Questions and tasks for selfcontrol.

1. Give the definition for “ion product of water”. What correlation exists between this value and constant of the water.

2. How is the numeral value of water ion product determined. Does it depend on temperature?

3. How can you determine the ion concentration of OH^- (H⁺) if the concentration of H⁺ (OH^-) ions is known?

4. Is the given statement right: “In the solution of strong acid (alkali) OH^- (H⁺) ions are contained”? Prove the answer.

5. What relationship exists between concentration of H⁺ (OH^-) ions in the solutions of strong and weak acids and bases and their molar concentration?

6. Give the definition for hydrogen ion exponent (pH) and hydroxide ion exponent (pOH). What is their sum equal to?

7. What values do the concentrations of ions H⁺ (OH^-) and pH (pOH) in different mediums accept?

8. Which electrolytes solutions will have the lowest value of pH if their molarities are the same: a) HCl or CH₃COOH b) HNO₂ or HClO c) HCl or KOH.

9. Give the definition for hydrolysis. Describe this process as: a) covalent and ionic; b) by cation and by anion c) simple and multistage d)reversible and unreversible. Give the examples.

10. How does the addition of hydrolyzed salt, such as ZnCl₂, K₂CO₃, Al₂S₃ influence on the equilibrium: H₂O ↔ H⁺ + OH^- and pH value

11. Choose the salt whose pH of solution will be more: a) NaNO₃ or NaNO₂ b)CH₃COOK or CH₃COONH₄ c) KClO or KClO₃.

12. Is it possible with the help of acid-base indicator to distinguish the salts solution: a)NaClO₄ and NaClO b) KI and NH₄I c) KClO₃ and KCl d) Na₂CO₃ and Zn(NO₃)_2, i) BeCl₂ and BaCl₂ F) MgSO₄ and K₂SO₄.

13. Give the definition of “ hydrolytic equilibrium”.

14. Give the definition for hydrolysis degree and hydrolysis constant. Do them depend on salt nature, solution concentration, temperature?

15. What formula expresses the relationship between constant and degree of hydrolysis.

16. What relationship exists between hydrolysis constant and dissociation constant of weak electrolyte?

17. What salt is hydrolyzed in a greater extent: a) KCN or KSCN (K_(HCN) = 7,9∙ 10^-10, K_(HSCN) = 1,4∙10^-1), b) Na₂S or NaHS (K_1a = 6∙10^-8, K_{2a =} 1∙10^-14). Prove the answer with calculations.

18. How can you explain the interincrease of hydrolysis of two salts when their solutions are mixed? In what cases this phenomenon is contemplated.

19. Solve for pH of 0,05% HNO₃ solution if ρ(s-n) = 1 g/ml; α(HNO₃) = 100%. The answer: 2,1

20. Solve for molarity of CH₃COOH solution whose pH is equal to 3.

    K_a = 1,75 · 10^-5. The answer: 0,06 mol/L.

21. Solve for pH of solution, obtained by dilution with water of 0,01L of 30% NaOH solution(ρ = 1,328 g/ml) till 0,75L. α = 70%. The answer: 13,0

22. Solve for pH of solution in 3 L of which 8,1·10^-4 Mol of OH^- ions are contained. The answer: 9,2.

23. Solve for pH of solution in 0,4 L of which 0,39 Mol of NH₃ are contained, if K_b = 1,77·10^-5.   The answer: 11,6.

24. Solve for pH of HCN solution if a) α = 0,0265%, b) 0,0084%.                

The answer: 5,53; 5,03.

25. Solve for [H⁺] and pOH of 0,2 N solution of weak one-basic acid whose dissociation degree is equal to 0,03.The answer: [H⁺] = 6,0·10^-3 mol/l;pOH = 11,78.

26. Solve for pH of solution obtained by mixing of 25 ml of 0,5 M HCl solution and 10 ml of 0,5 M NaOH solution and 15 ml of H₂O. The answer: 0,82.

27. Solve for pH of CH₃COOH solution in which it’s mass fraction is 0,6%. What volume of water must we add to this solution with volume of 1 L to obtain pH value equal to 3. The answer: 2,87; 0,8 L

28. Solve for hydrolysis constant of NH₄Cl if K_b = 1,77 · 10^-5.                                                                                         The answer:5,65·10^-10.

