### Introduction to hydrolysis calculations

Two types of hydrolysis calculations are usually taught in textbooks:

1) the salt of a weak acid and a strong base (example: sodium acetate, CH3COONa)
2) the salt of a weak base and a strong acid (example: ammonium chloride, NH4Cl)

Other types of hydrolysis are ignored or lightly treated. The other types are salts of a strong acid and a strong base (example: sodium chloride, NaCl) and the salt of a weak acid and a weak base (example: ammonium acetate, NH4CH3COO).

Hydrolysis calculations are best understood in terms of acid base calculations, because, well, that's what they are.

I. Salts of Weak Acids are Bases

Sodium acetate (formula = NaC2H3O2, when written in the inorganic style and CH3COONa when written in the organic style) is the prototypical salt of a weak acid. I will use the shorthand of NaAc, where Ac¯ stands for the acetate anion.

Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:

Ac¯ + H2O ⇌ HAc + OH¯

Notice that I have written it as an equilibrium AND that the Ac¯ is acting as a base (it accepted a proton from the water). This means I can write a Kb expression for Ac¯:

 [HAc] [OH¯] Kb = ––––––––––– [Ac¯]

Now we come to a VERY, VERY IMPORTANT point: we already know the value for this Kb! How? Well, we know that Kw = KaKb AND we already know the Ka for acetic acid (HAc) is 1.77 x 10¯5 and Kw = 1.00 x 10¯14.

So, without too much trouble, we calculate and find the Kb for the acetate ion (Ac¯) to be 5.65 x 10¯10

We can do this for any salt of a weak acid for which we know the Ka. Also, just below, you will see that we can do the same calculation for salts of weak bases. In that situation, we can look up the Kb of the weak base and, from that value and Kw, calculate the Ka of the salt of the weak base (since the salt is an acid).

II. Salts of Weak Bases are Acids

Ammonium chloride, NH4Cl, is the prototypical salt of a weak base. There is no shorthand for it, I will just use the formula.

Here is the chemical reaction (net ionic) for the hydrolysis of NH4Cl:

NH4+ + H2O ⇌ NH3 + H3O+

Notice that I have written it as an equilibrium AND that the NH4+ is acting as a base (it donated a proton to the water). This means I can write a Ka expression for NH4+:

 [NH3] [H3O+] Ka = –––––––––––– [NH4+]

Since we already know the Kb for NH3 (1.77 x 10¯5) and the value for Kw (1.00 x 10¯14), we can calculate the Ka for the ammonium ion to be 5.65 x 10¯10.

By the way, there isn't any chemical reason why the approximate value for Ka of acetic acid equals the approximate value for the Kb of ammonia. That's just the way it is.

Also, in the "What are Salts?" tutorial, you learned that the salt of a weak acid makes a basic solution. Well, that's because the salt is a base. Look once again at the reaction of the acetate ion with water. The acetate ion is accepting a proton and that's what bases do. In a similar manner, salts of weak bases form acidic solutions. That's because the salt of a weak base is an acid!! The ammonium ion donates a proton to the water and that's what acids do.

Consequently, we can use the same calculation techniques taught for use with weak acids and weak bases. Why? Becase the salt of a weak acid is a base and the salt of a weak base is an acid.