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Using the ICE method for neutralisation Mols Propanoic acid Mols NaOHMols Propanoate Initial (I)0.0113625 x10 -3 x.108 =.0027 0 Change (C)-0.0027.

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Presentation on theme: "Using the ICE method for neutralisation Mols Propanoic acid Mols NaOHMols Propanoate Initial (I)0.0113625 x10 -3 x.108 =.0027 0 Change (C)-0.0027."— Presentation transcript:

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9 Using the ICE method for neutralisation Mols Propanoic acid Mols NaOHMols Propanoate Initial (I)0.0113625 x10 -3 x.108 =.0027 0 Change (C)-0.0027 + 0.0027 Equilibrium (E) 0.00866 0 0.0027 Concentration = 0.100 L c propanoic acid = 0.0866 molL -1 c propanoate = 0.027 molL -1 pH = pKa + log [base] [acid] = 4.87 + log.027.0866 = 4.87 + log 0.31177 = 4.87 - 0.51 = 4.36

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16 Strong base- strong acid Strong base- weak acid Strong acid- weak base Weak base – weak acid pH at EP 7  8 8 8 8 5555 7777 Length of vertical region Long (  8 pH units) Short (  3 pH units) Not present pH at start 1111 3333  11 3333 Buffer zone Not present presentpresent 2 present

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