We´re asked to calculate the pH of a 0.1 M Ca(CH3COO)2 solution at 25 °C.
Calcium acetate is an ionic salt soluble in water.
In it, Ca(CH3COO)2 dissociates as follows:
Ca(CH3COO)2⇌ Ca2+(aq) + 2 CH3COO-(aq)
The expression of the equilibrium constant K is:
In solution, the acetate ion can react with water by accepting a proton to form acetic acid:
CH3CO2-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH-(aq)
Since acetate ion accepts a proton, it is a base according to the Bronsted-Lowry theory.
For a weak base, the equilibrium constant becomes the basic dissociation constant (Kb).
The expression for Kb is:
For the calcium acetate ion reaction, Kb adopts the form of:
Note that each concentration is raised by the stoichiometric coefficient: [CH3COOH], [HO-], and [CH3CO2-] are raised to 1.
Calcium acetate (Ca(CH3COO)2) is commonly used as a food additive, especially in candy. What is the pH of a 0.1 M Ca(CH3COO)2 solution at 25 °C
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Humphrey's class at TEXAS.