Examples of typical tasks solution.

Calculate the dissociation constant of the acetic acid if α for its 0,1 M solution equals 1,35%.      

 

     Data:                                                         Solution:

C(CH₃COOH) =0,1M              CH₃COOH ↔ CH₃COO^- + H⁺

α = 1,35% = 0,0135      K_a = α²·C; K_a = (1,35·10^-2)²· 0,1 = 1,82· 10^-5

The answer: K_a = 1,82· 10^-5

K_a =?                                                              

 

6,26· 10²¹ molecules and ions are contained in 1L of 0,01 M solution of acetic acid. Determine the isotonic coefficient and dissociation degree of CH₃COOH.

 

Data:                                                         Solution:

V_sol.= 1L                                       CH₃COOH ↔ CH₃COO^- + H⁺

C(CH₃COOH) =0,01 1) υ (CH₃COOH) = C · V_sol. = 0,01M

N(all particles)= 6,26· 10²¹ 2) N(of molecules) = 6,02·10²³·0,01  6,02· 10²¹

                                         3)      N(all particles)  6,02· 10²¹

i=?                               i = ——————— = ———— = 1,04

α=?                                    N(of molecules) 6,02·10²³

                                  4)     i-1   1,04 –1

                                        α = —— = ———— = 0,04 or 4%

                                               n-1    2-1

                                                             The answer: i = 1,04; α =4%

 

Calculate the ionic strength of the solution, containing 0,005 Moles of Cu(NO₃)₂ and 0,001 moles of Al₂(SO₄)₃ in 1L of water, and calculate the activity of Cu²⁺ ions in solution (ρ = 1g/ml).

 

Data:                                                         Solution:

ν(Cu(NO₃)₂) = 0,005 M   Cu(NO₃)₂ = Cu²⁺ + 2NO₃^- (1)

ν(Al₂(SO₄)₃) = 0,001 M   Al₂(SO₄)₃ = 2Al³⁺  + 3SO₄^2-(2)

ρ = 1 g/ml                                                _n

                                                      I = ½ ∑ C_i∙ Z_i²

α (Cu²⁺) -? I(sol.)-?                                                      ⁱ⁼¹

                                                                                               ν_i

                                                                                      C_i = —————

                                                             V(solution)

 

1) According to E_q (1): ν(Cu²⁺) = 0,005M;  ν(NO₃ ^-) = 2∙0,005 = 0,01M

2) According to E_q (2): ν(Al³⁺)= 2∙0,001M; ν(SO₄^2-) = 3∙ 0,001 = 0,003M

3) I(solution) = ½ (0,005∙ 2² + 0,001(-1)² + 0,002∙3² + 0,003(-2)²) = 0,03 (m/L)

4) α (Cu²⁺) = f·C(Cu²⁺); lgf = -0,5·Z² I = - 0,5· 4      0,03 = 1,66; f = 0,46

                 α (Cu²⁺) = 0,46· 0,005 =0,0023 (M/L).

                          The answer: I(sol.) = 0,03 M/L, α (Cu²⁺) = 0,0023 (M/L)

 

Determine the concentration of OH^- ions and dissociation degree of ammonia in 0,01 M solution, if K_b = 1,77·10^-5

Data:                                                         Solution:

C(NH₃) = 0,01 M/L               NH₃ + H₂O = NH₄⁺ + OH^-

K_b = 1,77·10^-5                                                               K_b

                                     [OH^-]=     K_b·C;   α = ——;

[OH^-] -? α -?                                                                 C

                                        1) [OH^-] =       1,77·10^-5 ·0,01 = 4,2 ·10^-4 (M/L)

                                   

                                     2) α =       1,77·10^-5/0,01 = 4,2· 10^-2 or

                                      α =[OH^-] / C = 4,2 ·10^-4 / 0,01 = 4,2 ·10^-2

                                             The answer: [OH^-] =4,2 ·10^-4 (M/L)

                                                                        α = 4,2 ·10^-2

 

Questions for self-control

1. What substances are called electrolytes? Tell the differences between aqueous electrolyte solutions and nonelectrolyte solutions.

2. How can you explain the electrolytic dissociation (ionization).

3. What values are the quantitative characteristics of electrolytic dissociation process (ionization).

4. What groups are all electrolytes conventionly divided by in dependence on their dissociation degrees. Give the examples.

