Relationship between pka, ka, and acidity strength

Acidity
Basicity
The Conjugate Seesaw

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Katie 1E
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Relationship between pka, ka, and acidity strength

Postby Katie 1E » Fri Dec 08, 2017 12:41 am

What is the relationship between pka, ka, and acid strength and how do they relate to pkb, kb, and base strength?

Vivian Nguyen
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Re: Relationship between pka, ka, and acidity strength

Postby Vivian Nguyen » Fri Dec 08, 2017 1:21 am

The higher the Ka the stronger the acid. Think about the calculation to get the equilibrium constant - [Products]/[Reactants]. As a result, the reaction would favor a more complete dissociation of the weak acid, thus giving it a stronger ability to donate a proton (strong acid).

pKa is the opposite though, the larger the pKa, the weaker the acid is. This can be shown by just calculating pKA = -logKa.

The same relationship goes for kb and pKb. The larger the kb, the stronger the base, and the larger the pKb, the weaker the base.

Putting them together, you get the "conjugate seesaw". So the stronger the acid, the weaker the conjugate base. The stronger the base, the weaker it's conjugate acid. This can be seen through the equation pKa + Pkb = pKw


I hope this could answer your questions!

Aliza Ajmal 1D
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Joined: Fri Sep 29, 2017 7:05 am

Re: Relationship between pka, ka, and acidity strength

Postby Aliza Ajmal 1D » Fri Dec 08, 2017 9:21 am

A way to remember that acids are stronger when pKa is smaller is to remember that the smaller/lower the pH, the stronger an acid is. I think Dr. Lavelle mentioned this way to remember that the low pKa = strong acid relationship is similar to pH during lecture

Sarah Brauer
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Re: Relationship between pka, ka, and acidity strength

Postby Sarah Brauer » Fri Dec 08, 2017 10:01 am

What is pKw?

Vivian Nguyen
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Re: Relationship between pka, ka, and acidity strength

Postby Vivian Nguyen » Sat Dec 09, 2017 4:17 pm

pKw =log Kw

Steven Chau 1B
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Re: Relationship between pka, ka, and acidity strength

Postby Steven Chau 1B » Sat Dec 09, 2017 7:58 pm

pKw = -log Kw = -log(1.0*10^-14)=14.00 at 25C


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