Problem 6D.15

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Prasanna Padmanabham 4I
Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

Problem 6D.15

Postby Prasanna Padmanabham 4I » Tue Jan 21, 2020 7:53 pm

Hi Everyone,
I am having a really hard time with this problem. I would appreciate if someone can please let me know what I'm doing wrong.

6D.15 Calculate the pH of (a) 0.19 m NH4Cl(aq); (b) 0.055 m AlCl3(aq).

For part B) is the equation: AlCl3 <==> Al (OH-)3 + 3HCl. The 3HCl then dissociates completetly into 3H+ ions and 3Cl- ions...is that correct?
If so, I set up my equation like this: Ka = ((3x^3)*x)/[AlCl3]. I said that x is the concentration of Al(OH-) or the change in AlCl3 concentration. I said that H+ concentration would be 3x because there are 3H+ ions being formed for every Al ion. I used the Ka=1.4 * 10^-5 value for Aluminum ion on page 478. After I find X, I multiply it with 3 and find the -log of that number.

Thank you!
Prasanna

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

Re: Problem 6D.15

Postby Brooke Yasuda 2J » Tue Jan 21, 2020 9:31 pm

So when AlCl3 dissociates in water, Cl- is the conjugate base of a strong acid so will not react at all with water. However, Al will coordinate with six water molecules. After it coordinates with water, it can actually give off protons, or acidic hydrogens, making it a weak acid. So you would have to find the equilibrium concentration and calculate the pH from this.

Prasanna Padmanabham 4I
Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

Re: Problem 6D.15

Postby Prasanna Padmanabham 4I » Thu Jan 23, 2020 12:16 pm

Follow up question: I am still very stuck on this problem, can someone please post a step-by-step of the solution?
Thank you so much!


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