Acetic acid has a K_a of 1.8 x 10^5. Three acetic acid/acetate buffer solutions,

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

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