Question

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

Answer 1

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.