Buffers Problem set #1 I. Which of the followringmixhures wosuld resal in bu fer

Question

Buffers Problem set #1 I. Which of the followringmixhures wosuld resal in bu fered soation 100.0 mL ofo0.100 MNH, (,-180 10') with 100.0 ml. of 0.1o ing 30.0 mL of 0.100 MHCIwith 100.0 mL of 0.100MNI east two of the abowe mixtures would result in a buffered solution 2. Which of the following will not produce a buffered solution? 100 mL of0.1 Af NACO, and 50ml.ofOI MIKI 100 ml. ofO.1 ANaCO, and 25 mL of02 MHCI DO tnl. of O 1M Na CO, and 75 ml?of 0.2 M HCl mL of 02 MNa CO, and 5 ml. of 10 MHCT 100 mL of 0.1 MNa,C0, and 50 ml. of 0.1 MNaO 3. What combination of subtances will give a buffered solution that has a pt of 5.05? (Assume orCHN ach pair of substaunces is disolved in 50L of water) ,fe NH,-18010 1.701? 1 0 mole Nil, and 1.5 mole NH,CI 1.5 mole NH, and 1.0 mole NH,CT 1 O mole CJLN and 1.5 mole C,H,NHC 1.3 mole C,H,N and 1.0 mole CH,NHC 4. Which of the following is true for a buffered solution? e solution will not change its pll very much even if a concentrated acid is added he solution will aot change its pH very much even if a strong base is added ny IT ions will react with a l of these hase of a weak acid already in solution. The following question refers to a 20-liter buffered solution created from 0.72 MNH, 8010)and 0.26 MNIHF. What is the pll of this solutioen? 6. The following question refers to a 2.0-liter buflered solution created from 0.31 MNIH, K-1#010-5) and 0.26 MNH,F. When 0.1° mol of H. ions is alled to the solution what is the pi?

Explanation / Answer

1. Mixture that would form buffer solution,

D)

--

2. Mixture that would not result into buffer solution,

E)

--

3. Buffer os pH 5.05 can be formed with,

C)

pH = pKa + log(C5H5N/C5H5NH+)

     = 5.23 + log(1.0/1.5)

     = 5.05

--

4. True statement about buffer solutions,

E)

--

5. pH of buffer

= pKa + log(NH4+/NH3)

= 9.25 + log(0.72/0.26)

= 9.70

B) 9.70

--

6. pH of buffer after adding HCl

pH = pKa + log(NH3/NH4+)

      = 9.25 + log[(0.31 M x 2 L - 0.10)/(0.26 M x 2 L + 0.10)]

      = 9.18

D) 9.18

--

7. pH = pKa + log(F-/HF)

          = 3.14 + log(0.95/0.50)

          = 3.42

B) 3.42

--

8. pH = pKa + log(KNO2/HNO2)

3.0 = 3.4 + log(KNO2/HNO2)

(KNO2/HNO2) = 0.4

let V1 be volume of KNO2 and V2 be volume of HNO2

then,

(0.2 M x V1/1.0 L)/(0.2 M x V2/1.0 L) = 0.4

V1 = 0.4V2

We have,

V1 + V2 = 1.0 L = 1000 ml

0.4V2 + V2 = 1000 ml

volume of HNO2 = V2 = 1000/1.4 = 714 ml

volume KNO2 = V1 = 1000 - 714 = 286 ml

D) 714 ml HNO2 and 286 ml KNO2

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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