What are the concentrations of HSO4- and SO42- for a 0.0050 M sulfuric acid solution?
We are trying to find x but not sure how to find it on the calculator, please help!!!
Ah, calculating concentrations, huh? Well, don't worry, I'm here to lend a helping hand, or should I say, a helping rubber nose?
For a 0.0050 M sulfuric acid solution, we need to keep in mind that sulfuric acid (H2SO4) is a strong acid that fully dissociates in water.
That means it breaks apart into two hydrogen ions (H+) and one sulfate ion (SO42-). Since there's a 1:2 ratio between H2SO4 and SO42-, we know that for every mole of H2SO4, we get two moles of SO42-.
Therefore, the concentration of HSO4- (which is the same as H+) is the same as the concentration of H2SO4, which is 0.0050 M.
As for the concentration of SO42-, since every H2SO4 molecule gives us one SO42- ion, the concentration will be twice that of H2SO4, which is 0.0100 M.
So, to sum it up:
HSO4- concentration: 0.0050 M
SO42- concentration: 0.0100 M
Hope that clears things up!
To find the concentrations of HSO4- and SO42- in a sulfuric acid solution, you need to consider the dissociation of sulfuric acid in water.
Sulfuric acid (H2SO4) is a strong acid that dissociates in water to release H+ ions and the bisulfate (HSO4-) ion:
H2SO4 -> 2H+ + SO42-
Since sulfuric acid is a strong acid, it completely dissociates in water, so the initial concentration of H2SO4 (0.0050 M) is equal to the concentration of H+ ions (0.0050 M).
The concentration of HSO4- is equal to the concentration of H+ ions because for every H+ ion released, one molecule of HSO4- is formed. Therefore, the concentration of HSO4- is also 0.0050 M.
The concentration of SO42- can be determined by subtracting the concentration of HSO4- from the initial concentration of sulfuric acid:
SO42- = 0.0050 M - 0.0050 M
= 0 M
Thus, the concentration of HSO4- is 0.0050 M, and the concentration of SO42- is 0 M in the given sulfuric acid solution.
To find the concentrations of HSO4- and SO42- in a sulfuric acid solution, you need to understand the dissociation reaction of sulfuric acid in water.
Sulfuric acid (H2SO4) is a strong acid that ionizes completely in water. The dissociation reaction is as follows:
H2SO4 -> 2H+ + SO42-
From this reaction, we can see that for every 1 molecule of sulfuric acid, 1 molecule of HSO4- (hydrogen sulfate ion) and 1 molecule of SO42- (sulfate ion) are formed.
Given that the initial concentration of sulfuric acid (H2SO4) is 0.0050 M, we can assume that the concentrations of HSO4- and SO42- are also 0.0050 M.
Thus, the concentrations of HSO4- and SO42- in a 0.0050 M sulfuric acid solution are both 0.0050 M.
Regarding your question about finding "x" on the calculator, could you please provide more context or clarify what you are trying to solve for?
..........H2SO4 ==> H^+ + HSO4^-
I........0.0050M....0......0
C.......-0.005...0.005...0.005
E..........0.....0.005...0.005
So for the first ionization, H^+ = 0.005M and HSO4^- = 0.005M. Then HSO4^- goes through k2.
........HSO4^- ==> H^+ + SO4^2-
I.....0.005M....0.005....0
C........-x.........x....x
E.....0.005-x...0.005+x...x
k2 = (H^+)(SO4^2-)/(HSO4^-)
0.012 = (0.005+x)(x)/(0.005-x)
I used k2 from memory; I suggest you look it up in your text and use that. Then simply solve the quadratic equation. Or you can solve it by reiteration.