HYDROLYSIS OF SALTS ASSIGNMENT

Example #1: What is the pH of a 0.0500 M solution of ammonium chloride, NH₄Cl. K_a of NH₄⁺ = 5.65 x 10¯¹⁰.

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of NH₄Cl:

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

2) Here is the K_a expression for NH₄⁺:

	[NH3] [H3O+]
Ka =	–––––––––––
	[NH4+]

3) We can then substitute values into the K_a expression in the normal manner:

	(x) (x)
5.65 x 10¯10 =	–––––––––
	0.0500 − x

4) Ignoring the minus x in the usual manner, we proceed to sove for the hydronium ion concentration:

x =

(5.65 x 10¯¹⁰) (0.0500)x = 5.32 x 10¯⁶ M = [H₃O⁺]

5) We then calculate the pH directly from the [H₃O⁺] value:

pH = −log 5.32 x 10¯⁶ = 5.274

Example #2: What is the pH of a 0.100 M solution of methyl ammonium chloride (CH₃NH₃Cl). K_a of the methyl ammonium ion (CH₃NH₃⁺ = 2.70 x 10¯¹¹)

Solution:

1) Here is the chemical reaction (net ionic) for the hydrolysis of CH₃NH₃⁺:

CH₃NH₃⁺ + H₂O ⇌ CH₃NH₂ + H₃O⁺

2) Here is the K_a expression for CH₃NH₃⁺:

	[CH3NH2] [H3O+]
Ka =	––––––––––––––
	[CH3NH3+]

3) We can then substitute values into the K_a expression in the normal manner:

	(x) (x)
2.70 x 10¯11 =	––––––––
	0.100 − x

4) Ignoring the minus x in the usual manner, we proceed to sove for the hydronium ion concentration:

x =

(2.70 x 10¯¹¹) (0.100)x = 1.64 x 10¯⁶ M = [H₃O⁺]

5) We then calculate the pH directly from the [H₃O⁺] value:

pH = −log 1.64 x 10¯⁶ = 5.784

Example #3: Given the pK_a for ammonium ion is 9.248, what is the pH of 1.00 L of solution which contains 5.45 g of NH₄Cl (the molar mass of NH₄Cl = 54.5 g mol¯¹.)

Solution:

1) Determine molarity of the solution:

5.45 g / 54.5 g mol¯¹ = 0.100 mol0.100 mol / 1.00 L = 0.100 M

2) Determine K_a for NH₄Cl:

K_a = 10¯^pK_a = 10¯^9.248K_a = 5.64937 x 10¯¹⁰

3) Write the equation for the hydrolysis of the ammonium ion and K_a expression:

NH₄⁺ + H₂O ⇌ H₃O⁺ + NH₃

	[H3O+] [NH3]
Ka =	––––––––––––
	[NH4+]

4) Insert values into K_a expression and solve:

	(x) (x)
5.64937 x 10¯10 =	––––––
	0.100

x = 7.51623 x 10¯⁶ M

5) Take negative log of this value (this value being the [H₃O⁺]) for the pH:

pH = −log 7.51623 x 10¯⁶ = 5.124

Example #4: Aniline is a weak organic base with the formula C₆H₅NH₂. The anilinium ion has the formula C₆H₅NH₃⁺ and its K_a = 2.3 x 10¯⁵.

(a) write the chemical reaction showing the hydrolysis of anilinium ion.
(b) calculate the pH of a 0.0400 M solution of anilinium ion.

Solution:

1) The chemical reaction for the hydrolysis is:

C₆H₅NH₃⁺ + H₂O ⇌ C₆H₅NH₂ + H₃O⁺

2) The calculations for the pH are:

	(x) (x)
2.3 x 10¯5 =	––––––––
	0.0400 − x

x =

(2.3 x 10¯⁵) (0.0400)x = 9.6 x 10¯⁴ M = [H₃O⁺]

pH = −log 9.6 x 10¯⁶ = 3.02

Example #5: Pyridine (C₅H₅N) is a weak base and reacts with HCl as follows:

C₅H₅N + HCl —> C₅H₅NH⁺ + Cl¯

What is the pH of a 0.015 M solution of the pyridinium ion (C₅H₅NH⁺)? The K_b for pyridine is 1.6 x 10¯⁹. (Hint: calculate the K_a for the pyridinium ion and use it in the calculation.)

