Latihan soal Larutan Buffer (Dlm Bahasa Inggris)

1.       Statements:

1)    Solution that the value of pH is exist on dilution

2)    Mixing of weak acid and strong base

3)    Solution that has weak base and conjugate acid

4)    Acid solution and salt solution

Those are buffer solutions:

a.       1, 2 and 3

b.       1 and 3

c.       2 and 4

d.       1, 3 and 4

e.       All of them

2.       pH solution bellow doesn’t change if is added water.

a.       NaCl and HCl

b.       H₂SO₄  and K₂SO₄

c.       NaOH and HNO₃

d.       NH₃ and NH₄Br

e.       NH₄OH and NaOH

3.       Solution bellow is buffer, if 100 mL KOH 0.1 M is added by 100 mL ….

a.       KCl 0.1 M

b.       NH₃ 0.1 M

c.       HCN 0.15 M

d.       HNO₃ 0.1 M

e.       CH₃COOH 0.1 M

4.       Mixing of 50 mL ethanoic acid 0.2 M and 100 mL sodium acetate 0.1 M. The value of Ka CH₃COOH = 2.10^-5, so the value of pH solution is….

1. 5
2. 5 - log 2
3. 5 + log 2
4. 9
5. 9 + log 2

5.       Ammonia gas is added NH₄NO₃ 0.01 M until 0.1 M. The value Kb of NH₃ is 10^-5, so the value of pH solution is….

a.       10

b.       9

c.       6

d.       5

e.       4

 

6.       How much crystal KOH (Mr = 56) must be added in 50 mL formate acid (HCOOH) 0.2 M to get pH = 4? (Ka HCOOH = 10^-4)

1. 560 mg
2. 280 mg
3. 5.6 g
4. 0.56 mg
5. 0.0028 g

7.       50 mL NH₃ solution and 100 sulphate acid 0.1 M. pH of mixing solution is 9. So, concentration of NH₃ solution is…. ( Kb NH₃ = 2.10^-5)

1. 0.8 M
2. 0.6 M
3. 0.2 M
4. 0.1 M
5. 0.05 M

8.       In buffer solution: CH₃COOH _(aq)                   CH₃COO^- _(aq)  + H⁺ _(aq)

Adding acid solution in that reaction will….

1. Move the equilibrium to right
2. Increase concentration of ion acetate
3.  Ratio of concentration of ethanoic acid and ion acetate is constant
4. Concentration of H⁺ is decrease (little)
5. pH solution decreases (little) and limited zero

9.       A solution is used to analyze, is made by dilute 1 mL HCl 0.1 M become 1 L. pH solution is not constant, so must be change with solution that has constant pH. That is made by dilute NaOH in 110 mL ethanoic acid 0.1 M (Ka = 10^-5). Mass of NaOH (Mr = 40) is….

1. 800 mg
2. 80 mg
3. 40 mg
4. 8 mg
5. 4 mg

10.   Volume ratio ammonia 0.1 M solution (Kb = 10^-5) and NH₄NO₃ to get the value of buffer solution has pH = 9 is….

1. 1 : 10
2. 10 : 1
3. 1 : 1
4. 1 : 5
5. 5 : 1

11.   Buffer system in blood is constant in pH = 7.4 because mixing of….

1. Carbonate and bicarbonate
2. Ethanoic acid and ion acetate
3. Water and salt
4. Chloride acid and sodium chloride
5. Hydrogen phosphate and ion phosphate

12.   Solutions bellows are buffer system to get weak base and conjugate acid or weak acid and conjugate base, except….

1. NH₃ + NH₄Cl
2. NaHCO₃ + Na₂CO₃
3. KH₂PO₄ + K₂HPO₄
4. HCOOH + HCOOK
5. NH₄ + KCN

13.   Crystal CaX₂ dilute perfect in HX solution (Ka = 5.10^-5) until in solution has 0.1 M HX and 0.1 M CaX₂. So, the value of pH solution is….

1. 9
2. 9 – log 5
3. 5 + log 2.5
4. 5
5. 5 – log 2.5

14.   What is solution to added in 200 mL HCOOH 0.1 M has pH = 2.5 becomes 4 is 100 mL ….

1. KOH 0.1 M
2. KOH 0.2 M
3. HCOOK 0.2 M
4. KOH 0.05 M
5. HCOOK 0.05 M

15.    Buffer solution with pH = 5 can be made by mixing of ethanoic acid 0.1 M (Ka = 10^‑5) and sodium hydroxide 0.1 M with volume….

1. 100 mL and 10 mL
2. 10 mL and 50 mL
3. 100 mL and 100 mL
4. 50 mL and 100 mL
5. 25 mL and 75 mL

B. ESSAY

1.       Buffer solution consist of 500 mL HCOOH 1 M and 500 mL HCOONa 1 M., is added 100 mL sulphate acid 0.05 M. Calculate pH before and after adding of sulphate acid! (Ka HCCOH = 2.10^-4)

2.       Calculate the pH of 200 mL HNO₂ 0.15 M and 150 mL KOH solution 0.10 M! (Ka = 1.10^-5)

3.       In 1 L solution consist of 0.01 mol NH₃ and 0.02 mol NH­₄⁺ from (NH₄)₂SO₄. If Kb NH₃ = 1.10^-5, calculate the value of solution pH!

4.       How much (gram) sodium hydroxide crystal must be added in 100 mL formate acod solution (HCOOH) 0.1 M to make buffer solution with pH = 4 – 2 log 2? (Ka HCOOH = 1.6 x 10^-4)

5.       Calculate the value of pH from mixing solution bellow!

a.       100 mL HCN 0.1 M + 50 mL NaCn 0.2 M. Ka HCN = 4.10^-5

b.       50 mL NH₃ 0.2 M + 100 mL NH₄Cl 0.1 M. Kb NH₃ = 1.10^-5

c.       150 mL CH₃COOH 0.1 M + 100 mL NaOH 0.1 M (Ka = 1.10^-5)

d.      200 mL NH₃ 0.1 M + 100 mL chloride acid 0.1 M (Ka = 1.10^-5)

Posted in: Latihan Soal Kelas XI