Question 16.10: A student prepared a 0.10 M solution of formic acid (HCOOH) ...
A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 °C to be 2.38. Calculate K_{a} for formic acid at this temperature.
Analyze We are given the molar concentration of an aqueous solution of weak acid and the pH of the solution, and we are asked to determine the value of K_{a} for the acid.
Plan Although we are dealing specifically with the ionization of a weak acid, this problem is similar to the equilibrium problems in Chapter 15. We can solve this problem using the method outlined in Sample Exercise 15.8, starting with the chemical reaction and a tabulation of initial and equilibrium concentrations.
Solve
The first step in solving any equilibrium problem is to write the equation for the equilibrium reaction. The ionization of formic acid can be written as:
The equilibrium expression is:
K_{a} =\frac{[H^{+}][HCOO^{-}]}{[HCOOH]}From the measured pH, we can calculate [H^{+}]:
pH = -log [H^{+}] = 2.38
log [H^{+}] = -2.38
[H^{+}]= 10^{-2.38} = 4.2 × 10^{-3} M
To determine the concentrations of the species involved in the equilibrium, we consider that the solution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acid into H^{+} and HCOO^{-}. For each HCOOH molecule that ionizes, one H^{+} ion and one HCOO^{-} ion are produced in solution. Because the pH measurement indicates that [H^{+}] = 4.2 × 10^{-3} M at equilibrium, we can construct the following table:
Notice that we have neglected the very small concentration of H^{+} (aq) due to H_{2}O autoionization. Notice also that the amount of HCOOH that ionizes is very small compared with the initial concentration of the acid. To the number of significant figures we are using, the subtraction yields 0.10 M:
(0.10 – 4.2 × 10^{-3}) M \simeq 0.10 MWe can now insert the equilibrium concentrations into the expression for K_{a}:
K_{a} =\frac{(4.2 × 10^{-3})(4.2 × 10^{-3})}{0.10}= 1.8 × 10^{-4}Check The magnitude of our answer is reasonable because K_{a} for a weak acid is usually between 10^{-2} and 10^{-10}.