Acid/Base question

[SilverBandCry!]

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Also what is the pH of a buffer that consists of 0.17M NH3 and 0.25M NH4Cl?

I know you can use the equation pH = pka + log (base/acid)
but I was only given that the kb of NH3 = 1.76E-5.

So how would I find Ka?

Thanks

 

[SilverBandCry!]

Bump does anyone know?

 

[kidthor]

Also what is the pH of a buffer that consists of 0.17M NH3 and 0.25M NH4Cl?

I know you can use the equation pH = pka + log (base/acid)
but I was only given that the kb of NH3 = 1.76E-5.

So how would I find Ka?

Thanks

Kb*Ka=Kw

Kw = 10^-14

As a side note: in the equation you listed above, it is log(conjugate base/acid) and not just "base."

 

[SilverBandCry!]

As a side note: in the equation you listed above, it is log(conjugate base/acid) and not just "base."[/QUOTE]


oh you're right. Thanks!

 

[SilverBandCry!]

Kb*Ka=Kw

Kw = 10^-14

As a side note: in the equation you listed above, it is log(conjugate base/acid) and not just "base."

oh damn. So can you use the concentrations 0.17M NH3 and 0.25M NH4Cl? How would you calculate the conjugate base and conjugate acid?

 

[lawj]

oh damn. So can you use the concentrations 0.17M NH3 and 0.25M NH4Cl? How would you calculate the conjugate base and conjugate acid?

NH4Cl dissociates completely in water, so in essence you have NH3 and NH4+. NH4 is the conjugate acid of NH3 (just follow the proton), so you plug those concentrations into your above equation, where the pKa will will be for NH4+ (which is 14 - pKb).

 

[EBI831]

Also what is the pH of a buffer that consists of 0.17M NH3 and 0.25M NH4Cl?

I know you can use the equation pH = pka + log (base/acid)
but I was only given that the kb of NH3 = 1.76E-5.

So how would I find Ka?

Thanks

Hey this is what I'd do. (1)use -log[1.76 x 10-5] = 5-.176= 4.82 this is pKb about (2) pKa + pKb =14, I think so pKa is 14-4.82 = 9.18 (3) so you solve that pH = 9.18 + log(.17/.25) = 9.01 for the pH.