G.Chem concentration question

[swooth01]

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This is just a question that I had and the numbers are just made up by me. If you were to ask to find the concentration of H+ in a 2.0 M solution of sulfuric acid (H2SO4) with a Ka = 1.0 x 10^-5

Since sulfuric acid has 2 H+ would you set up the question

Ka = [2x][x] / 2.0

solve for x and than would your concentration of H+? or would I have to multiple it by 2 after I solve for x?

or would I just solve Ka = [x][x] / 2.0
solve for x and than multiple that x by 2 ( for the 2 H+ from sulfuric acid)?

Thanks in advance.

 

[whoaaitzkyle]

I have an example written down in my notes and it says to solve for x and then multiply it by 2 before calculating the pH.

pH = -log 2(your value of [H+])

 

[Shunwei]

This is just a question that I had and the numbers are just made up by me. If you were to ask to find the concentration of H+ in a 2.0 M solution of sulfuric acid (H2SO4) with a Ka = 1.0 x 10^-5

Since sulfuric acid has 2 H+ would you set up the question

Ka = [2x][x] / 2.0

solve for x and than would your concentration of H+? or would I have to multiple it by 2 after I solve for x?

or would I just solve Ka = [x][x] / 2.0
solve for x and than multiple that x by 2 ( for the 2 H+ from sulfuric acid)?

Thanks in advance.

No you would not multiply it by two. Polyprotic acids have distinct Ka's for each of their distinct H's. Therefore, H2SO4, as given in your example, would have a Ka1 and Ka2, separated by about 10-4 magnitude. You would just calculate each dissociation separately and not as a whole.

And by the way, the case of H2SO4 as givein this example as a weak acid is wholly incorrect. You should know that it, along with sulfuric, percloric, hydrobromic, hydroiodic, and hydrochloric acids are the the prototypical strong acids, i.e. complete dissociation. A better example in this case would have been something like carbonic acid, H2CO3.

 

[jackbauer!]

No you would not multiply it by two. Polyprotic acids have distinct Ka's for each of their distinct H's. Therefore, H2SO4, as given in your example, would have a Ka1 and Ka2, separated by about 10-4 magnitude. You would just calculate each dissociation separately and not as a whole.

And by the way, the case of H2SO4 as givein this example as a weak acid is wholly incorrect. You should know that it, along with sulfuric, percloric, hydrobromic, hydroiodic, and hydrochloric acids are the the prototypical strong acids, i.e. complete dissociation. A better example in this case would have been something like carbonic acid, H2CO3.

👍 Ka values for strong acids and bases are extremely high b/c they dissociate completely... you're 1 x 10-4 Ka is a bit unrealistic.

jb!🙂

 

[swooth01]

Thanks for the help guys