Question

1)Design a buffer that has a pH of 10.12 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 bromide 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 bromide salt of conjugate acid =

grams weak base =

2)Design a buffer that has a pH of 10.91 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 1.00M in the weak base?

grams potassium salt of weak acid =  g

grams potassium salt of conjugate base =  g

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Answer 1

According to Henderson-Hasselbulch equation:

pH = pKa + Log([base]/[acid])

i.e. 10.12 = 10.62 + Log([CH3NH2]/[CH3NH3⁺])

i.e. Log([CH3NH3⁺]/[CH3NH2]) = 10.62 - 10.12 = 0.5

i.e. [CH3NH3⁺]/[CH3NH2] = 10^0.5 = 3.162

i.e. The percent of base (CH3NH2) in the buffer = 100/{1 + ([CH3NH3⁺]/[CH3NH2])}

= 100/(1+3.162) ~ 24%

Therefore, the percent of acid (CH3NH3⁺) in the buffer = 100 - 24 = 76%

1 M - 24%

i.e. 76% = (76/24) M = 3.167 M

Therefore, the mass of weak base (CH3NH2) = 1 mol * 31 g/mol = 31 g

And the mass of bromide salt of conjugate acid (CH3NH3Br) = 3.167 mol * 112 g/mol = 354.7 g