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Using Henderson-Hasselbach equation?

  1. Oct 5, 2007 #1
    pH = pKa + log [base/acid] (H-H equation)

    Using the H-H equation, calculate the theoretical volume of 0.2 M NaOH required to raise the pH of 0.1 M Tris HCl (pk = 8.1) to raise the pH to its pK value.

    The way I see it is to get pH = pK so using H-H equation:
    8.1 = 8.1 + log[0.2 M NaOH/0.1 M Tris HCl]

    8.1-8.1 = log[0.2 M NaOH/0.1 M Tris HCl]

    0 = log[0.2 M NaOH/0.1 M Tris HCl]

    Then taking anti log,
    1 = [0.2 M NaOH/0.1 M Tris HCl]

    1 = 0.5

    means

    pH = 1, has base:acid ratio of 1/2

    Thus if I have 0.1 mol/litre Tris HCL, I need half of 0.2 M NaOH.

    But how do I convert these to volumes?

    Please forgive me for asking this Q. I am really stuck. Am I on the right track? If so, how do I proceed to get the volumes. If not, what am I thinking that is wrong?

    Please give up a detailed understanding of what I did wrong?

    I would really appreciate it. Thanks.

    Then I am suppose to draw a curve but how can I? with only one point at the theoretical volume of 0.2 M NaOH. I just don't understand. Please help me understand what they want from me.
     
  2. jcsd
  3. Oct 6, 2007 #2
    Getting the Theoretical Vol required to raise pH of an acid sol to its pK value

    I think I erred in my reasoning. I should be using the c1*v1 = c2*v2 formula
    There is 50 mL of 0.1 M Tris HCl acid
    So Let v1 = 50 mL = 0.050 L
    c1 = 0.01 M Tris HCl

    Then c2 = 0.2 M NaOH
    v2 = ? mL

    c1*v1 = c2*v2
    (0.1 M Tris HCl)*(0.05 L) = (0.2 M NaOH) v2

    Isolate v2:
    v2 = c1*v1/c2
    v2 = [(0.1 M Tris HCl)(0.05 L)] / 0.2 M NaOH
    = 0.025 L
    = 25 mL

    Is this correct? Please let me know if I am wrong or on the wrong track. THANKS.
     
  4. Oct 6, 2007 #3
    Is the stoichiometry between NaOH and tris HCl 1 : 1 ?
     
  5. Oct 6, 2007 #4
    This is how I would solve the problem if the stoichiometry between NaOH and tris HCl is 1 to 1 . Incase not , you still apply the same method.

    First NaOH reacts with Tris HCl . The reaction is quasi-quantitative and complete. Clearly NaOH must be the limiting reactant. Then after the reaction is over , there must remain Tris HCL + its conj base so that pH = pK_a

    that is , the concentration of Tris-HCl remaining is :

    [tex] M_{acid} = \frac{0.1V_{T-HCl} - 0.2V_{NaOH}}{V_{T-HCl} + V_{NaOH}} [/tex]

    The concentration of the conj base is :

    [tex] M_{base} = \frac{0.2V_{NaOH}}{V_{T-HCl} + V_{NaOH}}[/tex]

    Since pH = pK_a and ofcourse assuming a buffer is formed,

    Then according to HH equation :

    [tex] Log_{10} \left( \frac{M_{base}}{M_{acid}} \right) = 0 [/tex]

    [tex] \frac{0.2V_{NaOH}}{0.1V_{T-HCl} - 0.2V_{NaOH}} = 1 [/tex]

    [tex] 0.1V_{T-HCl} - 0.2V_{NaOH} = 0.2V_{NaOH}} [/tex]

    [tex] V_{NaOH} = \frac{V_{T-HCl}}{4} [/tex]

    If the initial volume of tris HCl is known , the volume of NaOH can be calculated.
     
  6. Oct 6, 2007 #5
    Thank you

    Thank you so much. :smile:
     
  7. Oct 6, 2007 #6
    How can you prepare a buffer with HH with an acid with a weak acid?

    Need to prepare buffer with Henderson-Hasselbalch Equation?
    Use the H-H equation to calculate the volume of 0.1 M acetic acid and 0.1 M Na acetate required to prepare 50 ml 0.1 M acetate buffer pH 4.0.

    HH: pH = pk + log[base/acid]

    Acetic acid is an acid but acetate is not a base. Acetate is a weak acid but I put it as a base bc how am I suppose to use HH to get a ratio?

