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    Calculate the pH of the following aqueous solutions at 298K:
    (a) potassium hydroxide 0.406 mol dm-3 (3 marks)
    (b) nitric acid 0.0045 mol dm-3 (2 marks)
    (c) sulfuric acid 0.06 mol dm-3 (3 marks)
    (d) barium hydroxide 0.002 mol dm-3 (3 marks)
    Kw(298) = 1.00 x 10-14 mol2 dm-6


    I get

    a) 13.61
    b) 11.64
    c) 0.92
    d) 11.60

    are the wrong my answers
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    For B I get a pH of 2.3, how did you work it out?

    The rest are correct
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    (Original post by AtomicMan)
    For B I get a pH of 2.3, how did you work it out?

    The rest are correct
    cause for B, I got 2.3 but then used the pOH method and did 14-2.3=which gave 11.7

    because when the HNO3 dissociates, you get the concentration of H+ and then you use the Kw formula to get the concentration of OH- and then apply the pOH method by taking log of OH- and minus that from 14.
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    (Original post by otrivine)
    cause for B, I got 2.3 but then used the pOH method and did 14-2.3=which gave 11.7

    because when the HNO3 dissociates, you get the concentration of H+ and then you use the Kw formula to get the concentration of OH- and then apply the pOH method by taking log of OH- and minus that from 14.
    But HNO3 is a strong acid, so in water;

    HNO3 --> H+ + HNO3-

    So the concentration of HNO3 = the concentration of H+, as you said. If pH = -log[H+], then surely you can put 0.0045 directly into the equation as the concentration of [H+], without having to bring in Kw?
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    (Original post by AtomicMan)
    But HNO3 is a strong acid, so in water;

    HNO3 --> H+ + HNO3-

    So the concentration of HNO3 = the concentration of H+, as you said. If pH = -log[H+], then surely you can put 0.0045 directly into the equation as the concentration of [H+], without having to bring in Kw?
    of course!! thanks for telling me! its 2.35 not 2.3 right?
    and for d) you got 11.60 right
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    (Original post by otrivine)
    of course!! thanks for telling me! its 2.35 not 2.3 right?
    and for d) you got 11.60 right
    Yes, 2.35 to 3 significant figures.
    Yep, I got 11.60 for D.
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    (Original post by AtomicMan)
    Yes, 2.35 to 3 significant figures.
    Yep, I got 11.60 for D.
    A chemist has aqueous solutions of two weak acids, 0.1 mol dm-3 propanoic acid and 0.1 mol dm-3 carbonic acid, H2CO3. The pH of each solution was measured.
    Acid Concentration of acid pH
    CH3CH2COOH 0.1 mol dm-3 2.93
    H2CO3 0.1 mol dm-3 3.68

    (i) Which acid is the stronger acid? Give your reasons.
    (ii) Use the data in the table above to calculate the value of Ka for propanoi
    acid. State any assumptions that you make to simplify your calculation.




    i) I got CH3CH2COOH because the smaller PH --> stronger acid

    ii) I got Ka=1.38038 x 10-5
    assumed
    strong acid , H+=HA
    HA dissociates equal concentration of ions that have been dissociated so [H+][A-] --> [H+]2
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    (Original post by otrivine)
    A chemist has aqueous solutions of two weak acids, 0.1 mol dm-3 propanoic acid and 0.1 mol dm-3 carbonic acid, H2CO3. The pH of each solution was measured.
    Acid Concentration of acid pH
    CH3CH2COOH 0.1 mol dm-3 2.93
    H2CO3 0.1 mol dm-3 3.68

    (i) Which acid is the stronger acid? Give your reasons.
    (ii) Use the data in the table above to calculate the value of Ka for propanoi
    acid. State any assumptions that you make to simplify your calculation.




    i) I got CH3CH2COOH because the smaller PH --> stronger acid

    ii) I got Ka=1.38038 x 10-5
    assumed
    strong acid , H+=HA
    HA dissociates equal concentration of ions that have been dissociated so [H+][A-] --> [H+]2
    Everything youve said is right, except the first assumption is abit dodgy.

    Its a weak acid (says in the queston), so [H+] = [A-]. If it was a strong acid, [HA] = 0 since it would have all dissociated. You also assume that [HA] intially = [HA] at equilibrium.
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    (Original post by AtomicMan)
    Everything youve said is right, except the first assumption is abit dodgy.

