In what proportion does acetic acid dissolve in water
Answers
Acetic acid, formula CH3COOH, is a simple organic acid, with a pKa of 4.76. Pure water, H2O has a nominal pKa of 14 at 25*C. This means that a significant fraction of acetic acid deprotonates to the acetate ion (CH3COO-) in H2O, thereby generating an equal fraction of H3O+ (Hydronium) and CH3COO- (Acetate) ions. This, BTW, is what makes it a “Bronstead-Lowrey acid.”
Thus, CH3COOH + H2O ← → CH3COO- + H3O+ pKa = 4.76
By this definition the Acetate ion is the “conjugate base” of acetic acid.
Remember the definitions of pKa and Ka:
The definition of Ka is: [H+][B-] / [HB], where B- is the conjugate base of the acid, HB. The pKa value is defined from Ka, and can be calculated from the Ka value from the equation pKa = -Log10(Ka)
Here the equation is [H3O+] x [CH3COO-]/[CH3COOH] = Ka (dissociation constant)