Using the concentration of CH3COOH (0.8326M) and the equilibrium concentration of H3O+ (OJ = 31.6x10^-6 and milk = 3.16x10^-8), complete the reaction table for vinegar. Then calculate the acidity constant. Please show all work in solving the problem.
Reaction Table:
CH3COOH(aq) + H2O(l) <---> CH3COO-(aq) + H3O+(aq)
Initial
Change
Equilibrium
You need to clarify the post. What's the H3O^+ in OJ and milk to do with Ka for CH3COOH?
To complete the reaction table, we need to consider the initial concentration of CH3COOH and the equilibrium concentrations of H3O+ and CH3COO-.
Given:
- Concentration of CH3COOH: 0.8326 M
- Equilibrium concentration of H3O+: [H3O+] = 31.6 × 10^(-6) M (OJ)
- Equilibrium concentration of CH3COO-: [CH3COO-] = 3.16 × 10^(-8) M (milk)
Reaction Table:
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)
Initial: 0.8326 M 0 M 0 M 0 M
Change: -x -x +x +x
Equilibrium: 0.8326-x -x x x
Where "x" is the change in concentration in M.
To calculate the equilibrium concentration of H3O+ (x), we can use the given concentration of H3O+:
[H3O+] = 31.6 × 10^(-6) M
Therefore, x = 31.6 × 10^(-6) M
Let's plug in this value into the equilibrium row of the reaction table:
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)
Initial: 0.8326 M 0 M 0 M 0 M
Change: -x -x +x +x
Equilibrium: 0.8326- (31.6 × 10^(-6)) - (31.6 × 10^(-6)) 31.6 × 10^(-6) 31.6 × 10^(-6)
To calculate the acidity constant, also known as the Ka, we need to consider the expression:
Ka = ([CH3COO-][H3O+]) / [CH3COOH]
Using the equilibrium concentrations, plug in these values into the expression:
Ka = ((31.6 × 10^(-6))(31.6 × 10^(-6))) / (0.8326- (31.6 × 10^(-6)))
Simplify and calculate to find the value of Ka.