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More Titration Q - SG Chem

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  • Message 1. 

    Posted by MissChemTeacher_Bitesize (U15172316) on Thursday, 10th January 2013

    Hey!

    Here are some titration questions you were after. If you need help answering any of them get in touch.

    1) A titration found that 20.2cm³ of sodium hydroxide was neutralised by 15.5cm³ of sulphuric acid with a concentration of 0.1 mol/l.

    Use this information and the balanced equation to calculate the concentration of sodium hydroxide.

    H₂SO₄ + 2NaOH ----------> Na₂SO₄ + 2H₂O


    2) In a titration 10cm³ of potassium hydroxide solution was neutralised by 12cm³ of 1.2mol/l nitric acid.
    Use this information and balanced equation to calculate the concentration of potassium hydroxide.

    KOH + HNO₃ ----------> KNO₃ + H₂O


    Good Luck with these questions Miss_Chem smiley - cool

    Report message1

  • Message 2

    , in reply to message 1.

    Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 21st January 2013

    Hey Guys!
    Here are the solutions to these titration problems.

    H₂SO₄ + 2NaOH ----------> Na₂SO₄ + 2H₂O

    n of moles of H₂SO₄ = concentration x volume in litres
    n of moles of H₂SO₄ = 0.1 x 15.5/1000
    n of moles of H₂SO₄ = 0.00155 moles

    Use the ratio: H₂SO₄ ……:…….2NaOH
    Mole Ratio:………1………….:………..2
    n from above:..0.0015..:…..0.0031 moles

    concentration of NaOH = n of moles / volume in litres
    concentration of NaOH = 0.0031 / 0.01 litres
    concentration of NaOH = 0.153 mol/l

    Report message2

  • Message 3

    , in reply to message 2.

    Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 21st January 2013

    Whoops! The volume of NaOH should've been 0.0202l

    Report message3

  • Message 4

    , in reply to message 3.

    Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 21st January 2013

    Here is the solution to question 2

    KOH + HNO₃ ----------> KNO₃ + H₂O

    n of moles of HNO₃ = concentration x volume in litres
    n of moles of HNO₃ = 1.2 x 12/1000
    n of moles of HNO₃ = 0.0144 moles

    Use the ratio: HNO₃ : KOH
    Mole Ratio:…….1……:……1
    n from above: 0.0144: 0.0144

    concentration of KOH = n of moles / volume in litres
    concentration of KOH = 0.0144 / 0.01
    concentration of KOH = 1.44 mol/l

    Hope these solutions helped you out.

    Cheers Miss_Chem smiley - cool

    Report message4

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