Enough of a monoprotic acid is dissolved in water to produce a 0.0167 M solution

Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0167 M solution. The pH of the resulting solution is 6.03. Calculate the Ka for the acid

I got Ka= 5.21 x 10-11

It is telling me:

Inccorrect. You treated H+ concentration as 0. in this case, the initial 10-7 M H+ from water is not negligible. if [H+]initial=10-7 and [H+]final=9.3x10-7, by how much did hte concentration change? What does that say about how much A- was produced?

I keep reworking it but am cofused and cant get the right answer.

Explanation / Answer

well I think you followed the following procedure:
ka = ([H3O] 2) / (HA-H3O)
ka = [9.3325 * 10-7] 2) / (0.0167-9.3325 * 10-7)
ka = 5.215642306 * 10-11

I think what mejorte calls redondees or put many significant sifras, but besides that okay, sounds a bit incongruous that tell you that the concentration of product is 0.0167 and being told that their pH is 6.03, and is supposed to H3O and A- are the same, and raising H3O pH pulls negative

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