Acetic acid has a K_a of 1.8 x 10^5. Three acetic acid/acetate buffer solutions,
ID: 963290 • Letter: A
Question
Acetic acid has a K_a of 1.8 x 10^5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: Just as pH is the negative logarithm of [H3O^+], pK_a is the negative logarithm of K_a, [acetic acid] ten times greater than [acetate], [acetate] ten times greater than [acetic acid], and [acetate] = [acetic acid]. p K_a =-log K_a The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: Match each buffer to the expected pH. Notice that the pH of a buffer has a value close to the pK& of the acid, differing only by the logarithm of the concentration ratio [base]/[acid] - The Henderson-Hasselbalch equation in terms of pOH and pK_b is similar. How many grams of dry NH_4CI need to be added to 2.40 L of a 0.800 M solution of ammonia, NH_3, to prepare a buffer solution that has a pH of 8.86? K_b for ammonia is 1.8 x 10^-5.Explanation / Answer
A)
pH = 3.74
this means there is more acid than conjugate, since pKa = 4.75 and pH < pKa
pH = 4.74
pH = pKa, so conjugate(= acid
for pH 5.74
the conjguate base > acid, since pKa < pH
B)
pKb = 4.75
pOH = pKb+ log(NH4+/NH3)
pO H= 14-8.86 = 5.14
5.14= 4.75 + log(NH4+/0.8)
[NH4+] = 0.8*10^(5.14-4.75) = 1.96
then
V= 0.8 so
mol =MV = 1.96*0.8 = 1.568 mol
mass = mol*MW = 1.568*53.491 = 83.8738 g of NH4Cl