Topic: Autoprotolysis of water. Ion product of water.

Hydrogen and hydroxyl ion exponents. Salt hydrolysis.

 

   Electrostatic interaction of polar water molecules led to their self-ionisation (autoprotolysis): H₂O + H₂O ↔ H₃O⁺ + OH^-. Protons transfer from some molecules to others: H⁺ + H₂O ↔ H₃O⁺. The simplified version of water dissociation is following H₂O ↔ H⁺ + OH^-.

 At 25⁰C: [H⁺]∙[OH^-] = 1∙10^-14 and in pure water [H⁺]=[OH^-]= 10^-7 mol/L.

 In logarithmic form: pH + pOH = 14 and in pure water pH=pOH=7,

 where pH = -lg[H⁺], pOH = -lg[OH^-];

 [H⁺] –molarity of H⁺ ions, [OH^-] –molarity of OH^- ions.

In the solution of one-basic strong acids HAn (α =1)           

                             HAn → H⁺ + An^-

                             [H⁺] =C(HAn); pH = -lgC(HAn).

 In the solution of one-acidic strong bases KtOH (α =1)    

                              KtOH→Kt⁺ +OH^-

                              [OH^-] = C(KtOH),      pOH =-lgC(KtOH)

                                 pH = 14 –pOH = 14+ lgC(KtOH).

In the solution of one-basic weak acids:

                   [H⁺] = C∙α = √ K_a/C       pH = - lgC∙α = -lg√ K_aC

 

In the solution of one-acidic weak base:

                   [OH^-] = C∙α = √ K_b∙C     pOH = - lgC∙α = -lg √ K_b∙C

                                                                      

                          pH = 14 – pOH = 14 + lgC∙α =14 + lg K_b∙C

 If dissolution of any salt results in formation of weak electrolyte (weak acid or base or both simultaneously) we have ion hydrolysis as a result of which the equilibrium  H₂O ↔ H⁺ + OH^- shifts to right. In salt solution ions H⁺ can be accumulated ([H⁺]>[OH^-] - acidic medium) or OH^- ions ([H⁺]<[OH^-]- alkaline medium). In logarithmic form: pH <7 – acidic medium, pH>7 - alkaline medium.

   Quantitatively hydrolysis is characterized by hydrolysis constant (K_hyd) and hydrolysis degree (h):

                                  h² ∙ C

                    K_hyd = ————, when 1-h ≈ 1      K_hyd= h²∙C

               1- h

If the salt is formed by strong base and weak acid

                                             

                                              10^-14

                               K_hyd= ———;

                                              K_a

If the salt is formed by weak base and strong acid

                                               10^-14

                                K_hyd = ———;

                                                 K_b

If the salt is formed by weak acid and weak base

                                             10^-14

                                K_hyd= ———;

                                          K_a∙ K_b

 

Tasks with solutions

 

1. Solve for pH and pOH of 0,01 M HCl solution if α = 80%

 

Data:                                                         Solution:

    C(HCl)=0,01 mol/l                       HCl↔ H⁺ + Cl^-

α = 80% = 0,8           [H⁺] = C(HCl)∙ α; pH =-lg [H⁺]; pOH= 14- pH

 pH =? pOH =?        1) [H⁺] =0,01∙0,8 =8∙10^-3 (mol/l)                                                           

                                     2) pH= -lg=8∙10^-3 =3 – lg8 = 3-0,9 = 2,1

                                     3) pOH = 14 –2,1 = 11,9

                                                  Тhe answer: pH =2,1 pOH = 11,9.

 

2. Solve for pH of 0,001 M CH₃COOH solution if K_a = 1,8 ∙ 10^-5

 

Data:                                                         Solution

C(CH₃COOH) = 1∙10^-3 mol/l   CH₃COOH↔ CH₃COO^- + H⁺

 K_a = 1,8 ∙ 10^-5                           pH =-lg [H⁺]; 

pH=?                                    [H⁺] =        K_a∙C                                   

                                                   1) [H⁺] =         1,8 ∙ 10^-5∙ 10^-3 = 1,35 ∙ 10^-4

                                                 2) pH = -lg 1,35∙ 10^-4 = 4- 0,13 =3,87

                                                            The answer: pH =3,87.

 

3. Solve for pH of solution, containing 0,4 g of NaOH in 2 L (α = 100%)

 

Data:                                                         Solution:

m(NaOH) = 0,4 g               NaOH ↔ Na⁺ + OH^-

V(s-n) =2L                         pH= 14- pOH, pOH= -lg [OH^-];

α = 100                                                               m(NaOH)

                           [OH^-] = C(NaOH) = ———————

pH=?                                                               M(NaOH)∙ V(s-n) 

                                     0,4

           1) C(NaOH)= —— = 0,005 mol/l

                                   40∙2

             [OH] = 0,005 mol/l

            2) pOH = - lg 5∙10^-3 = 3- lg5 = 3-0,7 = 2,3

             3) pH = 14 – 2,3 = 11,7

                                                       The answer: pH =11,7.

 

4. Solve for pH of 0,1 M NH₃ solution if K_b = 1,8 ∙ 10^-5

Data:                                                         Solution

C(NH₃) = o,1 mol/l                  NH₃ + H₂O ↔ NH₄⁺ + OH^-

K_b= 1,8 ∙ 10^-5                          pH= 14- pOH, pOH= -lg [OH^-];

             pH=?                                 

[OH^-]= K_b∙C

                                                         

                                                 1) [OH^-]=        1,8 ∙ 10^-5∙ 0,1 =1,35 ∙ 10^-3

                                                 2) pOH = 3-lg 1,35 = 3- 0,13 =2,87

                                                 3) pH =14- 2,87 = 11,13

                                                            The answer: pH =11,13

 

5. Write down the equations of salts hydrolysis Na₃PO₄ ,Na₂HPO_4, NaH₂PO₄ Find the hydrolysis constants for them. Which salt is hydrolyzed in a greater extent?          K_a1 = 7,1 ∙ 10^-3, K_a2 = 6,2∙ 10^-8,  K_a3 =5∙ 10^-13

                             Solution:

1)Na₃PO₄ + H₂O ↔ Na₂HPO₄ + NaOH,  PO₄^3- +H₂O ↔ HPO₄^2- + OH^-

 

                          10^-14   1∙10^-14

              K_hyd1 = —— = ——— = 2∙10^-2

                           K_a3      5∙ 10^-13

 

2) Na₂HPO₄ + H₂O ↔ Na₂HPO₄ + NaOH, HPO₄^2- + H₂O↔ H₂PO₄^- + OH^-

 

                           10^-14   1∙10^-14

              K_hyd2 = —— = ——— = 1,6∙10^-7

                            K_a2      6,2∙ 10^-13

3) NaH₂PO₄ + H₂O ↔ H₃PO₄ + NaOH,  H₂PO₄^- + H₂O↔ H₃PO₄ + OH^-

 

                            10^-14   1∙10^-14

              K_hyd3 = —— = ——— = 1,4∙10^-18

                            K_a1      7,1∙ 10^-3

 

            The answer: Na₃PO₄ is hydrolyzed in greatest extent.