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A 30.0 cm3 solution of sodium hydroxide of concentration 8.0 g/L is used to neutralize a dibasic acid H2A. Calculate the relative formula mass of the dibasic acid.

Chemistry

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Answer

98

Step-by-step solution

6 steps

  1. Calculate the molar mass of sodium hydroxide (NaOH\text{NaOH}).

    Molar mass of NaOH=Na+O+H=23.0+16.0+1.0=40.0g/mol\text{Molar mass of NaOH} = \text{Na} + \text{O} + \text{H} = 23.0 + 16.0 + 1.0 = 40.0\text{g/mol}

  2. Calculate the concentration of NaOH\text{NaOH} in mol/dm3^3.

    Concentration of NaOH=Mass per litreMolar mass=8.0g/L40.0g/mol=0.2mol/dm3\text{Concentration of NaOH} = \frac{\text{Mass per litre}}{\text{Molar mass}} = \frac{8.0\text{g/L}}{40.0\text{g/mol}} = 0.2\text{mol/dm}^3

  3. Calculate the moles of NaOH\text{NaOH} used.

    Volume of NaOH=30.0cm3=30.01000dm3=0.030dm3\text{Volume of NaOH} = 30.0\text{cm}^3 = \frac{30.0}{1000}\text{dm}^3 = 0.030\text{dm}^3 Moles of NaOH=Concentration×Volume\text{Moles of NaOH} = \text{Concentration} \times \text{Volume} Moles of NaOH=0.2mol/dm3×0.030dm3=0.006mol\text{Moles of NaOH} = 0.2\text{mol/dm}^3 \times 0.030\text{dm}^3 = 0.006\text{mol}

  4. Write the balanced equation for the neutralization reaction.

    Let the dibasic acid be H2A\text{H}_2\text{A}.

    H2A(aq)+2NaOH(aq)→Na2A(aq)+2H2O(l)\text{H}_2\text{A}(\text{aq}) + 2\text{NaOH}(\text{aq}) \to \text{Na}_2\text{A}(\text{aq}) + 2\text{H}_2\text{O}(\text{l})

  5. Determine the moles of the dibasic acid (H2A\text{H}_2\text{A}).

    From the balanced equation, 1mol1\text{mol} of H2A\text{H}_2\text{A} reacts with 2mol2\text{mol} of NaOH\text{NaOH}.

    Moles of H2A=12×Moles of NaOH=12×0.006mol=0.003mol\text{Moles of H}_2\text{A} = \frac{1}{2} \times \text{Moles of NaOH} = \frac{1}{2} \times 0.006\text{mol} = 0.003\text{mol}

  6. Calculate the relative formula mass of the dibasic acid.

    Relative formula mass of H2A=Mass of H2AMoles of H2A\text{Relative formula mass of H}_2\text{A} = \frac{\text{Mass of H}_2\text{A}}{\text{Moles of H}_2\text{A}} Relative formula mass of H2A=0.294g0.003mol=98g/mol\text{Relative formula mass of H}_2\text{A} = \frac{0.294\text{g}}{0.003\text{mol}} = 98\text{g/mol}

    The relative formula mass of the dibasic acid is 98.

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