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    (Original post by TeachChemistry)
    So what don't you follow in the mark scheme?

    Moles of cobalt = (50 × 0.203)/1000 = 0.01015 mol
    Moles of AgCl = 4.22/143.4 = 0.0294
    Ratio = Cl- to Co = 2.9 : 1
    [Co(NH3)6]Cl3 (square brackets not essential)
    Difference due to incomplete oxidation in the preparation
    PRSOM
    Thank you so much I understand now! I was getting the moles of cobalt but got stuck after... (Completely ignored the AgCl part)
    Appreciate your help! And sorry for the late reply
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    Can anyone help me work out the pH of a contaminated buffer solution such as the one as follows:
    Calculate the change in pH of 100cm3 of a buffer solution containing o.5moldm-3 of ethanoic acid anf 0.5moldm-3 of potassium ethanoate when it is contaminated by:
    a. 1.0cm3 of 1.0moldm-3 of HCL
    b. 2cm3 of 1.5moldm-3 potassium hydroxide

    I sat there for 1 hour trying to get an answer but I have no clue how to work out all my class mates had different answers to each other..
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    (Original post by FemaleBo55)
    Can anyone help me work out the pH of a contaminated buffer solution such as the one as follows:
    Calculate the change in pH of 100cm3 of a buffer solution containing o.5moldm-3 of ethanoic acid anf 0.5moldm-3 of potassium ethanoate when it is contaminated by:
    a. 1.0cm3 of 1.0moldm-3 of HCL
    b. 2cm3 of 1.5moldm-3 potassium hydroxide

    I sat there for 1 hour trying to get an answer but I have no clue how to work out all my class mates had different answers to each other..
    Have you got a Ka for ethanoic acid in the question?
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    (Original post by TeachChemistry)
    Have you got a Ka for ethanoic acid in the question?
    Oh whoops for got to add that in
    Ka for ethanoic acid: [1.70x10^-5] moldm-3
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    (Original post by FemaleBo55)
    Oh whoops for got to add that in
    Ka for ethanoic acid: [1.70x10^-5] moldm-3
    Work out the pH before adding the HCl of the KOH.
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    (Original post by TeachChemistry)
    Work out the pH before adding the HCl of the KOH.
    Yeah which is 4.77 if rounded up to 2.d.p?
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    (Original post by FemaleBo55)
    Yeah which is 4.77 if rounded up to 2.d.p?
    Ok I wil take your word for that right now cos I'm watching telly. Now try the HCl part. The moles of potassium ethanoate goes down by the mole amount of HCl added whilst the moles of ethanoic acid goes up.
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    (Original post by djmans)
    do you find chemistry hard or easy?
    Well a mixture of both! I find memorizing all these reactions, equations and colours so hard since i am not that good at memorizing but chemistry remains easier than bio in terms of content and questions
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    What does the slash mean in synthetic route?

    e.g. LiAlH4/dry ether for reduction
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    have you guys finished all of your practicals yet?
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    Can someone help me write the full equation for:

    [Al(H2O)6]3+ with excess OH- to form [Al(OH)4]-
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    Hi Shiney101

    I think you have written your question incorrectly. You have written [Al(H2O)4]- but i believe you to mean [Al(OH)4]-. The equation would therefore be:

    [Al(H2O)6]3+ + 4OH- --> [Al(OH)4]- + 6H2O

    This is all balanced, and as you can see because of the 4 negative OH- ligands the othervall charge of the complex would be -1. Al is +3 and has four -1 ligands.

    I hope that helps

    Alex
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    (Original post by ajsullivan)
    Hi Shiney101

    I think you have written your question incorrectly. You have written [Al(H2O)4]- but i believe you to mean [Al(OH)4]-. The equation would therefore be:

    [Al(H2O)6]3+ + 4OH- --> [Al(OH)4]- + 6H2O

    This is all balanced, and as you can see because of the 4 negative OH- ligands the othervall charge of the complex would be -1. Al is +3 and has four -1 ligands.

    I hope that helps

    Alex
    Oh yhh you're right- I'll go and edit it
    Thank you!
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    Hello I need help for some part of this question.

