Salts of Weak Acids are Bases: Hydrolysis Calculations

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

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

Ac¯ + H₂O <==> HAc + OH¯

Here is the K_b expression for Ac¯:

We can then substitute values into the K_b expression in the normal manner:

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

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

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

CN¯ + H₂O <==> HCN + OH¯

Here is the K_b expression for CN¯:

We can then substitute values into the K_b expression in the normal manner:

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¯]

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.

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Practice Problems

1) Find the pH of a 0.30 M solution of sodium benzoate, C₆H₅COONa. The K_b for C₆H₅COO¯ (benzoate ion) is 1.55 x 10¯¹⁰.

2) Find the pH of a 0.20 M solution of sodium propionate (C₂H₅COONa), where the K_a of propionic acid = 1.34 x 10¯⁵. (Hint: K_w = K_aK_b)

Problems 3 & 4 are salt of weak base problems. Do them next.

However, if you must: go to the Practice Problem answers

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