Asked by Anonymous
For 520ml water calculate initial and final pH after adding 0.02ml HCl
Answers
Answered by
DrBob222
pure H2O:
(H^+)(OH^-) = Kw = 1E-14
(H^+) = sqrt Kw and pH = 7
Add0.02mL HCl. What strength HCl? 0.02mL doesn't tell me much. It is conc HCl that is 12M so new molarity would be
12M x 0.02/520 = ? and pH = -log(H^+).
(H^+)(OH^-) = Kw = 1E-14
(H^+) = sqrt Kw and pH = 7
Add0.02mL HCl. What strength HCl? 0.02mL doesn't tell me much. It is conc HCl that is 12M so new molarity would be
12M x 0.02/520 = ? and pH = -log(H^+).