Question

Preparing Buffer Solutions: Calculating the Volume of Weak Base Needed Use the table of K values given with this problem to choose the best weak base to start from for making a buffer that holds the pH of the solution at 4.70. Make your selection so that you maximize the capacity of the buffer. Assume that you want to make 100 mL of a buffer and you have already added enough of the conjugate acid salt to make the final buffer 0.30 M in this salt. How many mL of a 3.6 M solution of the weak base you selected must be added to the volumetric flask so that when filled to the mark, the pH of the buffer will equal 4.70. .09 Incorrect: Your answer is incorrect. mL Weak Base Kb NH3 (Ammonia) 1.76 X 10-5 C6H5NH2 (Aniline) 3.9 X 10-10 (CH3CH2)2NH (Diethyl amine) 6.9 X 10-4 C5H5N (Pyridine) 1.7 X 10-9 CH3CH2NH2 (Ethyl amine) 5.6 X 10-4 (CH3)3N (Trimethyl amine) 6.4 X 10-5 (CH3)2NH (Dimethyl amine) 5.4 X 10-4

Answer 1

pH = 4.70

pOH = 14 - pH = 14 - 4.70 = 9.30

From given weak bases, pK_b value for aniline is close to 9.30

K_a = 3.9 *10^-10

pK_a = -log(3.9 *10^-10) = 9.41

Now, Henderson Hasselbalch equation is

pOH = pK_b + log[salt]/[base]

9.30 = 9.41 + log ( 0.30/[base])

log ( 0.30/[base]) = -0.11

( 0.30/[base]) = antilog(-0.11) = 0.778

[base] = 0.30/0.778 = 0.3856 M

Now, Molarity equation is:

M₁V₁ = M₂V₂

where M₁ = 3.6 M , M₂ = 0.3856 M and V₂ = 100 mL

V₁ =  0.3856 M *100 mL/ 3.6 M = 10.71 mL