Hydrolysis calculations: salts of weak acids are bases

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Problem #1: What is the pH of a 0.100 M solution of sodium acetate? Kb = 5.65 x 10¯10. (I will use NaAc as shorthand)

Solution:

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

Ac¯ + H2O <==> HAc + OH¯

2) Here is the Kb expression for Ac¯:

  [HAc] [OH¯]
Kb = ----------------
  [Ac¯]

3) We can then substitute values into the Kb expression in the normal manner:

  (x) (x)
5.65 x 10¯10 = ----------------
  0.100 - x

4) Ignoring the minus x in the usual manner, we proceed to solve for the hydroxide ion concentration:

x = square root of [(5.65 x 10¯10) (0.100)]

x = 7.52 x 10¯6 M = [OH¯]

We then calculate the pH. Since this is a base calculation, we need to do the pOH first:

pOH = - log 7.52 x 10¯6 = 5.124

pH = 14 - 5.124 = 8.876


Problem #2: What is the pH of a 0.0500 M solution of KCN? Kb = 2.1 x 10¯5.

Solution:

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

CN¯ + H2O <==> HCN + OH¯

Here is the Kb expression for CN¯:

  [HCN] [OH¯]
Kb = ----------------
  [CN¯]

2) We can then substitute values into the Kb expression in the normal manner:

  (x) (x)
2.1 x 10¯5 = ----------------
  0.050 - x

3) Ignoring the minus x in the usual manner, we proceed to sove for the hydroxide ion concentration:

x = square root of [(2.1 x 10¯5) (0.050)]

x = 1.025 x 10¯3 M = [OH¯]

4) We then calculate the pH. Since this is a base calculation, we need to do the pOH first:

pOH = - log 1.025 x 10¯3 = 2.99

pH = 14 - 2.99 = 11.01


I certainly hope that you see that the two calculations are exactly alike, except for the numbers. In fact, when I wrote this tutorial in August 2002, I just copied and pasted the first example and changed the words and numbers. One final point to make: textbooks will often say "find the pH of the cyanide ion, CN¯." What they really mean is a complete chemical substance of which only the CN¯ is chemically relevant to the problem. Please keep in mind that the solution contains a cation (K+ in the example above, Na+ in the first example.


Problem #3: Find the pH of a 0.30 M solution of sodium benzoate, C6H5COONa. The Kb for C6H5COO¯ (benzoate ion) is 1.55 x 10¯10.

Problem #4: Find the pH of a 0.20 M solution of sodium propionate (C2H5COONa), where the Ka of propionic acid = 1.34 x 10¯5. (Use the equation Kw = KaKb to go from the Ka of the acid to the Kb of its conjugate base.)

Go to the answers for the above two problems.


Salts of weak bases are acids. Do them next.

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