what is the final pH after 1 drop(.05ml) of 6M HCl is added to 1.0L of freshly prepared pure water that was originally at a pH of 7.0. Is there a signigicant pH change?

Answer 1

moles HCl added = M x L = 6M x 0.00005 L = 3 x 10^-4 moles.
Neglecting the (H^+) from the ionization of water (because it is too small),
M = 3 x 10^-4 moles/1.00005 L or 3 x 10^-4.
pH = log (3 x 10^-4) = ??
I get about 3.5 or so but you need to go through the calculations

Answer 2

thank you sooo much :)