Calculate the pH of the 0.26 M solution of NH₄ClO₃ (K_b NH₃ = 1.8 x 10^-5).

Ammonium chlorate (NH₄ClO₃) is the salt of a weak base ammonia (NH₃) and a strong acid HClO₃, so it yields an acidic solution.

 

NH₃ + HClO₃ → NH₄ClO₃

 

So, let’s follow what we have discussed earlier and see why it produces an acidic solution.

Dissociate the salt to identify the weak component ion that will react with water:

 

NH₄ClO₃ → NH₄⁺ + ClO₃^–

 

NH₄⁺ is the ion with a weaker component (NH₃), so it is going to react with water producing hydronium ion which makes the solution acidic:

NH₄⁺(aq)  + H₂O(l) → NH₃(aq)  +  H₃O⁺(aq)  (acidic)

 

The next part is to determine the concentration of H₃O⁺ ions in order to calculate the pH.

 

We are given the K_b, which measures the base strength and it is based on the reaction of ammonia itself with water:

 

NH₃(aq) + H₂O(aq) ⇆ NH₄⁺(aq) + OH^–(aq)

 

This equation does not contain H⁺/H₃O⁺ ions, and we do not know the concentration of ammonia, therefore, we need to determine the K_a.

K_a and K_b can be calculated from one another using the relationship:

 

K_a · K_b = K_w = 10^-14

 

Rearranging this, we get an expression for K_a:

 

\[{K_{\rm{a}}}\; = \,\frac{{{K_{\rm{w}}}}}{{{K_{\rm{b}}}}}\; = \;\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ – 14}}}}}}{{{\rm{1}}{\rm{.8}}\; \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}}}\; = \;{\rm{5}}{\rm{.7 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

 

Once we know the K_a, it is the same procedure as we do when calculating the pH of a weak acid.

We assign x mol/l for ionization and set up an ICE table to determine the [H₃O⁺]:

 

NH₄⁺(aq)  + H₂O(l) → NH₃(aq)  +  H₃O⁺(aq)  (acidic)

	[NH4+]	[NH3]	[H3O+]
Initial	0.26	0	0
Change	-x	+x	+x
Equil	0.26 – x	x	x

 

Next, write the expression for K_a using the equilibrium concentrations in the ICE table:

 

\[{K_{{\rm{a}}}}\; = \,\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{][}}{{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}\; = \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.26}}\;{\rm{ – }}\;{\rm{x}}}}\; \approx \;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.26}}}} = \;{\rm{5}}{\rm{.7 x 1}}{{\rm{0}}^{{\rm{ – 10}}}}\]

Therefore, x = 1.2 x 10^-5

 

Check if the approximation was valid:

 

\[\% \, = \;\frac{{1.2\, \times \;{{10}^{ – 5}}}}{{0.26}}\; \times \;100\% \; = \;0.0047\% \]

 

The percentage of hydrolysis is less than 5%, and therefore, the approximation is valid.

 

The concentration of H₃O⁺ ions is 1.2 x 10^-5 M, so the pH is:

 

pH = -log 1.2 x 10^-5 = 4.9

 

This indicates an acidic solution which is what we expected to have for the solution of NH₄ClO₃.

 

Check Also

• Definitions of Acids and Bases
• Acid-Base Reactions
• Acid-Base Titrations
• Conjugate Acid and Conjugate Base
• Autoionization of Water and K_w
• The pH and Acidity
• Acid Strength, K_a, and pK_a
• Base Strength, K_b, and pK_b
• Ka, pKa, Kb, and pK_b Relationship
• The pH of a Strong Acid and Base
• pH + pOH = 14
• The pH of a Weak Acid
• The pH of a Weak Base
• The pH of Polyprotic Acids
• The acidity of a Salt Solution
• The pH of a Salt Solution
• The pH of Salts With Acidic Cations and Basic Anions
• pH Practice Problems
• Acids and Bases Practice Problems
  Acids and Bases Quiz