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Finding the Molar Mass of a Diprotic Acid - Chemistry Practical watch

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    “You are provided with a diprotic acid H2X, relative molecular mass between 110 and 130. The acid is a white crystalline solid which is soluble in water and can be completely neutralised with sodium hydroxide solution of a suitable convention. Phenolphthalein is a suitable indicator.”

    To work out the molar mass of the acid, I have to carry out a titration.

    I have written the balanced equation
    2NaOH + H2X -> Na2X + 2H2O

    And have decided to use alkali/sodium hydroxide of concentration 0.05moldm-3.

    However, I have to make a sensible suggestion for the concentration of acid I am going to use, and a calculation of the mass of acid and volume of solution I need to make up. How do I do this?
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    (Original post by jesss_)
    “You are provided with a diprotic acid H2X, relative molecular mass between 110 and 130. The acid is a white crystalline solid which is soluble in water and can be completely neutralised with sodium hydroxide solution of a suitable convention. Phenolphthalein is a suitable indicator.”

    To work out the molar mass of the acid, I have to carry out a titration.

    I have written the balanced equation
    2NaOH + H2X -> Na2X + 2H2O

    And have decided to use alkali/sodium hydroxide of concentration 0.05moldm-3.

    However, I have to make a sensible suggestion for the concentration of acid I am going to use, and a calculation of the mass of acid and volume of solution I need to make up. How do I do this?
    Suppose we want to use about 20cm^3 of NaOH in our titration - how many moles is that? How many moles of our diprotic acid would that react with? What mass of acid would that be, assuming an Mr of: 1) 110; 2) 130?

    If we use a similar mass of acid, we can then figure out how much NaOH we need to neutralise it in our titration (remember to add the indicator to the acid), and hence how many moles are actually there - now we have both the mass of acid we used and the number of moles of it.

    Since we havent talked at all about the acid’s concentration so far, it doesn’t matter what it is! We might want to use a similar volume, about 20cm^3 of water with whatever mass of acid we chose, but the theory works with 1000cm^3 of water, or even no water at all.
 
 
 
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