Question

If you prepared a solution by adding 0.05mL (1drop) of 0.5M HCL to 5 x 10^7 mL of "pure" water. Calc the resultant PH. The answer isn't ph= 9.3. you have to add all the hydronium ions from all sources... but how ?

Answer 1

I would do this but check my thinking.
(HCl) added = 0.5M x (0.05/5E7)= 5E-10M
..............H2O ==> H^+ + OH^-
initial...............0......0
change.................x......x
add 5E-10M...........5E-10....x
final..............x+5E-10.....x

Kw = (H^+)(OH^-) = 1E-14
(x+5E-10)(x)=1E-14
Multiply out to obtain a quadratic that looks like this.
x^2 + 5E-10x - 1E-14 = 0

Solve the quadratic and I get something like 9.975E-8 = x. (I know that's more significant figures than we can legally use; however, if we round everything down to one place (from the 0.5M and the 0.05 mL) it would be 7 anyway and there would be no need to do any of this calculation.
Then x+5E-10 = ?? and pH from there. Close to 7 but less than 7 as it should be. It looks logical anyway.