Answer and Explanation:
To determine the pH of a solution containing 0.18 M aqueous formate ion (HCO2-), we can use the dissociation constant (Ka) of formic acid (HCO2H), which is given as 1.8x10^(-4).
1. First, we need to recognize that formate ion (HCO2-) is the conjugate base of formic acid (HCO2H). When formic acid donates a proton (H+), it forms formate ion.
2. The dissociation reaction can be represented as follows:
HCO2H ⇌ H+ + HCO2-
3. Using the given Ka value, we can set up an equilibrium expression:
Ka = [H+][HCO2-] / [HCO2H]
4. Since the concentration of formate ion (HCO2-) is given as 0.18 M, we can assume that the concentration of formic acid (HCO2H) is also 0.18 M (assuming equal initial concentrations).
5. Substituting the known values into the equilibrium expression, we have:
1.8x10^(-4) = [H+][0.18] / [0.18]
6. Simplifying the equation, we find:
[H+] = 1.8x10^(-4)
7. Finally, we can calculate the pH using the formula:
pH = -log[H+]
Plugging in the value of [H+], we get:
pH = -log(1.8x10^(-4))
Evaluating this expression, we find:
pH ≈ 3.74
Therefore, the pH of the 0.18 M aqueous formate ion solution is approximately 3.74.
Additionally, since formic acid is a weak acid, it does not fully ionize in water. To determine the percent ionization of this reaction, we can use the equation:
% ionization = ([H+]/[HCO2H]) x 100
Using the known concentration of formate ion ([H+]), which is 1.8x10^(-4) M, and the assumed concentration of formic acid ([HCO2H]), which is also 0.18 M, we can calculate the percent ionization:
% ionization = (1.8x10^(-4) / 0.18) x 100
% ionization ≈ 0.1%
Therefore, the percent ionization for this reaction is approximately 0.1%.