[payata]

Hi,

In one of the tasks to practice, I can see that the author, calculated a task in two different way, somebody can explain to me why, I would calculate the same example A and B, exactly like in A (using [H+]= Ka x c weak acid), so why B equation  (which is [H+]^2= - Ka ± sq K²⁺_a + 4 x (Ka x c weak acid) / 2 ) is different?

Examples are in attach field

Thank you ina advance!

[Borek]

One approach uses approximations which makes calculations easier (but they are not always justified), see explanation here: http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[AWK]

Chemists like to count in a simplified way, but so that the calculation error does not exceed, e.g., 5%.
Insert c = 380K into both formulas. You will get [H +] = 19.5K or 19K, respectively - 1/20 just gives a 5% error. So for all concentrations for which c exceeds 380 K using the first simpler formula, we do not exceed 5% error in calculations, of course with correct rounding and without some other stupid error. Usually, we use the number 500 which is easy to divide c/K in memory (c divide by 1000, multiply the result by 2 and compare with K). In your case, 2x(0.1/1000) = 2x10^-4>K, so you can safely calculate with a simplified formula.

[payata]

Thank you for both explanations!