29. Solve for hydrolysis constant of following salts: HCOONa, HCOONH₄, NH₄NO₃ if K_a = 1,77·10^-4, K_b = 1,77·10^-5. The answer:5,64·10^-11, 3,20·10^-6; 5,65·10^-10.

30. Solve for hydrolysis degree and pH of 0,005 N KCN solution if K_a = 4,9·10^-10. The answer: 0,063; 10,5.

 

Sample of test card on the topic “Autoprotolysis of water.Hydrogen and Hydroxyl ion exponents. Salt hydrolysis”

Point to pH of 0,01 N HCl solution if α = 80%.

       a) 2,2       b) 11,9       c) 10^-2       d) 8·10^-3

 

2. Point to pH of 0,001 M CH₃COOH solution if K_a = 1,8 ·10^-5

 a) 3,87       b) 11,13       c) 1,8 · 10^-3       d) 1,35·10^-4

 

3. Calculate K_hyd. of KF, if    K_a = 6,5 ·10^-4

a) 1,54·10^-12      b) 1,54·10^-11      c) 1,54·10^-10         d) 1·10^-14

 

4 pH value and character of the medium in aqueous solution of K₂CO₃ are the following:

a) equal to 7, neutral                 c) more than 7, alkaline

b) less than 7, acidic                 d) less than 7, alkaline

 

5. Choose the formula for [H⁺] calculation:

10^-14                                                                          α

a) ——               b) (OH^-)             c) C∙ α²        d) ——

[OH^-]                                                                         K_a

 

The answers to tasks of card sample

Number of question              1             2            3          4         5

Code of answer                     a             a          b          c          a

 

Class 4

Topic: Buffer solutions. Heterogeneous equilibrium

Buffer solutions

The solutions of weak acids or weak bases with their salts possess the buffer properties: buffer solutions can be diluted without changing of H⁺ concentration (pH), buffer solution also tends to keep pH constant even when small amounts of strong acid or base are added to them. The expression for the concentration of H⁺ and pH in buffer solutions of weak acid HAn and its salt with strong base is applied:  

                                     C(HAn)

                 [H⁺] = K_a —————;

                                     C(An^-)

                              C(An^-)                C(Salt)

      pH = pK_a + lg ——— = pK_a + lg ————,

                              C(HAn)                 C(acid)

where pK_a = -lgK_a.

 

In the same way we can calculate the concentration of OH^- and pH in buffer solutions of weak bases (KtOH) and their salts with strong acid:

                          C(KtOH)

[OH^-] = K_b —————;   

                  C(Kt⁺)

 

                                      C(Kt⁺)                         C(Salt)

      pH = 14 – pK_b - lg ——— = 14 – pK_b -lg ————,

                        C(KtOH)                   C(base)

 

If the buffer solution consists of medium salt (Kt₂An) and acidic salt (KtHAn) of bi-basic acid H₂An, pH is calculated as:

                         C(An^2-)                    

pH = pK_a2 + lg ———,     where pK₂ = - lgK_2, exponent of K_afor

                       C(HAn^-)      H₂An for the second step.   

         

If the buffer mixture is formed by acidic salts (Kt₂HAn and KtH₂An) of three-basic acid H₃An, the following equation is used:

 

                             C(HAn^2-)                    

      pH = pK₂ + lg ————,      

                  C(H₂An^-)

Every buffer solution have its own particular capacity. Number of equivalents of acid or base necessary for changing of pH of 1 L of buffer solution by one unit, is called buffer capacity (B):

           ν(x)

B(x) = ————, when V(buf. sol.) = 1L, or the net equation

       pH₁ - pH₂

                   ν(x)

B(x) = —————————, where X –acid or base.

       V(buf. sol.) ∙│ΔpH│

 

Tasks with their solutions.

 

1. 15 ml of 0,03 M-solution HCOOH are added to 15 ml of 0,03 M-solution HCOONa. Calculate pH of derived buffer solution. K_a = 1,77∙10^-4.

