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    Home > Chemicals Industry > Chemical Technology > Calculation of Hydrolysis of Weak Acid and Strong Alkali Salt

    Calculation of Hydrolysis of Weak Acid and Strong Alkali Salt

    • Last Update: 2021-06-18
    • Source: Internet
    • Author: User
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    1.
    Weak acid and strong base salt

    NaCN is a weak acid and strong base salt, completely dissociated in water

    CN - combines with H + generated by dissociation of H 2 O to form a weak electrolyte HCN

    The combination of H + and CN - generates a weak electrolyte HCN, which makes the OH - concentration in the system greater than the H + concentration, and the solution appears alkaline
    .


    The hydrolysis reaction of CN - by adding the two formulas


    The equilibrium constant of the hydrolysis reaction is

    If the initial concentration of NaCN is c0, when the hydrolysis reaches equilibrium, there is

    Kh Θ is generally very small
    .


    When c0≥400Kb Θ , it can be calculated approximately, that is, c0-[OH-]≈c0, then


    Similar to the degree of dissociation, the degree of hydrolysis refers to the fraction of the hydrolyzed CN - accounting for the initial concentration of NaCN, usually expressed by h

    Salts of polybasic weak acids, such as Na 2 S, Na 2 CO 3 , Na 3 PO 4, etc.


    undergo hydrolysis reaction, and the solution is alkaline


    Since Kh1 Θ > Kh2 Θ , the OH- concentration in the system is mainly determined by the first step hydrolysis equilibrium
    .

    [Example 7-7] Calculation mol · DM 0.


    10 -3 of Na 2 S solution in the OH - , the HS - and H 2 S concentration


    The hydrolysis of Na 2 S in solution is carried out in two steps

    Since KH1 [Theta] "Kh2 [Theta] , so that the solution [OH - ] can be calculated first hydrolysis
    .

    Since Kh1 Θ is relatively large, it cannot be approximated
    .


    Solve the equation

    x=5.


    78×10 -2 mol.
    dm -3

    [OH - ]=5.


    78×10 -2 mol·dm -3

    Since Kh2 [Theta] is very small, so that the HS - the minimum degree of hydrolysis of the second step face can be neglected, i.


    e.
    [the HS - ] ≈ [OH - ]


    Since the [OH - ] ≈ [the HS - ] is

    [H 2 S]=Kh2 Θ =9.


    35×10 -8 mol·dm -3

     

     

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