Post a New Question


posted by on .

How much acid is needed to neutralize a 5L solution with a pH of 12.5. The acid
available is 1 N HCL.

  • Chemistry - ,

    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.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question