Data:                                                         Solution:

V₁ = V₂ = 15 ml        The first way:

 

C₁ = C₂ = 0,03 m/L [H⁺] = K_a^.C(HCOOH) / C(HCOO^-)

        pH =-lg[H⁺]

K_a = 1, 77∙10^-4.         1) [HCOO^-] _{in buf.sol.}= 15 ∙ 0,003/30 = 0,015M/L

                                 2) [HCOOH] _{in buf.sol.}= 15 ∙ 0,003/30 = 0,015M/L                            

pH -?                        3) [H⁺] = 1,77∙10^-4 ∙ 0,015/0,015 = 1,77∙10^-4

                                 4) pH = -lg1, 77∙10^-4 = 4 - lg 1,77 = 3,75

                           The second way:

It can be simplified as:             [HCOO^-]                15∙0,03

                         pH = pK_a + lg————— = 3,75 + lg——— =3,75

                                                 [HCOOH]               15∙0,03

                                                                     The answer: 3,75

2. How will pH of buffer mixture formed by 0,1M solution of CH₃COOH and by 0,1 M solution of CH₃COONa change if we add: a) 0,001M solution of HCl 

 b) 0,001M NaOH solution?                             K_a = 1,8∙10^-4.

Solution:

In buffer mixter                C(acid)                 0,1

                        [H⁺] = K_a ———— = 1,8 ∙10^-5∙—— = 1,8 ∙10^-5

                                          С(salt)                  0,1

                         pH = pK_a = - lg1,8∙10^-5 = 4,75

If the HCl acid is added:

      0,001        0,001  0,001

   CH₃COONa + HCl = CH₃COOH + NaCl

 

[CH₃COOH] = 10^-1 + 10^-3 = 1,01∙10^-1 (Mol/L)

[CH₃COONa] = 10^-1 - 10^-3 = 0,99∙10^-1 (Mol/L)

[H⁺] =1,8∙10^-5∙1,02 = 1,84 ∙10^-5(Mol/L)

If the NaOH is added:

   CH₃COOH + NaOH = CH₃COONa + H₂O

                        [H⁺] =1,8∙10^-5∙0,98 = 1,76 ∙10^-5(Mol/L)

                        pH = - lg1,76∙10^-5 = 5- 0,24 = 4,76

                                             The answer: pH is practically constant.

 

3. Calculate the buffer capacity of blood serum by acid, if 10 ml of 0,1 N HCl solution are consumed for titration of 50 ml of serum to change pH from 7,4 to 7,0.

Data:                                    Solution:

pH₁ = 7,4                                 ν(x)

pH₂ = 7                B(x) = —————————,

V(HCl_sol) =10 ml             V(buf. sol.) ∙│ΔpH│

C(HCl) = 0,1 M/L 1) ν(HCl) = 10∙10^-3∙0,1 = 10^-3

B-?                2) ΔpH = pH₁ – pH₂ = 7,4 –7,0 =0,4                                                       

3)                    10^-3

                                B(HCl) = ————— = 0,05 (M/L)

                                           50∙10^-3∙0,4

                                                          The answer: 0,05 (M/L)

 

Questions for self-control

1. What mixtures are called the “buffers”?

2. What is called conjugate acid, conjugate base? Give the examples of protolytic reactions.

3. What types are the buffer solution divided into? Give the examples.

4. Explain, how the buffer solutions work. As the example take the acetic buffer solution.

5. What is the mechanism of the buffer action, take for example ammonia buffer solution.

6. How should we calculate the concentration of H⁺ -ions in buffer solution of acidic type.

7. How should we calculate the concentration of H⁺ -ions in buffer solution of basic type.

8. How should you explain the fact: the addition of small amounts of strong acid and bases, and the dilution does not change the acidity of buffer solutions practically.

9. How should you calculate the pH –values in buffer solutions of acidic or basic type.

10. How do you understand: a) buffer capacity of solution b) buffer strength of solution.

11. 1 ml of 0,5M NH₃ solution was added to 20 ml of 1% NH₄NO₃  (ρ = 1 g/ml). Prepared solution was deluted in measured flask till 100 ml. Calculate the pH of obtained solution. K_b=1,8·10^-5, pK_b=4,75.  The answer: 8,56.

