Acetic acid has a Ka of 1.8 * 10-5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: acetic acid ten times greater than acetate, acetate ten times greater than acetic acid, and acetate=acetic acid.
Match each buffer to the expected pH. pH = 3.74 ; pH = 4.74 ; pH = 5.74
Part B: How many grams of dry NH4Cl need to be added to 2.30 L of a 0.600 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.71? Kb for ammonia is 1.8 * 10-5.
Express your answer with the appropriate units.

A)  acetic acid ten times greater than acetate pH = 3.74

B)  acetate ten times greater than acetic acid pH = 5.74

C) For solution 3: acetate=acetic acid pH = 4.74

Mass NH4Cl = 40.30 grams

Explanation:

Step 1: Data given

Ka is: 1.8 * 10^-5

Solution 1: acetic acid ten times greater than acetate

Solution 2: acetate ten times greater than acetic acid

Solution 3: acetate=acetic acid

Step 2: The pH formula

pH = pKa + log[CH3COO-]/[CH3COOH]

For solution 1:  acetic acid ten times greater than acetate this means ) [[CH3COO-]/[CH3COOH]) has a value of 1/10

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1/10)

pH = 4.74 -1

pH = 3.74

For solution 2: acetate ten times greater than acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(10)

pH = 4.74 + 1

pH = 5.74

For solution 3: acetate=acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1)

pH = 4.74 + 0

pH = 4.74

Part B: Calculate molarity

pOH = pKb + log [H

5.29= -log (1.8*10^-5)  + log [BH+]/[0.600 M]

5.29 = 4.744 + log [BH+]/[0.600 M]

0.546 = [BH+]/0.600

[BH+] = 0.3276 M

Moles NH4+ = 0.3276  M * 2.30 L

Moles NH4+ = 0.75348 moles

Moles NH4Cl = 0.75348 moles

Mass NH4Cl = 0.75348 moles * 53.49 g/mol

Mass NH4Cl = 40.30 grams

46.80 ml

Explanation:

V1 =50ml

T1 =313K°

T2 = 293K°

V2 =?

V1/T1 =V2/T2

V2=V1*T2/T1

V2=50*293/313

46.80ml