chemistry

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A 23.4 mL sample of 0.300 M trimethylamine, (CH3)3N, is titrated with 0.399 M perchloric acid.

After adding 6.67 mL of perchloric acid, the pH is

  • chemistry -

    mols (CH3)3N = M x L = ?
    mols HClO4 = M x L = ?
    mol(CH3)3N in excess = initial mols - mols HClO4 and mols (CH3)3N/total volume which is 23.4 mL + 6.67 mL (substitute as L).
    This is a weak base that react with H2O to give
    ...(CH3)3N + H2O==> (CH3)3NH^+ + OH^-

    Set up and ICE chart, solve for x = OH^- and convert to pH.

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