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    I am required to obtain an accurate value of molarity (concentraton) of a given solution of Sodium Hydroxide. The Standardisation is carrid out by titration against a known and accurate weight of the primary standard KHP (Given Mr: 204.22)

    Mass of KHP/g = 0.6002
    NaOH titre value 1/ml = 29.85 ml

    What s the Molarity of the NaOH solution?

    Any help provided will be appreciated.

    I keep getting 0.0008773 moldm-3... Says I'm wrong.Oddly I find these type of equations extremely simple, but for some reason I can't get this I feel like I'm making a stupid mistake somewhere...
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    (Original post by TonyGuacamole)
    I am required to obtain an accurate value of molarity (concentraton) of a given solution of Sodium Hydroxide. The Standardisation is carrid out by titration against a known and accurate weight of the primary standard KHP (Given Mr: 204.22)

    Mass of KHP/g = 0.6002
    NaOH titre value 1/ml = 29.85 ml

    What s the Molarity of the NaOH solution?

    Any help provided will be appreciated.

    I keep getting 0.0008773 moldm-3... Says I'm wrong.Oddly I find these type of equations extremely simple, but for some reason I can't get this I feel like I'm making a stupid mistake somewhere...
    did you write a balanced equation?
    did you use mol=mass/mr and n=v*c?
    did you convert your units correctly?
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    (Original post by thefatone)
    did you write a balanced equation?
    did you use mol=mass/mr and n=v*c?
    did you convert your units correctly?
    KHC8H4O4 + NaOH -----> KNaC8H4O4 + H2O

    I worked out the number of moles for KHP which is 0.6002/204.22 = 0.00293899

    I assumed this means C= 0.00293899 x (29.85/1000)?

    This equals 0.00008773, 0.0001 in 4 decimal places.
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    what volume KHP reacted with 29.85cm^3 of NaOH?


    ^^ ignore this
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    (Original post by TonyGuacamole)
    KHC8H4O4 + NaOH -----> KNaC8H4O4 + H2O

    I worked out the number of moles for KHP which is 0.6002/204.22 = 0.00293899

    I assumed this means C= 0.00293899 x (29.85/1000)?

    This equals 0.00008773, 0.0001 in 4 decimal places.
    also c=1000n/v assuming the volume is in cm^3
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    so using c=1000n/v
    i got 1000*2.9x10^-3/29.85= 0.098 molarity
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    (Original post by thefatone)
    so using c=1000n/v
    i got 1000*2.9x10^-3/29.85= 0.098 molarity
    Yeah I realised my mistake after a few second after posting this! Thank You anyway )
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    (Original post by thefatone)
    so using c=1000n/v
    i got 1000*2.9x10^-3/29.85= 0.098 molarity
    Also could I ask you another question if you don't mind? It's the same thing but for titre 2. This time the grams are 0.6066.
    KHP (Given Mr: 204.22)

    Mass of KHP/g = 0.6066
    NaOH titre value 1/ml = 29.85 ml



    What I'm getting is this:

    Moles = 0.6066/204.22
    = 0.00297033

    Concentration = 0.00297033/0.02985
    = 0.09950841
    to 4 d.p = 0.0995

    Am I wrong? The system is making me final lose marks because of it. Last error was due to a simple calculator mistake, but this time I'M SURE it's correct!
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    (Original post by TonyGuacamole)
    Also could I ask you another question if you don't mind? It's the same thing but for titre 2. This time the grams are 0.6066.
    KHP (Given Mr: 204.22)

    Mass of KHP/g = 0.6066
    NaOH titre value 1/ml = 29.85 ml



    What I'm getting is this:

    Moles = 0.6066/204.22
    = 0.00297033

    Concentration = 0.00297033/0.02985
    = 0.09950841
    to 4 d.p = 0.0995

    Am I wrong? The system is making me final lose marks because of it. Last error was due to a simple calculator mistake, but this time I'M SURE it's correct!
    yes that's exactly what i got too following your calculations. Only thing i see you lose marks for is no units on the concerntration but that's a small thing possibly another balanced equation? but i doubt they'll penalise you for not writing out the equation again
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    (Original post by thefatone)
    yes that's exactly what i got too following your calculations. Only thing i see you lose marks for is no units on the concerntration but that's a small thing possibly another balanced equation? but i doubt they'll penalise you for not writing out the equation again
    Thank You for your help! It was apparently a system error!
 
 
 
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