Question

Using the Henderson-Hasselbalch equation, calculate the percentage of an ammonia solution that would be in the ionic form at pH 9.40, assuming a pKa for the ammonium ion of 9.25.

Answer 1

Answer:

41.45% is the precentage of the ionic form

Explanation:

H-H equation is used to find pH of a buffer. The formula is:

pH = pKa + log₁₀ [X⁻] / [HX]

Where X⁻ is the conjugate base of the weak acid, HX.

In the ammonia buffer:

NH₄⁺ ⇄ NH₃ + H⁺

NH₄⁺ is the weak acid (Ionic form) and NH₃ is the conjugate base. Replacing:

9.40 = 9.25 + log₁₀ [NH₃] / [NH₄⁺]

1.4125 = [NH₃] / [NH₄⁺] (1)

In a percentage, [NH₃] + [NH₄⁺] = 100 (2)

Replacing (2) in (1):

1.4125 = 100 - [NH₄⁺] / [NH₄⁺]

1.4125[NH₄⁺] = 100 - [NH₄⁺]

2.4125[NH₄⁺] = 100

[NH₄⁺] = 41.45% is the precentage of the ionic form

Answer 2

The percentage of an ammonia solution is 41.5%

The Henderson Hasselbalch equation is an approximation that demonstrates the relation among a solution's pH or pOH, as well as the concentration ratio for the dissociated chemical species.

The dissociation of ammonium ion in the buffer is as follow:

$\mathbf{NH_4^+ \to NH_3 + H^+}$

The Henderson Hasselbalch equation for the dissociation of ammonium can be expressed by using the formula:

$\mathbf{pH = pKa + log \dfrac{[NH_3 ] }{[ NH_4^+] }}$

Given that:

• pH = 9.40
• pKa = 9.25

$\mathbf{9.40= 9.25 + log \dfrac{[NH_3 ] }{[ NH_4^+] }}$

$\mathbf{9.40- 9.25 = log \dfrac{[NH_3 ] }{[ NH_4^+] }}$

$\mathbf{0.15 = log \dfrac{[NH_3 ] }{[ NH_4^+] }}$

$\mathbf{10^{0.15} = \dfrac{[NH_3 ] }{[ NH_4^+] }}$

$\mathbf{1.41 = \dfrac{[NH_3 ] }{[ NH_4^+] }}$

$\mathbf{\dfrac{141}{100} = \dfrac{[NH_3 ] }{[ NH_4^+] }}$

∴

The percentage of NH₄⁺ $\mathbf{= \dfrac{141}{100+141}\times 100\%}$

= 58.5%

The percentage of NH₃ $\mathbf{= \dfrac{100}{100+141} \times 100\%}$

= 41.5%

Learn more about the Henderson-Hasselbalch equation here:

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