    Copper can be recovered from low-grade ores by ‘leaching’ the ore with dilute H2SO4,which converts the copper compounds in the ore into CuSO4(aq). The concentrationof copper in the leach solution can be estimated by adding an excess of aqueouspotassium iodide, and titrating the iodine produced with standard Na2S2O3(aq).2Cu2+ + 4I– 2CuI + I2I2 + 2S2O32– 2I– + S4O62– When an excess of KI(aq) was added to a 50.0 cm3 sample of leach solution, and theresulting mixture titrated, 19.5 cm3 of 0.0200 mol dm–3 Na2S2O3(aq) were required todischarge the iodine colour. Calculate the [Cu2+(aq)], and hence the percentage by mass of copper, in the leachsolution.

    Below is the mark scheme:

    n(thio) = 0.02 × 19.5/1000 = 3.9 × 10–4 mol (1)
    n(thio) = n(Cu2+), so n(Cu2+) in 50 cm3 = 3.9 × 10–4 mol
    so [Cu2+] = 3.9 × 10–4 × 1000/50 = (7.8 × 10–3 (mol dm–3)) (1)
    in 100 cm3, there will be 7.8 × 10–4 mol, which is 63.5 × 7.8 × 10–4 = 0.049 – 0.050% (1) [3]Allow ecf on 2nd and 3rd marks 0.5 gets 2 marks only


    I managed to calculate the first and second part but the third mark point is the one that I can't figure out. Why do I have to use 100cm^3?
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    In ethanedioate ions, C2O4 2- in my book it shows the the oxygen that are single bonded to the Carbon only have one lone pair but I don't really get why. If oxygen has 6 electrons in the outer shell and one of the electrons is used in the covalent bond with carbon then won't you be left with 5 electrons?

    Not sure if this makes much sense but if someone can help me out that would be really helpful, for some reason I've suddenly gotten really confused about these lone pairs...
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    Can someone help me with this multiple choice?

    Q: The sodium thiosulfate solution was prepared by dissolving 4.5 g of sodium thiosulfate in water and making the solution up to 250 cm^3 in a volumetric flask. The volumetric flask is accurate to 0.3 cm^3 so, to match this accuracy, the mass of the sodium thiosulfate should be accurate to at least

    A. 0.5 g
    B. 0.05 g
    C. 0.005 g
    D. 0.0005 g

    The answer is C. But why??? I have no clue at all
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    (Original post by mystreet091234)
    Can someone help me with this multiple choice?

    Q: The sodium thiosulfate solution was prepared by dissolving 4.5 g of sodium thiosulfate in water and making the solution up to 250 cm^3 in a volumetric flask. The volumetric flask is accurate to 0.3 cm^3 so, to match this accuracy, the mass of the sodium thiosulfate should be accurate to at least

    A. 0.5 g
    B. 0.05 g
    C. 0.005 g
    D. 0.0005 g

    The answer is C. But why??? I have no clue at all
    answered in other thread
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    I can't distinguish between the equivalence point and the end point and like, do these points occur within the vertical section or at the start/end of the pH curve ?
    Also why for strong acids/strong bases titration, why Phenolphthalein and Methyl orange indicators both work although approx ph of colour change for Phenolphthalein is 8.3-10 and Methyl orange is 3.1-4.4 ?

    Posted from TSR Mobile
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    (Original post by PlayerBB)
    I can't distinguish between the equivalence point and the end point and like, do these points occur within the vertical section or at the start/end of the pH curve ?
    Also why for strong acids/strong bases titration, why Phenolphthalein and Methyl orange indicators both work although approx ph of colour change for Phenolphthalein is 8.3-10 and Methyl orange is 3.1-4.4 ?

    Posted from TSR Mobile
    The equivalence point is the point at which you have mixed an equal number of moles of acid and base. The end point is the point at which the indicator changes colour. If you look at a pH curve, you will notice that there is a steep section for most titrations (except weak acid weak base) and you want your indicators end point to be somewhere in that steep part of the curve. For strong acid strong base, this steep section is approximately from 3-9 so either indicator would work
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    (Original post by PlayerBB)
    I can't distinguish between the equivalence point and the end point and like, do these points occur within the vertical section or at the start/end of the pH curve ?
    Do you own a Facer text? The end point and equivalence point are very different - the end point is the last point of the vertical region while the equivalence point is the mean of the starting point and the end point of the vertical region.

    Also why for strong acids/strong bases titration, why Phenolphthalein and Methyl orange indicators both work although approx ph of colour change for Phenolphthalein is 8.3-10 and Methyl orange is 3.1-4.4 ?

    Posted from TSR Mobile
    Both ranges fall in the vertical region (3 to 11), although I'd say phenolphthalein is a better.
 
 
 
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