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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.

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