How Do You Calculate the pH of a Mixed NH4OH and NH4Cl Solution?

[tua28494]

This isn't really homework, Its just a problem I'm trying to learn to solve. I'm not taking a chemistry course so I'm learning on my own.


Calculate the pH of a solution that is 0.65 M NH₄OH and 0.35 M NH₄Cl.

Well we can calulate the pOH = -log(OH^-) and 14-pOH = pH


I guess I need the equilibrium concentration of OH^-.

here is the chemical formula. I believe is necessary

NH₄OH -> NH_{4SUB]^{+/SUP] + OH-/SUP]
the Kb/SUB] =1.5 x 10-3SUP]}}

 

[Borek]

This is a buffer solution, so you should use Henderson-Hasselbalch equation.

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methods

 

[tua28494]

Without using the Henderson–Hasselbalch equation, Can I do this?
also different Kb value.

The NN4CL will completely dissociate adding 0.35 NH4

NH4OH -> NH4+ + OH-
0.65-x. 0.35+x. . x

H2O + NH3 -> NH4+ + OH-

Kb =1.8 x 10^-5 = [NH4+][OH-]/[NH3]= (0.35+x)(x)/ 0.65-x

x = [OH-]=0.000033 M

pOH = 4.5
pH = 14 - pOH = 14 - 4.5 =9.5

 

[Borek]

This is correct by accident.

You look at the dissociation of NH₄OH, but you ignore acidic dissociation of NH₄⁺.

NH₄⁺ <-> NH₃ + H⁺

Why?

My bet is that if you will try to do calculations using your approach and based on this reaction, you will get almost the same result, again accidentally.

For very small x (compared with 0.35 & 0.65), is (0.35+x)/(0.65-x) substantially different from 0.35/0.65?

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methods

 

[DeeNos]

Hello! It's great that you are learning about pH and trying to solve problems on your own. To calculate the pH of this solution, we need to first understand the chemical reaction that is occurring. NH4OH is a weak base, which means it partially dissociates in water to form NH4+ and OH- ions. The equation you wrote is correct:

NH4OH -> NH4+ + OH-

The equilibrium constant for this reaction is called Kb, and it is equal to the concentration of NH4+ ions multiplied by the concentration of OH- ions, divided by the concentration of NH4OH. In this case, the concentration of NH4OH is 0.65 M, so we can set up the equation:

Kb = [NH4+][OH-]/[NH4OH]

Substituting the values we know, we get:

1.5 x 10^-3 = (0.35)(x)/(0.65)

where x represents the concentration of OH- ions. Solving for x, we get x = 4.62 x 10^-4 M.

Now, we can use this value to calculate the pOH and pH of the solution:

pOH = -log(4.62 x 10^-4) = 3.34

pH = 14 - pOH = 14 - 3.34 = 10.66

Therefore, the pH of this solution is 10.66. I hope this helps you understand how to solve problems like this in the future. Keep learning and exploring the world of chemistry!