Hydrolysis - acidic, basic, and neutral salts

Skills to develop

Predict the acidity of a salt solution.

Calculate the pH of a salt solution.

Calculate the concentrations of various ions in a salt solution.

Explain hydrolysis reactions.

Acidic, Basic, and Neutral Salts

**Ions of Neutral Salts**
Cations
Na⁺	K⁺	Rb⁺	Cs⁺
Mg²⁺	Ca²⁺	Sr²⁺	Ba²⁺
Anions
Cl^-	Br^-	I^-,
ClO₄^-	BrO₄^-	ClO₃^-	NO₃^-

A salt is formed between the reaction of an acid and a base. Usually, a neutral salt is formed when a strong acid and a strong base is neutralized in the reaction: H⁺ + OH^- = H₂O The bystander ions in an acid-base reaction form a salt solution. Most neutral salts consist of cations and anions listed in the table on the right. These ions have little tendency to react with water. Thus, salts consisting of these ions are neutral salts. For example: NaCl, KNO₃, CaBr₂, CsClO₄ are neutral salts.

Acidic Ions
NH₄⁺	Al³⁺	Pb²⁺	Sn²⁺
Transition metal ions
HSO₄^-	H₂PO₄^-

Basic Ions
F^-	C₂H₃O₂^-	NO₂^-	HCO₃^-
CN^-	CO₃^2-	S^2-	SO₄^2-
HPO₄^2-	PO₄^3-

When weak acids and bases react, the relative strength of the conjugated acid-base pair in the salt determines the pH of its solutions. The salt, or its solution, so formed can be acidic, neutral or basic. A salt formed between a strong acid and a weak base is an acid salt, for example NH₄Cl. A salt formed between a weak acid and a strong base is a basic salt, for example NaCH₃COO. These salts are acidic or basic due to their acidic or basic ions as shown in the tables here.

Hydrolysis of Acidic Salts

A salt formed between a strong acid and a weak base is an acid salt. Ammonia is a weak base, and its salt with any strong acid gives a solution with a pH lower than 7. For example, let us consider the reaction: HCl + NH₄OH = NH₄⁺ + Cl^- + H₂O In the solution, the NH₄⁺ ion reacts with water (called hydrolysis) according to the equation: NH₄⁺ + H₂O = NH₃ + H₃O⁺. The acidity constant can be derived from K_w and K_b. [H₃O⁺] [NH₃] [OH^-]
K_a = ---------------- ------
[NH₄⁺] [OH^-]
= K_w / K_b
= 1.00e-14 / 1.75e-5 = 5.7e-10. Example 1

What is the concentration of NH₄⁺, NH₃, and H⁺ in a 0.100 M NH₄NO₃ solution? Solution
Assume that [NH₃] = x, then [H₃O⁺] = x, and you write the concentration below the formula in the reaction:

     NH₄⁺  +  H₂O  =  NH₃  +  H₃O⁺
    0.100-x           x       x

     K_a = 5.7E-10.

             x²
        = -------
          0.100-x

Since the concentration has a value much greater than K_a, you may use

     x = (0.100*5.7E(-10))^{1/2
       = 7.5E-6

     [NH₃] = [H⁺] = x = 7.5E-6 M
     pH = -log7.5e-6 = 5.12

     [NH₄⁺] = 0.100 M}

^{Discussion

Since pH = 5.12, the contribution of [H⁺] due to self ionization of water may therefore be neglected.}

^{Hydrolysis and Basic Salts}

^{A basic salt is formed between a weak acid and a strong base. The basicity is due to the hydrolysis of the conjugate base of the (weak) acid used in the neutralization reaction. For example, sodium acetate formed between the weak acetic acid and the strong base NaOH is a basic salt. When the salt is dissolved, ionization takes place:} ^{NaAc = Na⁺ + Ac^-} ^{In the presence of water, Ac^- undergo hydrolysis:} ^{H₂O + Ac^- = HAc + OH^-} ^{And the equilibrium constant for this reaction is K_b of the conjugate base Ac^- of the acid HAc. Note the following equilibrium constants:}

N o t e
Acetic acid	K_a=1.75e-5
Ammonia	K_b=1.75e-5

^{[HAc] [OH^-]
      K_b  =  -----------
                [Ac^-]

             [HAc] [OH^-]  [H⁺]
      K_b  =  -----------  ---
                 [Ac^-]    [H⁺]

