[ziagemini]

Hello All,

I would like to know how one can draw a buffer capacity curve from a titration curve. I have the titration data of HCl with NaoH. But i don't know how to make buffer capacity curve from it. Please help.

Thanks in advance.

Zen

[Borek]

What buffer capacity definition do you use?

[ziagemini]

Hi,

Basically i am trying to compare my simulation result with the actual one. For simulation i am using the following:
2.303*(10^(-pH)+(Ca*Ka*10^(-pH)/(10^(-pH)+Ka)^2)+(Cw*Kw*10^(-pH)/(10^(-pH)+Kw)^2))

And from the titration data we can also fin the buffer capacity: B = dCb/dpH= -dCa/dpH
Or more clearly by plotting the inverse of slope of titration curve versus pH. how to do that? Do i need to find the slope of my titration curve and then do the inverse and plot it?

Like slope of the curve is= 0.181
Inverse of slope= 1/0.181
Buffer capacity= IOS*pH

Hope to get some good reply soon.

[Borek]

I m not sure what your symbols mean, as you have not explained them.

Note that dC_b = C_B dV where C_B is analytical concentration of titrant. If you can express your titration curve as pH = f(V) it will be just a matter of finding derivative. Even if not, you can always find a value of derivative using numerical methods.

[ziagemini]

Hi,

Here Cb= concentration of strong base(M)
Ca= concentration of strong acid(M).

And for buffer capacity formula: dCb= differential quantity of strong base added

I still don't get your point. As i write before, i found in literature  ''ß as a function of pH maybe obtained by plotting the inverse of slope of titration curve v/s pH''. I did the titration of 0.1M of HCL with 0.1M of Naoh. I got the titration curve as ml of NaoH v/s pH. Now how should i proceed to get the corresponding buffer capacity curve?

Thanks a lot for your time.

[Borek]

Here Cb= concentration of strong base(M)
Ca= concentration of strong acid(M)

I guess these are not initial concentrations of the titrant and titrand, but of acid and base (even if they don't exist) in the mixture? So Ca doesn't change (as you seem to be ignoring volume of titrant added), and Cb changes?

I still don't get your point. As i write before, i found in literature  ''ß as a function of pH maybe obtained by plotting the inverse of slope of titration curve v/s pH''. I did the titration of 0.1M of HCL with 0.1M of Naoh. I got the titration curve as ml of NaoH v/s pH. Now how should i proceed to get the corresponding buffer capacity curve?

Titration curve plot shows pH as a functions of V of titrant added, so you have pH and V on axes. As number of moles of titrant added is directly proportional to V, you can also treat it as pH=f(Cb) (assuming I got your symbols right), in which case plot shows pH vs Cb. Slope is then dpH/dCb, or inverse of buffer capacity, like you wrote.

All you have to do is to calculate the slope and calculate the inverse - either algebraically or numerically.

For the record: buffer capacity page.

[ziagemini]

Hello,

I tried to find the buffer capacity as discussed before, but for slope i am getting very strage values and from that the buffer capacity. I am adding the pdf file of my calculation. Can somebody help me what i am doing wrong?

Hope to have some good answer soon.

Regards!

P.S. Slope= (y2-y1)/(x2-x1)

I also read somewhere that by doing so we can get buffer capacity:

ß=[(deltav/deltapH)*concentration of titrant]/[1000*(amount of acid in litres+amount of basein litres)]

Is it right???

 

[Borek]

Slope of the titration curve is just dpH/dV, even with the formula you listed I don't see how you got values present in the table. Explain how you got one of the slope values, say 9.75 (6 mL of titrant).

Note that your DeltaVml/deltapH column contains quite good approximation of the buffering capacity, with highest values around pH=pKa.

[ziagemini]

Thanks a lot for giving my post time and answering. i know that DeltaVml/deltapH gives very good approximation but i need a concrete buffer capacity curve. As you asked about the slope:

ml Naoh              ph                                       DeltaVml/deltapH
2                      3.671                                      6.270
4                      3.990                                      9.569

Slope between these two points= (9.569-6.270)/(3.990-3.671) = 10.344 if its OK how to proceed further with this slope?
I think that's where i am making mistake.

The second way i am trying to find the ß is:(lets say for 4 ml titrant used)
=(DeltaVml/deltapH*conc. Naoh*ml NaOH)/(1000*Total solution in litres)
=(9.569*0.01*4)/(1000*0.027)= 0.0132

Please tell me which way is right and what i am doing wrong. Thanks a lot for your help. Have a nice day.

Regards!

[Borek]

What you have calculated is not a slope. Slope of the titration plot is dpH/dV, or reciprocal of the DeltaVml/deltapH value you calculated in your table. These values are not exact because approximating derivative by finite differences. The smaller the difference between volumes for which you calculate pH, the better the approximation, that's numerical differentiation. For best results calculate pH for V-ΔV and V+ΔV (where ΔV is something small - say 0.001mL) and use these values to calculate your DeltaVml/deltapH - that will be very good estimate of buffer capacity at point V. Note that the distance between points will be 2*ΔV.

[ziagemini]

Hello All,

Still struggling to find the buffer capacity curve. Attached all the things i have done so far. But not able to find the perfact curve. Borek you are right that if the interval is small we will get refined values. Hope to hear some good opinion from your side.

Regards!

[Borek]

All your curves look in a way similar and all look similar to the correct buffer capacity curve, I have no idea what you mean by a "perfect curve".

β definition you give is a little bit strange. (conc. Naoh*ml NaOH)/(1000*Total solution in litres) is just molarity of NaOH added to the acid after being diluted. Then you multiply it by DeltaVml - CV is number of moles, which would make sense, however, multiplying V of titrant by C of titrant diluted gives just some random number, unrelated to anything.

[ziagemini]

Hello Everyone,

Hopefully last question on buffer capacity curve. The curves obtained from experimental as well as simulated seems to be OK (see file). The question is regarding the conversion of dV/dpH units i.e. ml Naoh/pH into moles/L.pH, as the units of Buffer Capacity.

lets say: dV/dpH= 30 ml NaoH/pH= so we multiply this with concentration of NaoH and divide by 1000. Do we have to divide this value with the total solution in litres?? ( i have done that)     

Hope you understand the question.Thanks a lot for your time.

Regards!