Consider the dissociation of formic acid in water HCHO2(aq) + H2O(l) ? H3O+ (aq) + CHO2 – (aq) Ka = 1.8×10–4 The equilibrium constant for an acid’s dissociation is referred to as the acid dissociation constant and is represented as Ka instead of just K. The subscript simply denotes that this particular equilibrium constant is for the dissociation of an acid in water. At room temperature, if the equilibrium concentration of [HCHO2] = 0.0987 M, what are the equilibrium concentrations of H3O+ and CHO2 –? What is the pH of the solution?
The Correct Answer and Explanation is :
To solve for the equilibrium concentrations of ( \text{H}_3\text{O}^+ ) and ( \text{CHO}_2^- ), and determine the pH of the solution, we need to use the acid dissociation constant (( K_a )) and the given concentration of formic acid, ( \text{HCHO}_2 ).
Step 1: Write the Dissociation Reaction
The dissociation reaction of formic acid in water is:
[
\text{HCHO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{CHO}_2^-(aq)
]
The acid dissociation constant (( K_a )) for this reaction is:
[
K_a = \frac{[\text{H}_3\text{O}^+][\text{CHO}_2^-]}{[\text{HCHO}_2]}
]
Given:
- ( K_a = 1.8 \times 10^{-4} )
- ( [\text{HCHO}_2] = 0.0987 \, \text{M} )
We need to find the equilibrium concentrations of ( [\text{H}_3\text{O}^+] ) and ( [\text{CHO}_2^-] ).
Step 2: Set Up the ICE Table
We set up an ICE (Initial, Change, Equilibrium) table to express the concentrations of each species.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HCHO2 | 0.0987 | -x | 0.0987 – x |
| H3O+ | 0 | +x | x |
| CHO2– | 0 | +x | x |
Step 3: Apply the ( K_a ) Expression
Now, substitute the equilibrium concentrations into the ( K_a ) expression:
[
K_a = \frac{[\text{H}_3\text{O}^+][\text{CHO}_2^-]}{[\text{HCHO}_2]}
]
[
1.8 \times 10^{-4} = \frac{x \cdot x}{0.0987 – x}
]
Since ( K_a ) is small, we can assume that ( x ) is small relative to 0.0987, so we can approximate ( 0.0987 – x \approx 0.0987 ). This simplifies the equation to:
[
1.8 \times 10^{-4} = \frac{x^2}{0.0987}
]
Solving for ( x ):
[
x^2 = (1.8 \times 10^{-4}) \times 0.0987
]
[
x^2 = 1.78 \times 10^{-5}
]
[
x = \sqrt{1.78 \times 10^{-5}} = 0.00422 \, \text{M}
]
Thus, the equilibrium concentrations are:
- ( [\text{H}_3\text{O}^+] = 0.00422 \, \text{M} )
- ( [\text{CHO}_2^-] = 0.00422 \, \text{M} )
Step 4: Calculate the pH
To calculate the pH, we use the formula:
[
\text{pH} = -\log[\text{H}_3\text{O}^+]
]
Substitute the value of ( [\text{H}_3\text{O}^+] ):
[
\text{pH} = -\log(0.00422)
]
[
\text{pH} = 2.38
]
Final Answer:
- The equilibrium concentrations are:
- ( [\text{H}_3\text{O}^+] = 0.00422 \, \text{M} )
- ( [\text{CHO}_2^-] = 0.00422 \, \text{M} )
- The pH of the solution is 2.38.
Explanation:
The dissociation of formic acid produces both ( \text{H}_3\text{O}^+ ) (hydronium ion) and ( \text{CHO}_2^- ) (formate ion). By using the given ( K_a ) value and assuming that the change in concentration of formic acid is small (due to the weak acid nature of formic acid), we were able to solve for the equilibrium concentrations of both products. The pH is determined from the concentration of ( \text{H}_3\text{O}^+ ) since pH is directly related to the concentration of hydronium ions.