5. What is the activity? How does it depend on the analytical concentration.

6. What is the activity coefficient, how is its value changing if the solution is being deluted.

7. What values define the activity factor for each ion and what formula can express this dependence.

8. What does it mean the ionic strength of solution and how is it determined.

9. Write down the relationship between ionic strength and molarity of diluted solutions for a) binary electrolytes of one charged and bi-charged ions (as KCl and ZnSO₄; b) three – and fore-ionic electrolytes such as Na₂SO₄ and FeCl_3.

10.  How and why does the degree of dissociation of weak electrolyte depend on the introduction into its solution of the same name ion or on the dilution of the solution.

11.  Why is the constant of dissociation (ionization) a more convenient characteristic of electrolyte than the degree of dissociation

12.  What formula expresses the law of dilution? Define the bounds of this law application.

13.  What factors does the constant of dissociation depend on?

14. In what cases K_dis = C(electrolyte) · α². Prove the answer.

15. If the electrolyte is strong, we have [Kt⁺] = C(KtAn), and if the electrolyte is weak: [Kt⁺] = C(KtAn)· α. Why is it so?

16.  How can we calculate the concentration of cation of weak electrolyte KtAn, if we know: a) dissociation constant and molarity of electrolyte b) the constant and degree of dissociation of electrolyte?

 

Tasks for self-solution

 

1. Calculate the ionic strength of K₂SO₄  solution whose molarity is equal to 0,02 M/L. The answer: 0,06 M/L.

2. Determine the ionic strength of the solution, containing 1,62g Ca(HCO₃)₂ in 250 ml of the solution. The answer: 0,28.

3. Calculate the ionic strength of solution, containing 2,08 g of BaCl₂ and 5,85 g of NaCl in 500 ml of solution. The answer: 0,26.

4. Calculate the ionic strength and activity of ions in solution, containing 0,01 M of Ca(NO₃)₂ and 0,01 M of CaCl₂ in 1L of solution.

5. Dissociation degree of acetic acid in 1N; 0,1 N; 0,01N solutions is equal to o,42%, 1,34%, and 4,25% respectively. Calculate K_a for these solutions. Prove that the constant of dissociation doesn’t depend on the concentration of the solution.

6. The dissociation constant of HNO₂ equals 5,1·10^-4. Calculate the degree of dissociation of its 0,01 M solution and concentration of H⁺ ions.                       

The answer: 22,6%; 2,26·10^-3 M/L.

7. The first dissociation constant of H₃PO₄ equals 7,11·10^-3. Calculate the concentration of H^{+ -}ions in 0,5 M solution if other steps of dissociation are neglected.   The answer: 5,5·10^-2 M/L.

8. What is the concentration of H⁺-ions in 1N solution of HCN if its K_a =4,9·10^-10? How many grams of ions are contained in 1,5L of acid solution? The answer: 2,21·10^-5 M/L; 5,75·10^-4 g.

9. Determine the dissociation degree and concentration of OH^- -ions in 0,1N solution of NH₄OH, if K_b = 1,77·10^-5.           The answer:1,33%, 1,33 ·10^-3 M/L.

10. How many times is the concentration of H⁺- ions in 1N HNO₃ (α = 0,82) as high as the concentration of H⁺-ions in 1N H₂SO₄ solution (α = 0,51)? The answer: 1,6 times.

11. How much water should we add to 300 ml of 0,2 M solution of CH₃COOH to double the dissociation degree of its acid?The answer: 50 ml.                                              

12.   6,82·10¹⁸ molecules and ions are contained in 1L of 0,001M HCOOH-solution. Determine dissociation degree of formic acid. The answer: 13,3%.

 

Example of a test card on the topic:

“The properties of electrolyte solution”

 

1.   What is the molarity of HNO₂ solution,if its dissociation degree equals 0,2? K_a= 4·10^-4.

                      a) 0,01              b) 0,01        c) 0,02

2. Choose the right formula for the calculation of K_a for acid with α<<0,03.

     a)   α²· C                           b) α²· C                     c) α· C

               ^———

                            1 – α

 3. What is the ionic strength of C-molarity solution of NaCl equal to?

             a) 0,5 C                            b) 1C                           c) 2C

 

 4. What factors does the dissociation constant depend on?

      a) temperature   b)concentration       c) solubility in water.

 

Calculate K_a of acetic acid, if dissociation degree of 0,01 M solution   equals 2% at 25⁰C.

a) 1,35·10^-4               b) 1,75·10^-5                   c) 9·10^-6

The answers to the tasks on the topic “The properties of electrolyte solutions”

 

Number of question           1             2          3           4           5

Code of answer                  b              b          b          a           c

Class 3.