question 3 Just as pH is the negative logarithm of [H3O+], p K a is the negative
ID: 694179 • Letter: Q
Question
question 3
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 A
Acetic acid has a Ka of 1.8×105. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations:
a [acetic acid] ten times greater than [acetate],
b [acetate] ten times greater than [acetic acid], and
c [acetate]=[acetic acid].
Match each buffer to the expected pH.
Drag the appropriate items to their respective bins.
pH = 3.74
pH = 4.74
pH = 5.74
Part B
How many grams of dry NH4Cl need to be added to 1.70 L of a 0.200 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.66? Kb for ammonia is 1.8×105.
Express your answer with the appropriate units.
Explanation / Answer
part A
pH=pKa+log[base][acid]
pka of aceticacid = -logka
= -log(1.8*10^-5)
= 4.74
a. pH = 4.74+log(1/10)
pH = 3.74
b. pH = 4.74+log(10/1)
pH = 5.74
c. pH = 4.74+log(1/1)
pH = pka = 4.74
part B
pOH=pKb+log[acid][base]
[acid] = NH4+ = x
[base] = NH3 = 1.7*0.2 = 0.34 mol
pkb = -logkb
= -log(1.8*10^-5)
= 4.74
14-8.66 = 4.74 +log(x/0.34)
x = 1.3536
no of mol of NH4Cl = x = 1.3536 mol
mass of NH4Cl = 53.5*1.3536
= 72.42 grams
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