12. 15 ml of 0,2 M K₂HPO₄ solution are added to 25 ml 0,2M KH₂PO₄ solution. Calculate the pH of obtained solution. K_a1= 7,6·10^-3, pK_a1= 2,12; K_a2= 6,3·10^-8, pK_a2 = 7,2; K_a3=1,3·10^-12, pK_a3=11,87.                                 The answer: 6,98.

13. 15 ml of 0,15M HCOOK solution are added to 12 ml 0,03 M HCOOH solution. Calculate the pH of obtained solution. K_a= 1,8·10^-4, pK_a= 3,75.

The answer: 4,35.

14. Calculate the pH of solution, 500 ml of which contain 1 g of HCOOH and 1 g of HCOOK. K_a= 1,8·10^-4, pK_a= 3,75. The answer: 3,49.

15. 0,5% solutions of ammonia and ammonium nitrate are mixed in equal amounts. Calculate the pH of the obtained solution whose density is 1 g/ml; K_b= 1,8·10^-5, pK_b= 4,75. The answer: 9,92.

16. How many ml of 0,2 M Na₂CO₃ solution can we add to 10 ml of 0,3 M NaHCO₃  solution to prepare the solution with pH = 10? K_a1= 4,4·10^-7, pK_a1= 6,36; K_a2=4,7·10^-11, pK_a2 =10,33. The answer: 7 ml.

17. How many grams of 0,2 M Na₂CO₃ should we add to 100 ml of 0,3 M NaHCO₃ solution to prepare the solution with pH 10? K_a1= 4,4·10^-7, pK_a1= 6,36; K_a2= 4,7·10^-11, pK_a2 =10,33. The answer: 1,49 g.

 

Heterogeneous equilibrium

 

In saturated solution of slightly soluble electrolyte KtAn chemical equilibrium is established:

                              KtAn_(solid) ↔ Kt⁺_(s-n) + An^-_(s-n)

The equilibrium is quantitatively characterized by the constant of heterogeneous equilibrium, which is called the solubility product (SP). SP(KtAn)=[KT⁺]∙[An^-] = IP(KtAn) ionic product.

In saturated solution SP = IP in unsaturated solution SP>IP, in supersaturated solution SP<IP. Concentration of electrolyte X in saturated solution is called solubility (S(x) in mol/l): S(x) = L(x)/M(X), where L(X) – solubility of electrolyte in gram/ l.

The relationship between the solubility and the solubility product of electrolyte in dependence on number of ions in a molecule of electrolyte can be presented in the following table:

 

Formula of electrolyte X (example)	Number of ions in molecule	SP(X)	S(X)
KtAn (AgCl)	2	S^2	^2         SP
KtAn2 (PbI2)	3	4S^3	^3 SP        4
KtAn3 (Fe(OH)3)	4	27S^4	^4 SP         27
Kt3An2 (Ca3(PO4)2)	5	108S^5	^5SP         108
KtnAnm	n+m	n^n m ^mS^n+m	^n+mSP        n^n∙m^m

 

Tasks with solutions

1. 2,88∙10^-6 g of AgI can be dissolved in 1 liter of water. Calculate the SP (AgI).

Data:                                                        Solution:

V(H₂O) = 1 L                     AgI_(s) = Ag⁺_(aq) + I^-_(aq)

m(AgI) = 2,88∙10^-6 g          SP(AgI) = [Ag⁺]∙[I^-]

                                                                         m(AgI)

SP(AgI) -?                [Ag⁺] = [I^-] = C(AgI) = —————— 

                                                                                           M(AgI)∙V(sol.)

 

                                  1)V(solution) = V(H₂O) = 1 L; M(AgI) = 23 g/M

                                     2) C(AgI) = 1,225∙10^-8 Mol/L

                                     3)SP(AgI) = (1,225∙10^-8)² =1,5∙10^-16

           The answer: 1,5∙10^-16

 

2. Does the precipitate of Fe(OH)₃  appear in the solution containing 1,5∙10^-16 Mol/L of FeCl₃ and 5∙10^-5 Mol/L of NaOH? SP = 3,8∙10^-38

Data:                                                        Solution:

C(FeCl₃) = 1,5∙10^-3M/L                Fe(OH)_{3 (s)}  = Fe³⁺_(aq) + 3OH^-_(aq)

C(NaOH) = 5∙10^-5 M/L                SP(Fe(OH)₃) = [Fe³⁺]∙[OH^-]³

SP(Fe(OH)₃) =3,8∙10^-38  1) [Fe³⁺] = C(FeCl₃) = 1,5∙10^-3M/L

                                                           2) [OH^-] = C(NaOH) = 5∙10^-5 M/L

IP><SP -?               3)Fe(OH)₃=[Fe³⁺]∙[OH^-]³=1,5∙10^-∙(5∙10^-5)=1,9∙10^-6

                                            4)IP>>SP, consequently, precipitate is formed.

