Multiple Choice
What is the pH of a solution that is 0.060 M in potassium propionate and 0.085 M in propionic acid, given that the Ka of propionic acid is 1.3 x 10^-5?
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.750 M solution of KH2PO4, considering it is a weak acid?
A
7.0
B
2.3
C
4.7
D
9.1
Verified step by step guidance1
Identify the relevant chemical species: KH2PO4 is a weak acid, and it can donate a proton to form H2PO4^-. The dissociation can be represented as: KH2PO4 ⇌ K^+ + H2PO4^-.
Recognize that H2PO4^- can further dissociate in water to form H^+ and HPO4^2-. The equilibrium expression for this dissociation is: H2PO4^- ⇌ H^+ + HPO4^2-.
Use the acid dissociation constant (Ka) for H2PO4^- to set up the equilibrium expression: Ka = [H^+][HPO4^2-] / [H2PO4^-].
Assume that the initial concentration of H2PO4^- is approximately equal to the initial concentration of KH2PO4, which is 0.750 M, since the dissociation is weak.
Set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of all species at equilibrium, and solve for [H^+]. Use the equation pH = -log[H^+] to find the pH of the solution.
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