Just as ph is the negative logarithm of [h3o+], pka is the negative logarithm of ka, pka=−logka the henderson-hasselbalch equation is used to calculate the ph of buffer solutions: ph=pka+log[base][acid] notice that the ph of a buffer has a value close to the pka of the acid, differing only by the logarithm of the concentration ratio [base]/[acid]. the henderson-hasselbalch equation in terms of poh and pkb is similar. poh=pkb+log[acid][base] part apart complete acetic acid has a ka of 1.8×10−5. three acetic acid/acetate buffer solutions, a, b, and c, were made using varying concentrations: [acetic acid] ten times greater than [acetate], [acetate] ten times greater than [acetic acid], and [acetate]=[acetic acid]. match each buffer to the expected ph. drag the appropriate items to their respective bins. view available hint(s) ph = 3.74 [acetic acid] ten times greater than [acetate] ph = 4.74 [acetate] = [acetic acid] ph = 5.74 [acetate] ten times greater than [acetic acid] previous answers correct part b how many grams of dry nh4cl need to be added to 2.20 l of a 0.400 m solution of ammonia, nh3, to prepare a buffer solution that has a ph of 8.89? kb for ammonia is 1.8×10−5