The answer: it is formed.

 

3. What concentration of Al³⁺ is necessary to form the precipitate of Al(OH)₃ from the solution with pH = 6?  SP(Al(OH)₃) = 5,1∙10^-33.

Data:                                                        Solution:

pH =6                              Al(OH)_3(s)  = Al³⁺_(aq) + 3OH^-_(aq)

SP(Al(OH)₃ = 5,1∙10^-33.            10^-14 

                                  [OH^-] = ———; SP(Al(OH)₃ )= [ Al³⁺]∙[OH^-]³

[Al³⁺] -?                                     [H⁺]

1) [H⁺]= 10^-pH= 10^-6 (Mol/L)

2) [OH^-]=10^-14/10^-6 = 10^-8 (Mol/L)

       5,1∙10^-33

3) [ Al³⁺] = ———— =5,1∙10^-9 (Mol/L)

           (10^-8)³           The answer: 5,1∙10^-9 (Mol/L)

Questions for self-control

1. What equilibrium is called heterogeneous?

2. Derive the expression of solubility product (SP) for slightly soluble salt AgCl.

3. Under what conditions does precipitate appear, dissolve, and exist in equilibrium with the solution. Compare IP (ion product) and SP (solubility product).

4. How does the solubility of a substance change in the presence in it’s saturated solution of the common ion? Prove the answer.

5. Derive the formulas for calculating SP if the solubility (S) is known for the cases: a) BaSO₄; b) Mn(OH)₂ c) Cr(OH)₃ d) Mg₃(PO₄) f) Kt_nAn_m

6. How many grams (g) of CaSO₄ are dissolved in 1 litre of water if

SP(CaSO₄) = 6,1∙10^-5? The answer: 1,062g.

7. 0,001215 g of Ca₃(PO₄)₂ are dissolved in 1 litre of water.                        Calculate SP (Ca₃(PO₄)₂). The answer: 1∙10^-25.

8. Compare the solubility of two salts: AgCl and Ag₂CrO₄. How many times is one salt as soluble as the another? SP(AgCl) = 1,1∙10^-10, SP(Ag₂CrO₄) = 1,6∙10^-12.

The answer: silver chromate is 7,3 times as soluble as silver chloride in moles, and 16 times in grams.

9. How many grams (g) of BaSO₄ are contained in 100 L of saturated solution, if SP(BaSO₄) = 1∙10^-10. The answer; 0,23 g.

10. How many grams (g) of AgBr are contained in 10 L of saturated solution, if SP(AgBr) = 4∙10^-13. The answer: 1,19∙10^-3 g.

11. How many grams (g) of NaOH should we add to 20 ml of the solution, containing 5∙10^-5 M MgCl₂ to precipitate Mg(OH)₂? SP(Mg(OH)₂) =1,2∙10^-11.The answer: 3,9∙10^-4 g.

12. How many milliliters of 2M HCl solution should we add to 20 ml of the

3∙10^-2 M Pb(NO₃)₂ solution, to precipitate PbCl₂? SP(PbCl₂) = 2,4∙10^-4.                                           The answer: 0,94 ml.

13. Is CaSO₄ precipitated if we add to 0,1 M CaCl₂ solution the equal volume of

0,1 M H₂SO₄ solution. SP(CaSO₄) = 6,1∙10^-5. The answer: IP = 4,26∙10^-4: CaSO₄ is precipitated.

14. What is the value of pH when Mg(OH)₂ is precipitated from 2∙10^-2 M MgSO₄ solution? SP(Mg(OH₂) = 1,2∙10^-11. The answer: 9,38.