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) ResetHelp 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