Solution:

1) First, we calculate the K_a:

K_aK_b = K_w

	1.00 x 10¯14
Ka =	––––––––––
	1.6 x 10¯9

K_a = 6.25 x 10¯⁶

2) Now, the solution follows the pattern outlined in the tutorial:

C₅H₅NH⁺ + H₂O ⇌ H₃O⁺ + C₅H₅N

	(x) (x)
6.25 x 10¯6 =	––––––
	0.015

x =

(6.25 x 10¯⁶) (0.015)x = 3.06 x 10¯⁴ M <— this is the [H₃O⁺]

pH = −log 3.06 x 10¯⁶ = 3.51

Bonus Example: The pH of a 0.160 M CH₃NH₃Cl solution is 5.500. What is the value of K_b for CH₃NH₂?

Solution technique: we will determine the K_a of CH₃NH₃Cl (since that is what we have data for). Based on the fact that CH₃NH₂ is the conjugate base, we will use K_aK_b = K_w to get the K_b.

Solution:

1) Here’s the K_a expression we are interested in:

	[H3O+] [CH3NH2]
Ka =	––––––––––––––––
	[CH3NH3+]

2) It comes from this reaction, the hydrolysis of CH₃NH₃⁺:

CH₃NH₃⁺ + H₂O ⇌ H₃O⁺ + CH₃NH₂

3) The concentrations of two components of the K_a expression come from the pH:

H₃O⁺ = 10¯^5.500 = 3.16228 x 10¯⁶ MThis is also the concentration of the CH₃NH₂.

4) Solve for the K_a:

	(3.16228 x 10¯6) (3.16228 x 10¯6)
Ka =	––––––––––––––––––––––––––––
	0.160

K_a = 6.25 x 10¯¹¹

5) Solve for the K_b:

K_aK_b = K_w(6.25 x 10¯¹¹) (K_b) = 1.00 x 10¯¹⁴

K_b = 1.6 x 10¯⁴

1) H₂CO₃ + H₂O  H₃O⁺ +HCO₃^–
a. Identify which of these is the conjugate base and which is the weak acid.
b. Does the weak acid hydrolyze?
2)
a. Write out the chemical equation for the hydrolysis HF.
b. Is water acting as a Bronsted-Lowry acid or Bronsted-Lowry base?
3)
a. Write out the equation for the dissociation of the salt NH₄Br.
b.Write out the hydrolysis of the cation that is produced from the dissociation of the ammonium bromide.
c. From what kinds of acids and bases is ammonium bromide (NH₄Br) made from? Strong acid/strong base? Strong acid/weak base?     Strong base/weak acid? Weak base/weak acid?
d. State whether salt hydrolyzes.
e. State whether solution is acidic or basic.
4). CH₃COO^– +H₂O  CH₃COOH + OH^–
What is the pH of 0.30 M of sodium acetate?
(Hint: First find Kb value)
Given: K_aof CH₃COOH= 1.8 x 10^-5
5)
a. Does sodium acetate (from previous problem) hydrolyze?
b. Is solution acidic or basic?

 

Solutions

1)
a. The conjugate base is the HCO₃^–. The weak acid is the H₂CO₃.
b. Yes it hydrolyzes.
₂₎
a.HF + H₂O  H₃O⁺ + F^–
    b.Water is acting as a Bronsted-Lowry base because it is accepting a proton (H⁺) from the HF.
3)
a. NH₄Br → NH₄⁺+ Br^–
b. Br^– does not hydrolyze; it is an ion.
NH₄⁺ +H₂O  H₃O⁺ +NH₃^–  <— Hydrolysis of NH₄+
c. HBr is a strong acid. Ammonia is a weak base. So NH₄Br  is made of a strong acid and weak base.
d. Yes it hydrolyzes.
e. Acidic
4)
CH₃COO^– +H₂O  CH₃COOH + OH^–

	CH3COO–	CH3COOH	OH–
I	0.30M
C	-x	x	x
E	0.30 – x	x	x

K_b=       K_w          
K_a of CH₃COOH
K_b= 1.0 x10^-14
1.8 x 10^-5

K_b= 5.6 x 10^-10
K_b= [CH₃COOH][OH^–]
[CH₃COO^–]
(Use the given K_b and the concentrations from the ICE table)
5.6 x 10^{-10 =     x2}      
0.30-x     (Assume x << 0.30)
x²= 1.68 x 10^-10
x=1.30 x 10^-5 = [OH^–]
pOH= -log (1.30 x 10^-5) = 4.89
pH = 14 – pOH
pH = 14.00 – 4.89 = 9.11
5)
a. Yes it hydrolyzes
b. Basic solution