Formic acid dissociates in water according to the equation: HCOOH(aq) → H+(aq) + HCOO-(aq) Given the following equilibrium concentrations, calculate the equilibrium constant: [HCOO-] = 0.550 M, [H+] = 0.060 M, [HCOOH] = 0.060 M.
The Correct Answer and Explanation is:
To calculate the equilibrium constant KaK_aKa for formic acid dissociation, we can use the equation for the dissociation of formic acid:HCOOH (aq)⇌H+(aq)+HCOO−(aq)\text{HCOOH (aq)} \rightleftharpoons \text{H}^+ (aq) + \text{HCOO}^- (aq)HCOOH (aq)⇌H+(aq)+HCOO−(aq)
The equilibrium expression for this reaction is:Ka=[H+][HCOO−][HCOOH]K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}Ka=[HCOOH][H+][HCOO−]
Step 1: Plug the given concentrations into the equation
From the problem, the following equilibrium concentrations are given:
- [HCOO−]=0.550 M[ \text{HCOO}^- ] = 0.550 \, \text{M}[HCOO−]=0.550M
- [H+]=0.060 M[ \text{H}^+ ] = 0.060 \, \text{M}[H+]=0.060M
- [HCOOH]=0.060 M[ \text{HCOOH} ] = 0.060 \, \text{M}[HCOOH]=0.060M
Now, substitute these values into the equilibrium constant equation:Ka=(0.060)(0.550)0.060K_a = \frac{(0.060)(0.550)}{0.060}Ka=0.060(0.060)(0.550)
Step 2: Simplify the equation
Ka=0.0330.060K_a = \frac{0.033}{0.060}Ka=0.0600.033
Step 3: Perform the division
Ka=0.55K_a = 0.55Ka=0.55
Conclusion
The equilibrium constant KaK_aKa for the dissociation of formic acid is 0.550.550.55.
Explanation
Formic acid is a weak acid, meaning it does not fully dissociate in water. The equilibrium constant KaK_aKa provides a measure of the extent of dissociation. The expression for KaK_aKa takes into account the concentrations of the products (H+ and HCOO-) and the undissociated reactant (HCOOH) at equilibrium. The calculated value of Ka=0.55K_a = 0.55Ka=0.55 suggests that formic acid is a weak acid, since it does not dissociate completely, and its KaK_aKa value is moderate, indicating partial dissociation. The lower the value of KaK_aKa, the weaker the acid, and the higher the value, the stronger the acid.
