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

Return to Acid Base Menu

Return to a listing of many types of acid base problems and their solutions

**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 K_{a} of salicylic acid?

**Solution:**

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

HSal ⇌ H^{+} + Sal¯

2) Write the K_{a} expression for HSal:

K_{a} = ([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 K_{a}:
K_{a} = [(3.71535 x 10¯^{3}) (3.71535 x 10¯^{3})] / 1.59278 x 10¯^{2}
K_{a} = 8.67 x 10¯^{4}

Comment: I did some Internet searching on the K_{a} 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 pK_{a} of salicylic acid to be 2.97; from that we see the K_{a} 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 K_{a} 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), CH_{2}CNCOOH has a pH of 1.77. What is the value of the K_{a}?

**Solution:**

1) Write the dissociation equation for HCya:

HCya ⇌ H^{+} + Cya¯

2) Write the K_{a} expression for HCya:

K_{a} = ([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 K_{a}:
K_{a} = [(1.69824 x 10¯^{2}) (1.69824 x 10¯^{2}] / 9.875 x 10¯^{2}
K_{a} = 2.92 x 10¯^{3}

Return to a listing of many types of acid base problems and their solutions

Return to Acid Base Menu