    Here is what I tried:

    pH = pka + log[0.1 M acetate/0.1 M acetic acid]
    4.0 - 4.76 = log[0.1 M Na acetate/0.1 M acetic acid]
    0.76= log[0.1M Na acetate/0.1 M acetic acid]

    5.75 = [0.1 M Na acetate/0.1 M acetic acid]

    5.75 = [0.1 NaOH]/[0.1 acetic acid]

    [acetic acid] = NaOH/5.74

    Vol of acetic acid = (50 ml) [NaOH]/5.74
    = 8.71 ml
    Vol of acetate = 50 ml - 8.71
    = 41.3 ml

    However, these are wrong answers.

    The correct answers should be:
    42.6 ml 0.1 M acetic acid + 7.4 ml 0.1 M Na acetate to give pH 4.0.

    Tell me what i did wrong and how to derive at the correction solutions above.

    Thanks.
     
  8. Oct 7, 2007 #7
    Sodium acetate is nothing but the conjugate base of acetic acid.

    According to HH equation,

    [tex] \frac{[NaOAC]}{[HOAC]} = 10^{pH - pK_a} [/tex]

    At equilibrium ,

    [tex] [NaOAC] = \frac{[NaOAC]_o V_{NaOAC}}{V_T} = \frac{(0.1)V_{NaOAC}}{V_T} [/tex]

    [tex] [HOAC] = \frac{[HOAC]_o V_{HOAC}}{V_T} = \frac{(0.1)(V_{HOAC})}{V_T} [/tex]

    [tex] \frac{[NaOAC]}{[HOAC]} = \frac{V_{NaOAC}}{V_{HOAC}} [/tex]

    By simple substitution ,

    [tex] \frac{V_{NaOAC}}{V_{HOAC}} = 10^{pH - pK_a} \ \ \ ... eq \ (1) [/tex]

    We need to relate the two volumes inorder to solve a system of 2 equations , 2 unknowns. Notice that,

    [tex] V_T = V_{HOAC} + V_{NaOAC} \ \ \ ... eq \ (2)[/tex]

    V_T , pk_a , and pH are ofcourse given.

    Solve to get :

    [tex] V_{NaOAC} = \frac{(V_T)(10^{pH - pK_a})}{ 1 + 10^{pH - pK_a} } [/tex]

    [tex] V_{HOAC} = \frac{V_T}{1 + 10^{pH - pK_a} } [/tex]

    Plug in numerical values to get the required values of both volumes.
     
  9. Oct 7, 2007 #8
    You have an initial conc of HOAC and NaOAC ( both 0.1 M ) but both volumes are unknown. The HH equation says that pH = pK_a + log ( [base] / [acid] ) where [base] and [acid] are the concentrations are equlibrium. So what did you do ?

    You plugged in 0.1 for each , which is wrong. That is because you must take into account the dilution factor. Remember that you are to add an unknown volume of 0.1 M of acid and an unknown volume of 0.1 M base. Hence , the amount of acid and base are , respectively :

    [tex] n_{HOAC} = 0.1 V_{HOAC} [/tex]

    [tex] n_{NaOAC} = 0.1 V_{NaOAC} [/tex]

    Since a buffer is to be formed, then the amount of acid and base is approximately constant. Their concentrations at equilibrium are then ,

    [tex] C_{HOAC} = \frac{n_{HOAC}}{V_T} = \frac{0.1 V_{HOAC}}{V_{HOAC} + V_{NaOAC}} [/tex]

    [tex] C_{NaOAC} = \frac{n_{NaOAC}}{V_T} = \frac{0.1 V_{NaOAC}}{V_{HOAC} + V_{NaOAC}} [/tex]

    Once you've realized that , the rest becomes pure basic algebra.

    You start by plugging the expressions of the concentrations in the HH equation and solve to find each V.
     
  10. Oct 7, 2007 #9
    Thanks so much. I've been ill and still am so I can't get my head to think straight. I really appreciate your help Hunt_.
     
  11. Oct 7, 2007 #10
    Preparing a sample buffers using weak acid/strong base?

    Preparing a sample buffers using weak acid/strong base?
    You will be provided with 0,5 M acetic acid as a source of acetic acid and 0.2 M NaOH with which to prepare the same 0.1 M buffers.
    a) Calculate the number of moles of acetic acid required to prepare 50 ml of a 0.1 M buffer. Then calculate the volume of 0.5 M acetic acid needed to prepare 50 ml of 0.1 M buffer.
    b) Use the Henderson-Hasselbalch equation to calculate the number of moles of NaOH required to bring the pH up to 4.0 and 5.0 Then calculate the volume of 0.2 M NaOH that contains the number of moles.

