AcO^(-)])/([AcOH])=x^2/(A_0-x)`

Which we can solve as a quadratic equation,

`[H^(+)]=x=-K_a/2+-sqrt(K_a^2+4K_aA_0)/2`

`=-(1.77xx10^(-5))/2+-sqrt((1.77xx10^(-5))^2+4xx1.77xx10^(-5)xx0.5)/2`

`[H^(+)]=2.97xx10^(-3)moldm^(-3)`

The resulting proton concentration is significantly higher than that of pure water (cf. 1x10-7moldm-3), and so,

`pH=-lg([H^(+)])=-lg(2.97xx10^(-3))=2.53`

pH of Weak Bases

To estimate the pH of a solution of a weak base we follow a similar method to that of a weak acid. That is, we solve the equilibrium expression for hydroxide concentration, and then through the dissociation product of water, Kw, express the proton concentration. As with strong bases, when a weak base is made up in aqueous solution, it abstracts a proton from water releasing the conjugate-acid and hydroxide,

`B(aq)+H_2O(aq)earr BH^(+)(aq)+OH^(-)(aq)`

The equilibrium is represented by the base-dissociation constant,

`K_b=([OH^(-)(aq)][BH^(+)(aq)])/([B(aq)][H_2O(l)])`

As before, the water has a unit concentration, thus,

`K_b=([OH^(-)(aq)][BH^(+)(aq)])/([B(aq)])`

Example VI:

Calculate the hydroxide concentration, and subsequently the pOH and pH, of an aqueous solution made up to 0.5mold-3 of sodium acetate. The pKb of sodium acetate is 9.24.

We first convert the pKb to a Kb,

`K_b=10^(-pK_b)=10^(-9.24)=5.75xx10^(-10)`

Then express the equilibrium equation,

`K_b=([OH^(-)(aq)][AcOH(aq)])/([AcO^(-)(aq)])`

...which we solve for hydroxide using the ICE method,

  `NaOAc` `stackrel(K_b) earr AcOH` `+OH^(-)`
I `A_0` 0 0
C `-x` `+x` `+x`
E `A_0-x` `x` `x`

Using the reaction variable x, we re-construst the base-dissociation equilibrium,

`K_b=(x^2)/(A_0-x)`

Which we can solve for x as a quadratic equation,

`[OH^(-)]=x=-K_b/2+-sqrt(K_b^2+4K_b A_0)/2`

`=-(5.75xx10^(-10))/2+-sqrt((5.75xx10^(-10))^2+4xx5.75xx10^(-10)xx0.5)/2`

Thus,

`[OH^(-)]=1.69xx10^(-5)moldm^(-3)`

The pOH is simply the cologarithm of the OH-(aq) concentration,

`pOH=-lg([OH^(-)(aq)])=4.77`

Then using the relationship between the dissociation product of water and pOH, the pH can be found,

`pK_w=pH+pOH`

Hence,

`pH=pK_w-pOH=14-4.77`

`pH=9.23`