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
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:
[tex]\mathbf{NH_4^+ \to NH_3 + H^+}[/tex]
The Henderson Hasselbalch equation for the dissociation of ammonium can be expressed by using the formula:
[tex]\mathbf{pH = pKa + log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
Given that:
[tex]\mathbf{9.40= 9.25 + log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{9.40- 9.25 = log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{0.15 = log \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{10^{0.15} = \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{1.41 = \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
[tex]\mathbf{\dfrac{141}{100} = \dfrac{[NH_3 ] }{[ NH_4^+] }}[/tex]
∴
The percentage of NH₄⁺ [tex]\mathbf{= \dfrac{141}{100+141}\times 100\%}[/tex]
= 58.5%
The percentage of NH₃ [tex]\mathbf{= \dfrac{100}{100+141} \times 100\%}[/tex]
= 41.5%
Learn more about the Henderson-Hasselbalch equation here:
https://brainly.com/question/13651361
