ABSE_pH of Dilute Strong Acids Lesson

pH of Dilute Strong Acids

If you do the calculation to find -log (1.5 x 10^-10), you get an answer of 9.8. Always stop to ask yourself if an answer makes sense. Does it?  It doesn't!

The solution is made only of water and HBr, an acid. So, it is not a basic solution even though the math showed a pH greater than 7. We must be missing something. What we are missing here is the contribution of H⁺ ions from the auto-ionization of water. In an acidic or basic solution, there are two sources of H⁺ ions.

[H⁺]_Total = [H⁺]_{from solute}   +   [H⁺]_{from water}

In the calculations on acid or base solutions on the previous page, we neglected any contribution of either H⁺ or OH^- ions to the overall solution because those values are not significant. We assumed the equation to be:

 [H⁺]_Total = [H⁺]_{from solute} +  zero from water

 In the "Think About It" example above, the concentration of the solution is so small that the contribution of H⁺ ions from the auto-ionization of water is significant. So, we must add both sources of H⁺ to get the total and calculate the pH.

We only need to approach a pH (or pOH) calculation in the manner shown above when we are dealing with very dilute solutions of strong acids or bases.  Recall the list of strong acids and strong bases that you learned back in module 3 on chemical reactions.

Calculating pH or pOH of these strong acids and strong bases are normally straight forward. We use these equations as needed:

pH = -log[H+]

pOH = -log[OH^-]

pk_w = -log k_w

pH + pOH = pk_w

pH + pOH = 14

kw = [H₃O⁺][OH^-] = 1.0 x 10^-14

Remember as we calculate pH, that the terms strong or weak do not refer to concentration, rather they refer to complete dissociation in water.  

Weak Acids and Bases

While strong acids and bases ionize completely, weak ones do not. Below is an illustration of this. Notice that the strong acid forms many ions and the weak one does not.

 

Weak acids and bases therefore exist in equilibrium with the ions formed by their reaction with water.

The general ionization equation of a weak acid in water is:

HA   +   H₂O ⇌  H₃O⁺   +   A-

The equilibrium expression for this reaction is:

$k_{eq}=\frac{[H_3O^+][A^-]}{[H_2O][HA]}$

Moving [H₂O] to the side with k_eq gives us:

$k_a=\frac{[H_3O^+][A^-]}{[HA]}$

You must be able to write the k_a equation and equilibrium expression for any weak acid.  

The general ionization equation of a weak base in water is:

B   +   H₂O ⇌  BH⁺   +   OH^-

The equilibrium expression for this reaction is:$k_{eq}=\frac{[BH^+][OH^-]}{[H_2O][B]}$

Moving [H₂O] to the side with k_eq gives us:$k_{eq}=\frac{[BH^+][OH^-]}{[B]}$

You must be able to write the k_b equation and equilibrium expression for any weak base.

pk_a and pk_b

The value of k_a and k_b are usually small numbers and can be expressed in log form (like pH). pk_a = -log k_a

Here are some common acids with their k_a and pka values.

The larger the k_x, the stronger the acid or base.

The larger the pk_x, the weaker the acid or base.

There is an interesting relationship between the equilibrium constants for a conjugate acid-base pair. Let's write the equilibrium expressions for HCHO₂ and CHO₂^- to see this relationship.

Because of this relationship, the following is also true:

pk_a   +   pk_b   =   pk_w   = 14

pk_a   = -log k_a

pk_b   = -log k_b

 

Remember to work on the module practice problems as you complete each section of content.  

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