what is the OH- concentration of 0.08 M solution of CH3COONa
Call CH3COONa = NaAc and HAc is acetic acid; i.e., CHCOOH. It's the Ac^- that is hydrolyzed.
.......Ac^- + HOH ==> HAc + OH^-
I....0.08..............0.....0
C......-x..............x.....x
E....0.08-x............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.08-x). Solve for x = OH^-
To determine the concentration of hydroxide ions (OH-) in a solution of a given compound, you need to consider the dissociation of the compound in water. In the case of CH3COONa, it dissociates into acetate ions (CH3COO-) and sodium ions (Na+).
Since acetate ions are a weak base, they hardly react with water to produce hydroxide ions. Therefore, we can assume that the contribution of hydroxide ions from the dissociation of CH3COONa is negligible. Hence, the OH- concentration in a 0.08 M solution of CH3COONa is approximately 0 M.
If you want to determine the OH- concentration in a solution, it would be more relevant to consider compounds that are strong bases, such as sodium hydroxide (NaOH), potassium hydroxide (KOH), etc.
To find the OH- concentration of a solution of CH3COONa, we need to consider the dissociation reaction of CH3COONa in water:
CH3COONa --> CH3COO- + Na+
From this reaction, we can see that CH3COONa dissociates into CH3COO- and Na+ ions. Since Na+ ions do not contribute to the OH- concentration, we can ignore them in this calculation.
Therefore, the OH- concentration will be equal to the concentration of the CH3COO- ions formed from the dissociation. According to stoichiometry, the concentration of CH3COO- will be equal to the concentration of CH3COONa.
So, the OH- concentration of a 0.08 M solution of CH3COONa is also 0.08 M.