Asked by Ryan
How much acid is needed to neutralize a 5L solution with a pH of 12.5. The acid
available is 1 N HCL.
available is 1 N HCL.
Answers
Answered by
DrBob222
pH = -log(H^+)
12.5 = -log(H^+)
(H^+) = 3.16E-13 M
Then (OH^-) = 1E-14/3.16E-13 = 0.0316
OH^- + HCl ==> H2O + Cl^-
mols OH = M x L = 0.0316*5L = 0.158 mols
mols H^+ needed from HCl= 0.158
M HCl = mol HCl/L HCl. You have mols and M, solve for L and convert to mL if needed.
12.5 = -log(H^+)
(H^+) = 3.16E-13 M
Then (OH^-) = 1E-14/3.16E-13 = 0.0316
OH^- + HCl ==> H2O + Cl^-
mols OH = M x L = 0.0316*5L = 0.158 mols
mols H^+ needed from HCl= 0.158
M HCl = mol HCl/L HCl. You have mols and M, solve for L and convert to mL if needed.