23. In a titration of a weak acid by a strong base a. two equivalents of base ar

Question

23. In a titration of a weak acid by a strong base

a. two equivalents of base are always needed to neutralize all the acid present

b. the equivalence point cannot be defined exactly

c. there is a region in which the pH changes slowly

d. the equivalence point depends on the nature of the added base

24. A solution at pH 7 contains a weak acid, HA. The pKa of the acid is 6.5. What is the ratio of [A]:[HA]?

a. 1:3

b. 1:1

c. 3:1

d. 10:1

25. An ammonia buffer contains NH3:NH4+ in a ratio of 0.4 moles:0.6 moles (pK = 9.75). What will be the pH if you add 0.01 moles of HCl to this buffer?

a. 8.98

b. 9.04

c. 9.25

d. 9.46

e. 9.52

26. Buffering capacity refers to

a. the effectiveness of commercial antacids

b. the extent to which a buffer solution can counteract the effect of added acid or base

c. the pH of a buffer solution

d. the molecular weight of the substance used as a buffer

23. In a titration of a weak acid by a strong base a. two equivalents of base are always needed to neutralize all the acid present b. the equivalence point cannot be defined exactly c. there is a region in which the pH changes slowly d. the equivalence point depends on the nature of the added base 24. A solution at pH 7 contains a weak acid, HA. The pKa of the acid is 6.5. What is the ratio of a. 1:3 c. 3:1 d. 10:1 25. An ammonia buffer contains NH3NH1" in a ratio of 0.4 moles:0.6 moles (pK = 9.75). What will be the pH ifyou add 0.01 moles of H to this buffer? a. 8.98 b. 9.04 C. 9.25 d. 9.46 e. 9.52 26. Buffering capacity refers to a. the effectiveness of commercial antacids b. the extent to which a buffer solution can counteract the effect of added acid or base c. the pH of a buffer solution d. the molecular weight of the substance used as a buffer

Explanation / Answer

23 There is a region where ph changes slowly, this is called buffer region

24. Apply hendersson hasselbach equation

PH = PKa + log (A / HA)

7 = 6.5 + log (A / HA)

0.5 = log (A / HA)

(A / HA) = 10 0.5 = 3.16

ratio is 3.16 : 1

if you round it then 3:1

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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