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OCR Salters Chemistry F335 12th June 2013 Exam revision thread Watch

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    (Original post by ActaNonVerba)
    Mark scheme definitions for chromophore and pharmacophore anyone?
    Chromophore - the area of a molecule that gives it its colour
    Pharmacophore - the area of a molecule responsible for giving a drug its activity
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    Could someone explain how to work out the pH of a buffer for me? So stuck!
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    (Original post by ameliestia)
    Can anyone explain KPa to me?? I am really confused about when we need to use it...?
    do you mean pKa ?
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    Can someone please help me with this question it's from the Jan 2013 paper question 3d(i):

    When 11.8g of hydrated calcium nitrate(V) Ca(NO3)2 . xH2O, are gently heated, 3.6g of water are lost, leaving the anhydrous salt, Ca(NO3)2.

    Give the value of x in the formula of the hydrated salt.

    The mark scheme says the answer is 4 and shows the calculations 8.2/164.1=0.05 and 3.6/18=0.2 but i'm not sure how they get to 4????

    Thanks!
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    (Original post by Tikara)
    do you mean pKa ?
    Oh wow yeah.. I am really not going well hahaha
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    (Original post by ameliestia)
    Oh wow yeah.. I am really not going well hahaha
    don't worry haha

    Sal Khan actually made me remember this but 'p' infront of anything is just
    -log(of something)
    [like pH = -log(H+)]

    Ka is the acidity constant. This usually is a really small number and they cover a large range of values, it's converted to pKa solely for ease of use and so it can be easily compared to pH - coarse tuning remember!
    So if you wanted a pH of 4, you'd pick the acid with the closest pKa value to 4 and then fine tune with the salt:acid ratio

    hope that helps
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    (Original post by Tikara)
    don't worry haha

    Sal Khan actually made me remember this but 'p' infront of anything is just
    -log(of something)
    [like pH = -log(H+)]

    Ka is the acidity constant. This usually is a really small number and they cover a large range of values, it's converted to pKa solely for ease of use and so it can be easily compared to pH - coarse tuning remember!
    So if you wanted a pH of 4, you'd pick the acid with the closest pKa value to 4 and then fine tune with the salt:acid ratio

    hope that helps
    Ah, that's it helpful! I remember doing a question where we had to say which pKa was the most acid, so it would just be the lowest value?
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    (Original post by ameliestia)
    Ah, that's it helpful! I remember doing a question where we had to say which pKa was the most acid, so it would just be the lowest value?
    yeah - and if it was which Ka is the most acidic, as it's minus log - it would be the highest Ka value
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    (Original post by Tikara)
    yeah - and if it was which Ka is the most acidic, as it's minus log - it would be the highest Ka value
    Brilliant, thanks
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    (Original post by chocolate102030)
    Can someone please help me with this question it's from the Jan 2013 paper question 3d(i):

    When 11.8g of hydrated calcium nitrate(V) Ca(NO3)2 . xH2O, are gently heated, 3.6g of water are lost, leaving the anhydrous salt, Ca(NO3)2.

    Give the value of x in the formula of the hydrated salt.

    The mark scheme says the answer is 4 and shows the calculations 8.2/164.1=0.05 and 3.6/18=0.2 but i'm not sure how they get to 4????

    Thanks!
    From here on its simpler than it looks, you are shown that there's only 1 mole of Ca(NO3)2 so you divide it by itself to get 1, to work out the number of moles of water, you divide 0.2 by the number of moles in the calcium nitrate 0.05 which gives you 4 (it's basically 0.2 * 20 = 4)
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    Hi I'm really confused with paper January 2012 - last page of questions... They're all about pH...
    I'd appreciate if someone could help as not really sure...
    Btw, in buffer solutions, does Ka = [H+] all the time?
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    (Original post by hannahhannah1)
    Could someone explain how to work out the pH of a buffer for me? So stuck!
    Ka = [H+] * [salt]/[acid]

    This is found on Pg. 191 of the Chemical Ideas book
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    Stuck on NMR! Any notes on it?
    I dont get how to work out the structure by just looking at the NMR peaks :/
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    (Original post by DoeADeer)
    Hi I'm really confused with paper January 2012 - last page of questions... They're all about pH...
    I'd appreciate if someone could help as not really sure...
    Btw, in buffer solutions, does Ka = [H+] all the time?
    no it doesn't

    Ka will only equal [H+] if the concentration of salt is the same as the concentration of acid

    as Ka = [H+] * [salt]/[acid]
    so if [salt] = [acid]

    [salt]/[acid] just = 1

    so Ka = [H+] * (1)
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    Now the real question is... how much cramming can be done in 5 hours tomorrow???
    We will see... sleep well my friends, for we rise at daybreak ... the knights of OCR emerge from the mist at 2:00pm tomorrow, and we must not withhold our lances or let our wits reach their end. We will be victorious.
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    remember guys

    Strong Acid
    Assume: [HA] = [H+]

    Weak Acid
    Assume 1: [H+] = [A-]

    ^--- this usually leads to having to square root to get the [H+] and then -log it to get pH

    Assume 2: [HA] at equilibrium = [HA] put into solution.

    This usually results in the greatest inaccuracy in calculations
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    What's the difference between Pharmacophore and chromophore?

    And how do you identify it on a molecule? for example when you're asked to circle it.

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    (Original post by DoeADeer)
    Hi I'm really confused with paper January 2012 - last page of questions... They're all about pH...
    I'd appreciate if someone could help as not really sure...
    Btw, in buffer solutions, does Ka = [H+] all the time?
    January 2012

    Question 5e)ii)

    Ka = 9.3x10-4moldm-3

    We're told that concentration of A- is twice that of HA. Using Ka = [H+] * [A-]/[HA], we know that ratio is equal to 2. Rearrange to give (0.5)Ka = [H+] therefore the answer is -log((0.5)*9.3x10-4moldm-3) = pH 3.33

    Question 5f)i)


    27cm3 of 0.050moldm-3 HA solution is added and that 0.10moldm-3 A-
    solution is added and we want to calculate volume to give 1/3 neutralisation.

    => 27cm3 of 0.050moldm-3
    => 9cm3 is 1/3 of 27cm3
    => Molarity of A- is double that of HA so final volume is 4.5cm3


    Question 5f)ii)


    Using Ka = [H+] * [A-]/[HA], we understand from prior question that one third of the acid is neutralised; so in our calculations we must account of that.

    [salt] = 0.0045dm3 * 0.1moldm-3= 0.00045mol
    [acid] = (2/3) * 0.027dm3 * 0.05moldm-3 = 0.00090mol

    Therefore ratio of [salt]/[acid] = 1/2

    => 2 * Ka = [H+]
    => 2 * 9.3x10-4moldm-3 = 1.86x10-3
    => -log(1.86x10-3) = pH 2.73

    May be a bit late but hope this helps!
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    We only need to be able to calculate pH of STRONG bases, right?
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    does anyone else here draw their hydrogen bonds of their molecules on first to make sure the H-O||||H bonds are straight?
 
 
 
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