[H+] calculation

[Oh_Gee]

16. i understand how to find [H+] from Ka1 but i don't understand how the got [H+] from Ka2. if the [H+] value in Ka2= [H+][SO42-] / [HSO4-] is supposed to be 1, then why do they add 1(from Ka1) + .012? how can [H+] from Ka2 be both 1 and .012

17. what's a good math trick to find square root of 7.5 x 10 ^-3

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[Chrisz]

The first deprotonation is complete, 1M H+ contributed by H2SO4, and 1M HSO4-
You know hso4- is a weak acid, so you would expect it to contribute to the H+ concentration as well

Ka2=H(SO42-)/HSO4-
=(1+x)(x)/(1-x) Initially, we have 1M H and 1M HSO4-, so they have to be incoporated into the equation
Since it is a weak acid and the initial concentrations of HSO4- and H is large (1M each), we can do an approximation
=1(x)/1
S0, Ka2=x=0.012 which is the concentration of H+ contributed by dissociation of HSO4-
Our approximation is valid because it is within 5% of 1M
Now, we have to calculate the total concentration contribute by both the first deprotonation and second deprotonation
1+0.012=1.012M

 

[Oh_Gee]

The first deprotonation is complete, 1M H+ contributed by H2SO4, and 1M HSO4-
You know hso4- is a weak acid, so you would expect it to contribute to the H+ concentration as well

Ka2=H(SO42-)/HSO4-
=(1+x)(x)/(1-x) Initially, we have 1M H and 1M HSO4-, so they have to be incoporated into the equation
Since it is a weak acid and the initial concentrations of HSO4- and H is large (1M each), we can do an approximation
=1(x)/1
S0, Ka2=x=0.012 which is the concentration of H+ contributed by dissociation of HSO4-
Our approximation is valid because it is within 5% of 1M
Now, we have to calculate the total concentration contribute by both the first deprotonation and second deprotonation
1+0.012=1.012M

so in calc of Ka2, [H+] is both 1 and .012?

also in the other problem i uploaded, they didn't even consider ka2 in that cause it's "negligible" what is the cutoff for that?

 

[Chrisz]

In fact, I would simply ignore the second pka, if it is extremely small. for example pka=10^-5
If I was you, i would simply ignore the second dissosciation, simply knowing the second dissociation contributes to a tiny amount of H+, and pick up B as the answer. If you have taken the aamc practice, the difference between answers are usually large, and do not require complicated calculations. In the second case, the difference between the answers are large enough, and you do not need to consider the second pKa and and third pKa.