Question

a) How many grams of solid potassium cyanide should be added to 1.00 L of a 0.158 M hydrocyanic acid solution to prepare a buffer with a pH of 8.688 ?

b) How many grams of solid ammonium chloride should be added to 2.00 L of a 0.171 M ammonia solution to prepare a buffer with a pH of 9.940 ?  

c)

Design a buffer that has a pH of 7.98 using one of the weak acid/conjugate base systems shown below.

Weak Acid	Conjugate Base	Ka	pKa
HC2O4-	C2O42-	6.4 × 10-5	4.19
H2PO4-	HPO42-	6.2 × 10-8	7.21
HCO3-	CO32-	4.8 × 10-11	10.32

How many grams of the potassium salt of the weak acid must be combined with how many grams of the potassium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams potassium salt of weak acid = __________

grams potassium salt of conjugate base = ___________

Answer 1

Ans 1) We know

pH=pKa+log[KCN/HCN]

Here pH=8.68

pKa for cyanide ion=9.31

8.68=9.31+log[KCN/HCN]

[KCN/HCN] =0.234

Now moles of HCN = 1L x 0.158M

                            =0.158mol

moles of KCN =0.158x0.234

                     = 0.0369

Mass of KCN = moles of KCN x molar mass of KCN

                     = 0.0369x65.12

                       =2.4g

Ans 2) We know

pH =pka+log[NH₃/NH₄Cl]

Here pH = 9.9

pka for ammonium ion= 9.3

9.9=9.3+log[NH₃/NH₄Cl]

[NH₃/NH₄Cl]=3.98

[NH₃]=0.171

[NH₄Cl]=0.042

moles of NH₄Cl= 2 x 0.042

                        = 0.0859

Mass of NH₄Cl = moles of NH₄Cl x molar mass

                      = 0.0859 x 53.5

                       = 4.59g

Ans 3 ) let us take HCO₃^- as weak acid and conjugate base is CO₃^2- .

Given pH= 7.98

pka=10.32

Now

pH=pka+ log [CO₃^2- / HCO₃^-]

7.98=10.32+log[CO₃^2-/HCO₃^-]

-2.34 = log[CO₃^2-/HCO₃^-]

Now given volume = 1L

[HCO₃^-]= 1 x 1 =1M

10^-2.34 = [CO₃^2-]/1

[CO₃^2-]=0.00457M

moles of CO₃^2- =0.00457 x1

                       =0.00457

mass of CO₃^2- = moles x molar mass

                        = 60x0.00457

                        =0.274g

moles of [HCO₃^-]=1

mass of HCO₃^- = 1 x 61

                        = 61g

Therefore mass of weak acid= 61g

mass of conjugate base = 0.274g