### Calculate the Ka of a weak acid given the pH and other concentration data (not molarity)

Example #1: A student prepared a solution of salicylic acid (a monoprotic weak acid, MW = 138.123 g mol¯1) and measured the pH of the solution to be 2.430. Then, she evaporated 100.0 mL of the solution and collected 0.220 g of dry salicylic acid. What is the Ka of salicylic acid?

Solution:

1) Write the dissociation equation for salicylic acid, HSal:

HSal ⇌ H+ + Sal¯

2) Write the Ka expression for HSal:

Ka = ([H+] [Sal¯]) / [HSal]

3) Use the pH to determine [H+] and [Sal¯]:

[H+] = 10¯pH = 10¯2.20 = 3.71535 x 10¯3 M (some guard digits)

4) Determine molarity of HSal:

0.220 g / 138.123 g mol¯1 = 1.59278 x 10¯3 mol

1.59278 x 10¯3 mol / 0.100 L = 1.59278 x 10¯2 M

5) Calculate Ka:
Ka = [(3.71535 x 10¯3) (3.71535 x 10¯3)] / 1.59278 x 10¯2

Ka = 8.67 x 10¯4

Comment: I did some Internet searching on the Ka of salicylic acid and found some striking differences in reported values. For example, I found "Quantitative Chemical Analysis" (by Daniel C. Harris) on Google Books. On page 183, he reports the pKa of salicylic acid to be 2.97; from that we see the Ka to be 1.07 x 10¯3, almost a 20% difference.

This is not to say any one number is wrong, just that various sources give different values for the Ka of salicylic acid. I just haven't fond an authoritative source for the value. Yet!

Example #2: An aqueuous solution containing 8.40 g/L of Cyanoacetic acid (abbreviated HCya), CH2CNCOOH has a pH of 1.77. What is the value of the Ka?

Solution:

1) Write the dissociation equation for HCya:

HCya ⇌ H+ + Cya¯

2) Write the Ka expression for HCya:

Ka = ([H+] [Cya¯]) / [HCya]

3) Use the pH to determine [H+] and [Cya¯]:

[H+] = 10¯pH = 10¯1.77 = 1.69824 x 10¯2 M (I kept some guard digits)

4) Determine molarity of HCya:

8.40 g / 85.0618 g mol¯1 = 9.875 x 10¯2 mol

1.59278 x 10¯3 mol / 1.00 L = 9.875 x 10¯2 M

5) Calculate Ka:
Ka = [(1.69824 x 10¯2) (1.69824 x 10¯2] / 9.875 x 10¯2

Ka = 2.92 x 10¯3