Asked by Nicollette
suppose that after measuring the pH of 1.05 M HCl solution, three drops (0.20 mL) of the solution remained behind on the pH electrode.
a) when this electrode is then dipped into 100 mL of pure distilled water, show by calculation that the pH of the water is not pH= 7
a) when this electrode is then dipped into 100 mL of pure distilled water, show by calculation that the pH of the water is not pH= 7
Answers
Answered by
DrBob222
HCl = 1.05 x 0.2/100 = about 0.0021M
..........H2O ==> H^+ + OH^-
I...............0.0021...0
C.................+x.....+x
E.............0.0021-x....x
Kw = (H^+)(OH^-)
1E-14 = (0.0021+x)(x)
Solve for x and convert H^+ to pH.
I don't think the ionization of H2O even enters into it.
..........H2O ==> H^+ + OH^-
I...............0.0021...0
C.................+x.....+x
E.............0.0021-x....x
Kw = (H^+)(OH^-)
1E-14 = (0.0021+x)(x)
Solve for x and convert H^+ to pH.
I don't think the ionization of H2O even enters into it.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.