Substituting the equilibrium concentrations into the equilibrium (Kb) expression
ID: 501734 • Letter: S
Question
Substituting the equilibrium concentrations into the equilibrium (Kb) expression and solving for x gives the concentration of OH-: ____M
What is the pOH of this solution?
What is the pH of at the equivalence point (find pH from pOH in previous step):
CH3COO- acts as a weak base (ag) OHaq) The equilibrium-constant (Ka) expression for this reaction is: CH3COOH) DOH Kb CH COO The Kb for CH3COO- can be calculated from the Ka of acetic acid. The concentration of CH3COO- can be plugged into an ICE table to solve for the concentration of OH OH CH3COO H2O CH3COOH (ag) (aq) Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M)Explanation / Answer
pOH is define as negative logarithm to the base 10 of concentration of hydroxyl ion.
i.e., pOH = - log([OH-])
Therefore, by substituting the value of concentration of OH calculated into above equation gives you pOH of the solution.
For pH, subtract pOH from 14. i.e., pH = 14 - pOH