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    Question:
    Chocolate mousse contains gelatine and a compound to promote fast setting of the mousse.
    Compound A is such a setting agent. It has two acidic hydrogen atoms per molecule and is one of the six acids listed below.

    oxalic acid malonic acid succinic acid glutaric acid adipic acid pimelic acid
    HOOCCOOH
    HOOCCH2COOH
    HOOC(CH2)2COOH
    HOOC(CH2)3COOH
    HOOC(CH2)4COOH
    HOOC(CH2)5COOH

    The student analysed a sample of compound A by titration.
    The student dissolved 2.82 g of compound A in water and made the solution up to 250
    cm3 in a volumetric flask. He titrated 25.0 cm3 of this solution with 0.175 mol dm–3 NaOH.
    22.05 cm3 of NaOH were required for complete neutralisation.
    Use the results of the student’s analysis to identify compound A from the list above. Show all of your working.

    Answer:
    amount of NaOH in titration = 0.175 x 22.05/1000 or 3.86 × 10–3 (1) (calc: 3.85875 x 10–3)
    amount of A in 25.0 cm3 = 0.5 × mol NaOH or 1.93 × 10–3 (1) (calc: 1.929375 × 10–3)
    amount of A in 250cm3 =10×1.93×10–3 or1.93×10–2 (1) 1.93×10–2 molA has a mass of 2.82g
    molar mass of A = 2.82/1.93 × 10–2 = 146 g mol–1 (1) (or Mr of A is 146)
    Therefore A is adipic acid / HOOC(CH2)4COOH (1)

    why do i have to halve (*0.5) the moles of naoh. is this because there are two acidic h+ per molecule of A
    why is this?
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    (Original post by esmeralda123)
    Question:
    Chocolate mousse contains gelatine and a compound to promote fast setting of the mousse.
    Compound A is such a setting agent. It has two acidic hydrogen atoms per molecule and is one of the six acids listed below.

    oxalic acid malonic acid succinic acid glutaric acid adipic acid pimelic acid
    HOOCCOOH
    HOOCCH2COOH
    HOOC(CH2)2COOH
    HOOC(CH2)3COOH
    HOOC(CH2)4COOH
    HOOC(CH2)5COOH

    The student analysed a sample of compound A by titration.
    The student dissolved 2.82 g of compound A in water and made the solution up to 250
    cm3 in a volumetric flask. He titrated 25.0 cm3 of this solution with 0.175 mol dm–3 NaOH.
    22.05 cm3 of NaOH were required for complete neutralisation.
    Use the results of the student’s analysis to identify compound A from the list above. Show all of your working.

    Answer:
    amount of NaOH in titration = 0.175 x 22.05/1000 or 3.86 × 10–3 (1) (calc: 3.85875 x 10–3)
    amount of A in 25.0 cm3 = 0.5 × mol NaOH or 1.93 × 10–3 (1) (calc: 1.929375 × 10–3)
    amount of A in 250cm3 =10×1.93×10–3 or1.93×10–2 (1) 1.93×10–2 molA has a mass of 2.82g
    molar mass of A = 2.82/1.93 × 10–2 = 146 g mol–1 (1) (or Mr of A is 146)
    Therefore A is adipic acid / HOOC(CH2)4COOH (1)

    why do i have to halve (*0.5) the moles of naoh. is this because there are two acidic h+ per molecule of A
    why is this?
    yes
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    (Original post by charco)
    yes
    but the mistake i would make is to double.
    why halve and not double since there are two moles of h+
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    (Original post by esmeralda123)
    but the mistake i would make is to double.
    why halve and not double since there are two moles of h+
    because you are working out how much of the acid you are reacting the base with ... which since it is diprotic is half the amount of NaOH
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