Question

write then form of the acid that will exist in a pH 5.5 solution?

More basic solutions than the pKa will leave acid in its

Ch3COOH pKa 4.76

Ch3CH2NH3+ pKa 11.0

Ch3CH2OH pKa 15.9

please do each of these 3 and explain, thanks in advance

Answer 1

For a general reaction:

The Henderson-Hasselbalch equation is:

pKa = pH + log [HA] / [A^-]

1. When the pH = pKa then [HA] = [A^-] i.e. equal amount of protonated form(HA) and deprotonated form(A^-) exist.
2. When pH < pKa, then log [HA] / [A^-] > 0 or [HA] > [A^-] i.e the protonated form(HA) will be in excess.
3. When pH > pKa then log [HA] / [A^-] < 0 or [HA] < [A^-] i.e. the deprotonated form(A^-) will be in excess

So,pKa tells us if a given molecule is going to either give a proton ,or remove a proton at a certain pH.

• For acetic acid :    Protonated form deprotonated form pk_a = 4.76 pH of solution = 5.5 Since, pH > pk_a , so deprotonated form will be in excess and acetic acid will exist as acetate ion.
• For ethylamine:   CH₃-CH₂-NH₃⁺   CH₃-CH₂-NH₂  + H⁺   Protonated form deprotonated form pK_a = 11.0 pH of solution = 5.5   Since, pH < pk_a , so protonated form(CH₃-CH₂-NH₃⁺₎will be in excess.
• For Ethanol:   Protonated form deprotonated form pK_a = 15.9 pH of solution = 5.5 Since, pH < pk_a , so protonated form(CH₃CH₂OH) will be in excess.