HW problem 12.35

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Paul Adkisson 1D
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HW problem 12.35

Postby Paul Adkisson 1D » Mon Nov 28, 2016 5:25 pm

Hey, so I know that the textbook mentions certain trends: For oxoacids, increasing the number of Oxygen atoms increases the relative strength of the acid. Also, for 2 oxoacids of different central atoms with the same number of Oxygen atoms, the central atom with the greater electronegativity is stronger. Now, on to the problem, the textbooks lists the relative strengths as H2SeO3 < H3PO4 < H3PO3 < HSeO4-
Why are they ordered like this even though H3PO4 has more Oxygens than H3PO3, and Electronegativity(Se) = 2.5 and Electronegativity(P) = 2.2?

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Re: HW problem 12.35

Postby Sarah_Heesacker_3B » Mon Nov 28, 2016 6:21 pm

Since the problem asked you to find the Ka values (which can be found by taking the antilog of the pKa value given), you can use those to determine the strength of the acid. A larger Ka value means the acid is stronger. Since H3PO3 has a Ka value of .01 and H3PO4 has a Ka value of .0076, H3PO3 has a larger Ka value and is therefore a stronger acid. Hope this helps!

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Re: HW problem 12.35

Postby Chem_Mod » Mon Nov 28, 2016 6:24 pm

This question gives you pKa values and asks you to find the Ka value and then put them in order of increasing strength. The given pKa are: H3PO4 = 2.12, H3PO3 = 2.00, H2SeO3 = 2.46, and HSeO4- = 1.92. So the lowest pKa is HSeO4-, then H3PO3, then H3PO4, and lastly H2SeO3. So based solely on the pKa values given and the general rule that the lower the pKa the stronger the acid, then the answer given is correct. The question isn't asking you to predict which one is stronger using the correlations of structure and acid strength, it's testing if you know that the smaller the pKa (and the larger the Ka) the stronger the acid.

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