PH calculation

  • Thread starter Guillermo
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  • #1
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Here is the problem:

Given pKa=9.3 what is pH of 0.005 mol/l ammonia solution.

The problem is, I don't get this pKa part. I mean, I know t means Ka=5.0119*10^-10 but ammonia is a base, not acid.

G
 

Answers and Replies

  • #2
Borek
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pKa + pKb = pKw
 
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  • #4
Borek
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Please do not quote signatures.

No. What you have calculated is pOH, not pH. Besides, check the degree of dissociation - are you sure you can use simplified equation?
 
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  • #5
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Ah, so the pH is 10.5!

G
 
  • #6
Borek
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NO!

Equation you have used - [OH] = sqrt(C*Kb) is valid only if the degree of dissociation is below 5%. In this case it is higher:

3.16e-4(dissociated ammonia concentration)/0.005(ammonia concentration) * 100% = 6.3%

so you have to use full quadratic equation.

Smarkotan oz gluthozmaz
 
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  • #7
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Borek said:
NO!

so you have to use full quadratic equation.

:cry: I missed your last post and my result was wrong :cry:

G

PS Check my new problem, please...
 
  • #8
GCT
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Here is the problem:

Given pKa=9.3 what is pH of 0.005 mol/l ammonia solution.

The problem is, I don't get this pKa part. I mean, I know t means Ka=5.0119*10^-10 but ammonia is a base, not acid.

alright, take it step by step...to go from pKa to Ka, simply negative inverse log of pKa. KaKb=Kw, Kb=Kw/Ka (Kw is the autodissociation of water).

[tex]Kb=[OH-][NH4+]/[NH3],~Kb=[x][x]/[initial~conc.~NH3-x][/tex] solve for x, and that will be [0H-]. pOH=-log[0H-], pH+p0H=14, solve for pH.

that's it, you shouldn't be having so much trouble with this
 

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