    Answers:
    10 ml (0.5 M acetic acid) + 3.7 ml (0.2 M NaOH to give pH 4.0)
    10 ml (0.5 M acetic acid + 15.9 ml (0.2 M NaOH to give pH 5.0)

    I got 10 ml of 0.5 acetic acid by calculating:
    c1 = 0.1 M, v1 = 50 ml = 0.050 L
    c2 = 0.5 M, v2 = ?
    c1*v1 = c1*v2
    v2 = cv*v1/c2
    = 0.1 M * 0.050 L/0.50 M.
    = 0.01 L
    = 10 mL

    But do I using H-H eq, to get the following:
    3.7 ml (0.2 M NaOH to give pH 4.0)
    15.9 ml (0.2 M NaOH to give pH 5.0)

    Additional Details

    6 hours ago
    I tried:

    pH = pka + log [Mbasic/Macid]

    Mbasic = 0.2 NaOH
    Vacetic acid + VNaOH


    4.0 = 4.76 + log[0.2 M NaOH/0.5 M acetic acid]
    4.0 - 4.76 = log[0.2 M NaOH/0.5 M acetic acid]
    0.76 = log[0.2 M NaOH/0.5 M acetic acid]
    10^5.75 = log[0.2 NaOH/0.5 M acetic aid]

    5.75 = [0.2 M NaOH/0.5 M acetic acid]

    Am I on the right track? How do I proceed to get to the answer of 3.7 ml 0.2 M NaOH to give pH 4.0
     
  12. Oct 8, 2007 #11
    Preparing a sample buffers using weak acid/strong base?

    Preparing a sample buffers using weak acid/strong base?
    You will be provided with 0,5 M acetic acid as a source of acetic acid and 0.2 M NaOH with which to prepare the same 0.1 M buffers.
    a) Calculate the number of moles of acetic acid required to prepare 50 ml of a 0.1 M buffer. Then calculate the volume of 0.5 M acetic acid needed to prepare 50 ml of 0.1 M buffer.
    b) Use the Henderson-Hasselbalch equation to calculate the number of moles of NaOH required to bring the pH up to 4.0 and 5.0 Then calculate the volume of 0.2 M NaOH that contains the number of moles.

    Answers:
    10 ml (0.5 M acetic acid) + 3.7 ml (0.2 M NaOH to give pH 4.0)
    10 ml (0.5 M acetic acid + 15.9 ml (0.2 M NaOH to give pH 5.0)

    I got 10 ml of 0.5 acetic acid by calculating:
    c1 = 0.1 M, v1 = 50 ml = 0.050 L
    c2 = 0.5 M, v2 = ?
    c1*v1 = c1*v2
    v2 = cv*v1/c2
    = 0.1 M * 0.050 L/0.50 M.
    = 0.01 L
    = 10 mL

    b)

    Mbasic = [0.2 NaOH]/[Vacetic acid + VNaOH]

    Macid = [0.5 V acetic acid - 0.2 VNaOH]/[Vacetic acid - VNaOH]

    Mbasic/Macid = [0.2 V NaOH]/[V acetic acid - V NaOH]

    pH = pka + log [Mbasic/Macid]

    4.0 = 4.76 + log[Mbasic/Macid]
    4.0 - 4.76 = log[0.2 V NaOH/0.5 V acetic acid - 0.2 V NaOH]
    0.76 = log[0.2 V NaOH/(0.5 V acetic acid - 0.2 V NaOH)]

    10^5.75 = log[0.2 V NaOH/(0.5 V acetic acid - 0.2 V NaOH)]


    5.75 = log[0.2 V NaOH/(0.5 V acetic acid - 0.2 V NaOH)]

    5.75 (0.5 Vacetic acid - 1.15 VNaOH) = 0.2 V NaOH
    2.88 V acetic acid - 1.15 V NaOH = 0.2 V NaOH
    0.2 V NaOH + 1.15 V NaOH = 2.88 V acetic acid
    1.35 V NaOH = 2.88 V acetic acid
    V NaOH = 2.88 V acetic acid/1.35 V
    = 2.13 mL

    HOwever, this is not the right answer. The correct answer should be VNaOH= 3.7 ml 0.2 M NaOH to give pH 4.0.
    and V NaOH = 15.9 ml (0.2 M NaOH to give pH 5.0)

    Do you see where I erred?

