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 1.8×10−5. 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 1.8×10−5.

Express your answer with the appropriate units.

Answer 1

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