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Just as pH is the negative logarithm of [H3O+], pKa is the negative logarithm of Ka, pKa=−logKa The...

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Chemistry

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.

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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.

mass of NH4Cl =

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PLEASE HELP ME ANSWER PART B

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