    Its a weak acid (says in the queston), so [H+] = [A-]. If it was a strong acid, [HA] = 0 since it would have all dissociated. You also assume that [HA] intially = [HA] at equilibrium.
    so in total should there 2 assumptions? was my second assumption right
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    (Original post by otrivine)
    so in total should there 2 assumptions? was my second assumption right
    Your second assumption is right. Another one is that you ignore the H+ produced from the dissociation of water.
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    (Original post by AtomicMan)
    Your second assumption is right. Another one is that you ignore the H+ produced from the dissociation of water.
    and why is that?
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    (Original post by otrivine)
    and why is that?
    Becuase then [H+] wouldnt equal [A-], complicating things. Besides, the amount of H+ produced from the dissociation of water is tiny, so it can be ignored.
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    (Original post by AtomicMan)
    Becuase then [H+] wouldnt equal [A-], complicating things. Besides, the amount of H+ produced from the dissociation of water is tiny, so it can be ignored.
    oh ok thanks

    ) Write down the equation for the reaction that occurs when an aqueous solution of propanoic acid is mixed with an aqueous solution of carbonic acid, H2CO3 and label the conjugate acid-base pairs

    this one would it be

    CH3CH2COOH + H2CO3 equilibrium sign CH3CH2COO- +H3CO3+

    CH3CH2COOH = ACID 1
    CH3CH2COO- = CONJUGATE BASE 1

    H2CO3 = BASE 2
    H3CO3+ = CONJUGATE ACID 2
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    (Original post by otrivine)
    oh ok thanks

    ) Write down the equation for the reaction that occurs when an aqueous solution of propanoic acid is mixed with an aqueous solution of carbonic acid, H2CO3 and label the conjugate acid-base pairs

    this one would it be

    CH3CH2COOH + H2CO3 equilibrium sign CH3CH2COO- +H3CO3+

    CH3CH2COOH = ACID 1
    CH3CH2COO- = CONJUGATE BASE 1

    H2CO3 = BASE 2
    H3CO3+ = CONJUGATE ACID 2
    All correct
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    (Original post by AtomicMan)
    All correct
    thanks

    Lactic acid, CH3CH(OH)COOH, is a weak acid and has a
    Ka = 1.37 x10-4 mol dm-3 . Calculate the pH of a 0.01 mol dm-3 solution of lactic acid. State any assumptions you make in your calculation


    for this one , is the PH= 2.93 and are the assumptions the same as before?
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    (Original post by otrivine)
    thanks

    Lactic acid, CH3CH(OH)COOH, is a weak acid and has a
    Ka = 1.37 x10-4 mol dm-3 . Calculate the pH of a 0.01 mol dm-3 solution of lactic acid. State any assumptions you make in your calculation


    for this one , is the PH= 2.93 and are the assumptions the same as before?
    Yes, your pH is correct and yes the assumptions are the same.
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    (Original post by AtomicMan)
    Yes, your pH is correct and yes the assumptions are the same.
    A reaction between A and B is second order. Write three different rate law equations that might possibly apply to the reaction.
    ?
    not suree
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    (Original post by AtomicMan)
    Everything youve said is right, except the first assumption is abit dodgy.

    Its a weak acid (says in the queston), so [H+] = [A-]. If it was a strong acid, [HA] = 0 since it would have all dissociated. You also assume that [HA] intially = [HA] at equilibrium.
    This is much more true for strong acids than weak acids. In reality it is never true but a decent approximation the stronger the acid (or to be more precise, the higher the [H+]). It is not even entirely true for completely strong acids (i.e. even if Ka=infinity the only way [H+]=[A-] is if Ca=infinity too). Can you tell me why?
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    (Original post by otrivine)
    A reaction between A and B is second order. Write three different rate law equations that might possibly apply to the reaction.
    ?
    not suree
    This must mean second order overall. So if the order with respect to A is nA and with respect to b is nB, nA+nB=2.

    Thereafter you can pick literally any combination of real numbers that add up to 2 as your orders and then write them in the form r=k*[A]nA*[B]nB. For example one possible rate equation is r=k*[A]-2*[B]4, as -2+4=2, although that is a bit crazy and presumably doesn't exist so I'd find simpler ones to please your teacher.
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    (Original post by Big-Daddy)
    This is much more true for strong acids than weak acids. In reality it is never true but a decent approximation the stronger the acid (or to be more precise, the higher the [H+]). It is not even entirely true for completely strong acids (i.e. even if Ka=infinity the only way [H+]=[A-] is if Ca=infinity too). Can you tell me why?
    Sorry I dont really understand your question I am just saying what Edexcel define as the assumption for Ka calculations. In reality things are probably different, but I imagine for A-Level they dont want to overcomplicate things for us.
 
 
 
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