Just as pH is the negative logarithm of [H_3 O^+], PK_a is the negative logarith

Question

Just as pH is the negative logarithm of [H_3 O^+], PK_a is the negative logarithm of K_a, pK_a = - log K_a The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: pH = pK_a + log [base]/[acid] Notice that the pH of a buffer has a value close to the pK_a 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. pOH = pK_b + log [acid]/[base] Acetic acid has a K_a of 1.8 times 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 Drag the appropriate items to their respective bins. How many grams of dry NH_4 Cl need to be added to 2.00 L of a 0.400 Absolution of ammonia, NH_3, to prepare a buffer solution that has a pH of 8.77? K_b for ammonia is 1.8 times 10^-5. Express your answer with the appropriate units.

Explanation / Answer

Part A : pH

pH = 3.74 : acetic acid 10 times greater than [acetate]

pH = 4.74 : [acetate] = [acetic acid]

pH = 5.74 : [acetate] ten times greater than [acetic acid]

Part B : Using Hendernsen-Hasselbalck equation

let x moles of NH4Cl is added

pH = pKa + log(base/acid)

8.77 = 9.25 + log((0.4 M x 2 L - x)/x)

0.331x = 0.8 - x

x = 0.8/1.331

   = 0.60 mol

grams of NH4Cl needed to prepare buffer = 0.60 mol x 53.491 g/mol

                                                                   = 32.095 g

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.