Question

A) Design a buffer that has a pH of 9.93 using one of the weak base/conjugate acid systems shown below.

Weak Base	Kb	Conjugate Acid	Ka	pKa
CH3NH2	4.2×10-4	CH3NH3+	2.4×10-11	10.62
C6H15O3N	5.9×10-7	C6H15O3NH+	1.7×10-8	7.77
C5H5N	1.5×10-9	C5H5NH+	6.7×10-6	5.17

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

grams chloride salt of conjugate acid = _____

grams weak base = _____

B) Design a buffer that has a pH of 9.63 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 sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams sodium salt of weak acid =_____

grams sodium salt of conjugate base = _____

Answer 1

Ans. #A. The buffering capacity of a weak base – conjugate acid buffer is maximum when pKb of the weak base is closes to the specified pOH.

            pOH of buffer = 14.00 – pOH = 14.00 – 9.93 = 4.07

# pKb of CH₃NH₂ = 14.00 – pKa = 14.00 – 10.62 = 3.38

pKb of C₆H₁₅O₃N = 14.00 – 7.77 = 6.23

pKb of C₅H₅N = 14.00 – 5.17 = 8.83

Since pOH of CH₃NH_2­ is closest to the pOH of buffer, it is the most suitable option.

# Using Henderson-Hasselbalch equation for base-

            pOH = pKb + log ([BH⁺] / [B])

            Where,

                        B = Base = Brucine

                        BH⁺ = Conjugate acid

Putting the values in above equation-

            Or, 4.07 = 3.38 + log ([BH⁺] / [B])                                                 

            Or, 4.07 – 3.38 = log ([BH⁺] / [B])                                                                                      

            Or, [BH⁺] / [B] = antilog (0.69)                                                                                          

Or, [BH⁺] / [B] = 4.898                    

Or, [BH⁺] = 4.898 [B]

# Given, [B] = 1.0 M

So,       [BH⁺] = 4.898 x 1.0 M = 4.898 M  

# Given, Total volume of buffer solution (culture medium) = 1.00 L

Now,

Mass of [CH₃NH₂] required = (Molarity of CH₃NH₂ x vol. of solution in L) x Molar mass

                                                            = (1.0 M x 1.00 L) x (31.05744 g/mol)

                                                            = 31.057 g

Mass of [NaCH₃NH₃] required = (Molarity of Na-CH₃NH₃ x vol. of soln. in L) x Molar mass

                                                            = (4.898 M x 1.0 L) x (55.055148 g/mol)

                                                            = 269.660 g

#B. The buffering capacity of a weak acid – conjugate base buffer is maximum when pKa of the weak acid is closes to the specified pH.

Since pKa of HCO₃^- is closest to the specified pH, it is the weak acid of choice.

Now, using Henderson- Hasselbalch equation for weak acid

            pH = pKa + log ([A^-] / [AH])                    - equation 1

                        where, [A^-] = conjugate base

[AH] = weak acid

Putting the values in equation 1-

            9.63 = 10.32 + log ([A^-] / [AH])

            Or, [A^-] / [AH] = antilog (9.63 – 10.32) = antilog (- 0.69)

            Or, [A^-] = 4.898 [AH]

#. Given, concentration of weak base, [AH] = 1.0 M

So, [A^-] = 4.898 x 1.0 M = 4.898 M

# Given, Total volume of buffer solution (culture medium) = 1.00 L

Now,

Mass of [HCO₃^-] required = (Molarity of HCO₃^- x vol. of solution in L) x Molar mass

                                                            = (1.0 M x 1.00 L) x (61.01714 g/mol)

                                                            = 61.017 g

Mass of [Na₂CO₃] required = (Molarity of Na₂CO₃ x vol. of solution in L) x Molar mass

                                                            = (4.898 M x 1.0 L) x (105.988736 g/mol)

                                                            = 519.133 g