# Using Henderson-Hasselbach equation?

1. Oct 5, 2007

### Little Bear

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. Oct 6, 2007

### Little Bear

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.

3. Oct 6, 2007

### Hunt_

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

4. Oct 6, 2007

### Hunt_

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.

5. Oct 6, 2007

### Little Bear

Thank you

Thank you so much.

6. Oct 6, 2007

### Little Bear

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

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.

7. Oct 7, 2007

### Hunt_

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.

8. Oct 7, 2007

### Hunt_

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.

9. Oct 7, 2007

### Little Bear

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_.

10. Oct 7, 2007

### Little Bear

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.

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)

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

11. Oct 8, 2007

### Little Bear

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.

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.

12. Oct 10, 2007

### Hunt_

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 :

$$V_{NaOH} = \frac{n_o}{0.2} \ \ \frac{10^{pH - pK_a}}{1 +10^{pH - pK_a}}$$

13. Oct 10, 2007

### Hunt_

I will look later and see where you made your mistake . Sorry but there's No time now...

14. Oct 11, 2007

### Hunt_

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.

15. Aug 1, 2010

### kissandbang

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

16. Aug 1, 2010

### Redbelly98

Staff Emeritus