This Method Is Nearly Identical To The One Presented In The Example ...

What is the pH of a 2.0x10-7 M HCl solution?

ALTERNATE METHOD 1:

This method is nearly identical to the one presented in the example, except that we allow the water in solution to dissociate as though pure before using Kw to calculate the correct equilibrium concentrations of H+ and OH-.

As before, we start off with 2.0x10-7 M H+ from the HCl. However, now we will allow the water to dissociate as if the HCl were not there, giving 1.0x10-7 M H+ and 1.0x10-7 OH- from the H2O, resulting in a total [H+] of 3.0x10-7 M. Since the concentrations of H+ and OH- are now too high for equilibrium according to Kw, we now allow the water dissociation equilibrium to readjust:

	H2O	H+	OH-
INITIAL	-	3.0x10-7 M	1.0x10-7 M
CHANGE	-	-x M	-x M
EQUILIBRIUM	-	(3.0x10-7 - x) M	(1.0x10-7 - x) M

Kw = [H+]eq*[OH-]eq 1.0x10-14 = (3.0x10-7 - x) * (1.0x10-7 - x) 1.0x10-14 = x2 - (4.0x10-7 * x) + 3.0x10-14 0 = x2 - (4.0x10-7 * x) + 2.0x10-14

Using the quadratic formula, we get x = 3.41x10-7 M or x = 5.86x10-8 M. Since [H+] = (3.0x10-7 - x) M and we cannot have a negative [H+], x = 5.86x10-8 M.

So, [H+]eq = (3.0x10-7 - 5.86x10-8) M = 2.41x10-7 M, giving a pH of 6.62.

ALTERNATE METHOD 2:

Instead of considering the two equations sequentially, we can solve them simultaneously. This is a more general approach that does not require the assumption that one of the two reactions is unaffected by the other.

The pH of the solution is controlled by the two reactions:

HA H+ + A- H2O H+ + OH-

To find the equilibrium [H+] by simultaneous solution, we need to solve for all the unknowns in both reactions when the system is at equilibrium. There are four unknowns: [HA], [A-], [H+] and [OH-]. However, since HA is a strong acid, it will dissociate completely such that [HA]=0 and [A-]=Ca, the concentration of the strong acid solution. Thus, we are left with only two unknowns at equilibrium: [H+] and [OH-].

To solve for two unknowns, we need two independent equations relating them. One equation is the one describing the equilibrium dissociation of water:

(1) Kw = [H+] * [OH-]

The other equation can be derived using the principle of charge balance, which states that the net charge on a system does not change in a chemical reaction. Before any ionization takes place, only water and HA are present. Since both are neutral, the system initially has no net charge. This means that, after ionization, the system must also have no net charge.

[total anions] = [total cations]

Since the only charged species produced are H+, OH- and A-, we have

(2) [A-] + [OH-] = [H+]

Using [A-]=Ca in equation (2),

Ca + [OH-] = [H+] [OH-] = [H+] - Ca

Now, substituting into equation (1), we have

Kw = [H+]eq*[OH-]eq Kw = [H+] * ([H+] - Ca) Kw = [H+]2 - Ca*[H+]

This general equation gives the exact concentration of [H+] in a strong acid solution when the dissocation of water is continued.

Using Ca = 2.0x10-7 M and [H+] = x,

Kw = [H+]2 - Ca*[H+] 1.0x10-14 = x2 - (2.0x10-7 * x) 0 = x2 - (2.0x10-7 * x) - 1.0x10-14

Using the quadratic formula, we get x = 2.41x10-7 M or x = -4.14x10-8 M. Since we cannot have a negative [H+], x = 2.41x10-7 M.

So, [H+]eq = 2.41x10-7 M, giving a pH of 6.62.

Although all three methods work equally well in the strong acid case and any can be used, the simultaneous solution method is important because it is the only method that can be used for the equivalent weak acid calculation, as we shall see.