## Using Henderson-Hasselbach equation?

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.

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 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.
 Is the stoichiometry between NaOH and tris HCl 1 : 1 ?

## Using Henderson-Hasselbach equation?

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 :

$$M_{acid} = \frac{0.1V_{T-HCl} - 0.2V_{NaOH}}{V_{T-HCl} + V_{NaOH}}$$

The concentration of the conj base is :

$$M_{base} = \frac{0.2V_{NaOH}}{V_{T-HCl} + V_{NaOH}}$$

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

Then according to HH equation :

$$Log_{10} \left( \frac{M_{base}}{M_{acid}} \right) = 0$$

$$\frac{0.2V_{NaOH}}{0.1V_{T-HCl} - 0.2V_{NaOH}} = 1$$

$$0.1V_{T-HCl} - 0.2V_{NaOH} = 0.2V_{NaOH}}$$

$$V_{NaOH} = \frac{V_{T-HCl}}{4}$$

If the initial volume of tris HCl is known , the volume of NaOH can be calculated.

 Thank you so much.
 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.
 Sodium acetate is nothing but the conjugate base of acetic acid. According to HH equation, $$\frac{[NaOAC]}{[HOAC]} = 10^{pH - pK_a}$$ At equilibrium , $$[NaOAC] = \frac{[NaOAC]_o V_{NaOAC}}{V_T} = \frac{(0.1)V_{NaOAC}}{V_T}$$ $$[HOAC] = \frac{[HOAC]_o V_{HOAC}}{V_T} = \frac{(0.1)(V_{HOAC})}{V_T}$$ $$\frac{[NaOAC]}{[HOAC]} = \frac{V_{NaOAC}}{V_{HOAC}}$$ By simple substitution , $$\frac{V_{NaOAC}}{V_{HOAC}} = 10^{pH - pK_a} \ \ \ ... eq \ (1)$$ We need to relate the two volumes inorder to solve a system of 2 equations , 2 unknowns. Notice that, $$V_T = V_{HOAC} + V_{NaOAC} \ \ \ ... eq \ (2)$$ V_T , pk_a , and pH are ofcourse given. Solve to get : $$V_{NaOAC} = \frac{(V_T)(10^{pH - pK_a})}{ 1 + 10^{pH - pK_a} }$$ $$V_{HOAC} = \frac{V_T}{1 + 10^{pH - pK_a} }$$ Plug in numerical values to get the required values of both volumes.

 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]
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 :

$$n_{HOAC} = 0.1 V_{HOAC}$$

$$n_{NaOAC} = 0.1 V_{NaOAC}$$

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

$$C_{HOAC} = \frac{n_{HOAC}}{V_T} = \frac{0.1 V_{HOAC}}{V_{HOAC} + V_{NaOAC}}$$

$$C_{NaOAC} = \frac{n_{NaOAC}}{V_T} = \frac{0.1 V_{NaOAC}}{V_{HOAC} + V_{NaOAC}}$$

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.

 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_.
 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
 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.
 The reaction occuring is : $$NaOH + HOAC \rightarrow H_2 O + NaOAC$$ 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 : $$[HOAC] + [NaOAC] = 0.1 M$$ Multiplying the total volume , V = 50 ml. $$n_{HOAC} + n_{NaOAC} = 5 mmol$$ We proved above that : $$n_{HOAC} + n_{NaOAC} = n_o$$ 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 , $$\frac{n_{NaOAC}}{n_{HOAC}} = 10^{pH - pK_a}$$ By substitution , $$\frac{n_{NaOAC}}{n_o - n_{NaOAC}} = 10^{pH - pK_a}$$ Rearrange to get : $$n_{NaOAC} = n_o \frac{10^{pH - pK_a}}{1 +10^{pH - pK_a}}$$ since n_{NaOAC} formed = n_{NaOH} reacted then : $$n_{NaOH} = n_o \frac{10^{pH - pK_a}}{1 +10^{pH - pK_a}}$$ To find its volume , use n = CV. It follows : $$V_{NaOH} = \frac{n_o}{0.2} \ \ \frac{10^{pH - pK_a}}{1 +10^{pH - pK_a}}$$
 I will look later and see where you made your mistake . Sorry but there's No time now...

Your method is correct but you have few calculation errors.

 b) Mbasic = [0.2 V 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 - 0.2 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]

correct

 0.76 = log[0.2 V NaOH/(0.5 V acetic acid - 0.2 V NaOH)]
4 - 4.76 = - 0.76 NOT 0.76

 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)]
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)]

 1.35 V NaOH = 2.88 V acetic acid V NaOH = 2.88 V acetic acid/1.35 V = 2.13 mL
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.

 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 :))