A) Design a buffer that has a pH of 9.93 using one of the weak base/conjugate ac

Question

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

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.

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 = _____

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

Explanation / Answer

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 CH3NH2­ 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

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