               [HAc]    [OH^-][H⁺]
      K_b  =  ---------- ---------
             [Ac^-] [H⁺]

         =  K_w / K_a
         = 1.00e-14 / 1.75e-5 = 5.7e-10.}

^Thus, ^{K_a K_b = K_w} ^or ^{pK_a + pK_b = 14} ^{for a conjugate acid-base pair. Let us look at a numerical problem of this type.}^{Example 2}
^{Calculate the [Na⁺], [Ac^-], [H⁺] and [OH^-] of a solution of 0.100 M NaAc (at 298 K). (K_a = 1.8E-5)}^{Solution

Let x represent [H⁺], then}

^{H₂O  +  Ac^-  =  HAc  +  OH^-
            0.100-x    x       x

      x²
    ---------  =  (1E-14)/(1.8E-5)  =  5.6E-10
    0.100-x}

^{Solving for x results in}

^{x = sqrt( 0.100*5.6E-10)
      = 7.5E-6

    [OH^-]  =  [HAc]  =  7.5E-6

    [Na⁺] = 0.100 F}

^{Discussion

This corresponds to a pH of 8.9 or [H⁺] = 1.3E-9.}
^{Note that K_w / K_a = K_b of Ac^-, so that K_b rather than K_a may be given as data in this question.}

^{Salts of weak acids and weak bases}

^{A salt formed between a weak acid and a weak base can be neutral, acidic, or basic depending on the relative strengths of the acid and base.} ^{If K_a(cation) > K_b(anion) the solution of the salt is acidic.} ^{If K_a(cation) = K_b(anion) the solution of the salt is neutral.}
^{If K_a(cation) < K_b(anion) the solution of the salt is basic.}
^{Example 3}
^{Arrange the three salts according to their acidity. NH₄CH₃COO (ammonium acetate), NH₄CN (ammonium cyanide), and NH₄HC₂O₄ (ammonium oxalate).} ^{K_a(acetic acid) = 1.85E-5,

K_a(hydrogen cyanide) = 6.2E-10,

K_a(oxalic acid) = 5.6E-2,

K_b(NH₃) = 1.8E-5.} ^Solution
^{ammonium oxalate -- acidic, K_a(o) > K_b(NH₃)

ammonium acetate -- neutral K_a = K_b

ammonium cyanide -- basic K_a(c) < K_b(NH₃)}

^{Confidence Building Questions}

^{The reaction of an acid and a base always produces a salt as the by-product, true or false? (t/f)}^{Answer true

Consider...

Water is the real product, while the salt is formed from the spectator ions.}
^{Is a solution of sodium acetate acidic, neutral or basic?}^{Answer basic

Consider...

Acetic acid is a weak acid that forms a salt with a strong base, NaOH. The salt solution turns bromothymol-blue blue.}
^{Are solutions of ammonium chloride acidic, basic or neutral?}^{Answer acidic

Consider...

Ammonium hydroxide does not have the same strength as a base as HCl has as an acid. Amonium chloride solutions turn bromothymol-blue yellow.}

^{Calculate the pH of a 0.100 M KCN solution.

K_a(HCN) = 6.2e-10, K_b(CN^-) = 1.6E-5.}^{Answer 11.1

Consider...}

^{KCN  =  K⁺  +  CN^-
                   CN^-  +   H₂O  =  HCN  +  OH^-
                (0.100-x)           x       x

     x = (0.100*1.5E-5)^1/2
       = 1.2E-3
   pOH = 2.9
    pH = 11.1}

^{The symbol K_b(HS^-) is the equilibrium constant for the reaction:

a. HS^- + OH^- = S^2- + H₂O

b. HS^- + H₂O = H₂S + OH^-

c. HS^- + H₂O = H₃O⁺ + S^2-

d. HS^- + H₃O⁺ = H₂S + H₂O}^{Answer b

Consider...

Write an equation for K_b yourself. Do not guess. The b. is the closest among the four.}
^{What symbol would you use for the equilibrium constant of} ^{HS^- = H⁺ + S^2-} ^{Answer K_a

Consider...

This is the ionization of HS^-; K_a for HS^-, or K_a2 for H₂S.}

Chemistry

Hydrolysis - acidic, basic, and neutral salts

Skills to develop

Acidic, Basic, and Neutral Salts

Hydrolysis of Acidic Salts

^{Hydrolysis and Basic Salts}

^{Salts of weak acids and weak bases}

^{Confidence Building Questions}

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