Just as pH is the negative logarithm of [H3O+], pKa is thenegative logarithm of Ka, pKa=−logKa The Henderson-Hasselbalch equation is used to calculate the pH ofbuffer solutions:pH=pKa+log[base][acid] Notice that the pH of a buffer has a value close to thepKa of the acid, differing only by the logarithm of theconcentration ratio [base]/[acid]. The Henderson-Hasselbalchequation in terms of pOH and pKb is similar.pOH=pKb+log[acid][base] | Part A Acetic acid has a Ka of 1.8×10−5. Three aceticacid/acetate buffer solutions, A, B, and C, were made using varyingconcentrations:[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. SubmitHintsMyAnswersGive UpReviewPart Correct Part B How many grams of dry NH4Cl need to be added to 1.50 L of a0.400 M solution of ammonia, NH3, to prepare a buffersolution that has a pH of 8.57? Kb for ammonia is1.8×10−5. Express your answer with the appropriate units. SubmitHintsMyAnswersGive UpReviewPart Incorrect; Try Again; 5 attempts remaining PLEASE HELP ME ANSWER PART B |