Just as is the negative logarithm of , is the negative logarithm of , The Hender

Question

Just as is the negative logarithm of , is the negative logarithm of ,

The Henderson-Hasselbalch equation is used to calculate the of buffer solutions:

Notice that the of a buffer has a value close to the of the acid, differing only by the logarithm of the concentration ratio . Part A Acetic acid has a of . Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations:

1. ten times greater than ,
2. ten times greater than , and
3. .

Match each buffer to the expected .

Drag each item to the appropriate bin. SubmitHintsMy AnswersGive UpReview Part Correct Part B How many grams of dry need to be added to 2.10 of a 0.500 solution of ammonia, , to prepare a buffer solution that has a of 8.76? for ammonia is . Express your answer numerically in grams to three significant figure

Explanation / Answer

pOH = 14 - 8.76 = 5.24

pKb = 4.74

5.32 - 4.74 = 0.5

10^0.5 = 3.16 = [NH4+]/ 0.500

[NH4+] = 0.500 x 3.16=1.58 M

moles NH4+ = 1.58 x 2.10 L=3.318

moles NH4Cl = 3.318 x 53.492 g/mol=177.486 g

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