Understanding of pKa and Ammonia Solution

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SUMMARY

The discussion centers on calculating the pH of a 0.005 mol/L ammonia solution given a pKa of 9.3. Participants clarify that pKa corresponds to a Ka of 5.0119 x 10^-10, and emphasize that ammonia, being a weak base, requires the use of the full quadratic equation for accurate pH calculation due to a degree of dissociation exceeding 5%. The correct pH is determined to be 10.5 after applying the appropriate calculations, including the relationship between Ka, Kb, and Kw.

PREREQUISITES
  • Understanding of pKa and pKb concepts
  • Knowledge of weak base dissociation
  • Familiarity with the quadratic equation for chemical equilibria
  • Basic skills in logarithmic calculations
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  • Learn about the relationship between Ka, Kb, and Kw in acid-base chemistry
  • Study the quadratic formula application in chemical equilibrium problems
  • Explore the concept of degree of dissociation in weak acids and bases
  • Investigate the use of chemical calculators for pH and concentration calculations
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Chemistry students, educators, and professionals involved in acid-base chemistry, particularly those focusing on weak bases and pH calculations.

Guillermo
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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
 
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pKa + pKb = pKw
 
Last edited by a moderator:
Please do not quote signatures.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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Ah, so the pH is 10.5!

G
 
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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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...
 
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).

Kb=[OH-][NH4+]/[NH3],~Kb=[x][x]/[initial~conc.~NH3-x] 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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