    I am apologetic if I seem slow or unable to think. I know you don't know me but I am very ill/bedridden. School is all I have. I really need your help, not asking for sympathy but hope that you don't lose patience with me. I really don't want to trouble you. I appreciate the help you've given me so far. Thank you for your patience.
     
  13. Oct 10, 2007 #12
    The reaction occuring is :

    [tex] NaOH + HOAC \rightarrow H_2 O + NaOAC [/tex]

    For part 1 :

    If n_o is the initial amount of the acid , and x is the amount of NaOH. Then after the rxn is complete, n_o - x is the remaining amount of the acid and x is the amoun formed of its conjugate base.

    Notice the total amount of moles between the acid and its conj base is :

    Initiall : n_o + 0 = n_o

    Finally : n_o + x - x = n_o

    The number of particles is conserved between the two. ( Thay is why C1V1 = C2V2 works well here . )

    Accoriding to the given , we have a 0.1 M buffer. At eq , the conc of the acid + it conj base :

    [tex] [HOAC] + [NaOAC] = 0.1 M [/tex]

    Multiplying the total volume , V = 50 ml.

    [tex] n_{HOAC} + n_{NaOAC} = 5 mmol [/tex]

    We proved above that :

    [tex] n_{HOAC} + n_{NaOAC} = n_o [/tex]

    So , n_o = 5 mmol ( initial amount of acid ) . It follows that the volume of acetic acid needed is 5 mmol / 0.5 M = 10 ml.

    For part 2 :

    From HH equation ,

    [tex] \frac{n_{NaOAC}}{n_{HOAC}} = 10^{pH - pK_a} [/tex]

    By substitution ,

    [tex] \frac{n_{NaOAC}}{n_o - n_{NaOAC}} = 10^{pH - pK_a} [/tex]

    Rearrange to get :

    [tex] n_{NaOAC} = n_o \frac{10^{pH - pK_a}}{1 +10^{pH - pK_a}} [/tex]

    since n_{NaOAC} formed = n_{NaOH} reacted then :

    [tex] n_{NaOH} = n_o \frac{10^{pH - pK_a}}{1 +10^{pH - pK_a}} [/tex]

    To find its volume , use

    n = CV. It follows :

    [tex] V_{NaOH} = \frac{n_o}{0.2} \ \ \frac{10^{pH - pK_a}}{1 +10^{pH - pK_a}} [/tex]
     
  14. Oct 10, 2007 #13
    I will look later and see where you made your mistake . Sorry but there's No time now...
     
  15. Oct 11, 2007 #14
    Your method is correct but you have few calculation errors.


    correct

    4 - 4.76 = - 0.76 NOT 0.76

    Where did you get 5.75 from ? You take the anti-log of -0.76 :

    10^(-0.76} = [0.2 V NaOH/(0.5 V acetic acid - 0.2 V NaOH)]



    the correct expression should be : V (NaOH) = 0.37 V (HOAC)
    It follows V (NaOH) = 3.7 ml

    P.S.

    Notice that Vacetic acid + VNaOH < 50 ml ( total volume ) in both cases. this is to show you that some water has been added to the solution to reach 50 ml although this is not mentioned in the given.
    Therefore , when writing down the Molarity of , say the conjugated base :

    Mbasic = [0.2 V NaOH]/[Vacetic acid + VNaOH]

    You made a mistake by not including the volume of water added. Instead you can just write V_T ( or Vacetic acid + VNaOH + V water ) or 50 ml ( since it is known ). It is not a fatal mistake because once you take the ratio [base]/[acid] , the denominator is the same and so would cancel.

    Also notice that instead of doing this numerical tedious work , you can derive a general equation and then finally plug in the given constants. This way , you get to solve the 2 similar cases of part b much faster , you only need to change the pH value and V(NaOH) is determined instantly with less probability of calculation error.
     
  16. Aug 1, 2010 #15
    heya. hi :)) i'm new here?
    can i use some help too? i'm having a hard time
    to this :

    -using Henderson-Hasselbalch equation, calculate
    the volume of 0.2 M aceic acid and 0.2 M sodium acetate,
    needed to prepare 50 ml of 0.1 M acetate buffer solution (pH=4.5).
    pKa of acetic acid is 4.74. please i need a help and a solution for this.
    thanks so much :))
     
  17. Aug 1, 2010 #16

    Redbelly98

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