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maacastrobr

Answer:

0.692

Explanation:

To solve this problem we can use Henderson-Hasselbach's (H-H) equation:

  • pOH = pKb + log [tex](\frac{[NH_4Cl]}{[NH_3]})[/tex]

In which pOH can be calculated from the pH:

  • pOH + pH = 14
  • pOH = 14 - pH = 14 - 9.09
  • pOH = 4.91

Then we use the H-H equation to calculate[tex]\frac{[NH_4Cl]}{[NH_3]}[/tex]:

  • 4.91 = 4.75 + log [tex](\frac{[NH_4Cl]}{[NH_3]})[/tex]
  • 0.16 = log [tex](\frac{[NH_4Cl]}{[NH_3]})[/tex]
  • [tex]10^{0.16}[/tex] = [tex]\frac{[NH_4Cl]}{[NH_3]}[/tex]
  • [tex]\frac{[NH_4Cl]}{[NH_3]}[/tex]= 1.44

As the problem asks for the ratio of ammonia to ammonium chloride, we invert [tex]\frac{[NH_4Cl]}{[NH_3]}[/tex]:

  • [tex]\frac{[NH_3]}{[NH_4Cl]}=\frac{1}{1.44}[/tex] = 0.692

The mole ratio of ammonia to ammonium chloride in a buffer with a pH of 9.09 is 0.692.

How we write Henderson Hasselbalch equation?

Henderson Hasselbalch equation for the weak base (NH₃) and for their conjugate acid (NH₄Cl) will be written as:

pOH = pKb + log[NH₄Cl]/[NH₃], where

pKb = base dissociation constant = 4.75

pOH will be calculated from the given value of pH as:

pH + pOH = 14

pOH = 14 - 9.09 = 4.91

Now we these values in the above equation and we get,

4.91 = 4.75 + log[NH₄Cl]/[NH₃]

log[NH₄Cl]/[NH₃] = 0.16

[NH₄Cl]/[NH₃] = 10⁰°¹⁶

[NH₄Cl]/[NH₃] = 1.44

[NH₃]/[NH₄Cl] = 1/1.44

[NH₃]/[NH₄Cl] = 0.692

Hence, required mole ratio is 0.692.

To know more about Henderson Hasselbalch equation, visit the below link:

https://brainly.com/question/26746644

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