I'm measuring buffer capacity by titrating $\ce{NaOH}$ into a buffer of $\ce {H3PO4 + NaH2PO4}$. I'm measuring buffer capacity by recording the amount of $\ce{NaOH}$ required to increase the $\mathrm{pH}$ of the buffer by 2 units, and then taking the ratio.

My problem is that I have no idea how much $\ce{NaOH}$ I'll need to change the $\mathrm{pH}$ by that much. If I've underestimated the strength of my buffer, I might have to reduce the drop in $\mathrm{pH}$ to just one unit. I need to have a rough idea so I can organize how much of each I'll need to prepare.

Basically, is there any way I can predict (roughly) how much $\ce{NaOH}$ I'll need?


  • $1\:\mathrm{M}$ $\ce{H_3PO_4}$ and $1\:\mathrm{M}$ $\ce{NaH_2PO_4}$
  • Buffer volume: $50\:\mathrm{mL}$
  • $1\:\mathrm{M}$ $\ce{NaOH}$
  • $\begingroup$ If you can calculate the theoretical buffer capacity, you might be able to come to an approximation. But theoretical capacities usually differ from the experimental ones. $\endgroup$
    – user16096
    May 15, 2015 at 7:03
  • $\begingroup$ How would I go about that though? I can get the pH with the Henderson-Hasselbalch equation, but where do I go from there? $\endgroup$
    – deusy
    May 15, 2015 at 9:14

2 Answers 2


You can calculate the quantity of sodium hydroxide, required to change the $\mathrm{pH}$ of your system by two unites, by calculating the concentration of sodium ion in the charge balance equation:

$$\ce{[Na+] +[H+]=[OH- ] +[H2PO4^{-}]} \,\,\,\,(1)$$ In fact, you know the initial concentration of sodium ion in the buffer system before adding sodium hydroxide (from the known concentration of $\ce{NaH2PO4}$), and you calculate the final concentration of sodium ion after adding sodium hydroxide by the above equation. So you can deduce the required quantity of sodium hydroxide.

As for the other ions in the charge balance equation, they are calculated as follows: At first the $\mathrm{pH}$ of the buffer system $\mathrm{pH=pK}_a=2.15$.

  • If you want to change the $\mathrm{pH}$ of your system by two unites, then the final $\mathrm{pH}$: $$\mathrm{pH}= 4.15\,\,\,\,(2)$$

    Now you can estimate the concentrations of ion hydronium and ion hydroxide.

  • The use of Henderson-Hasselbalch equation gives: $$\ce{[H2PO4^- ]= 100[H3PO4 ]}\,\,\,\,(3)$$

  • The equation of matter conservation gives: $$\ce{[H2PO4^- ] + [H3PO4 ]= C_0=1 \mathrm{M}}$$ As the buffer is $25\,\mathrm{mL}$ of $1\,\mathrm{M}$ of each of the components, so there's a total of $1\,\mathrm{M}$ in the $50\,\mathrm{mL}$ solution due to dilution (not as you mentioned in your comment).

    Now, you can calculate the concentration of ion $\ce{H2PO4^{-}}$ from equation (3) and (4):$$\ce{[H2PO4^- ] = \frac{100}{101}\mathrm{M}}$$

If we substitute the numerical values in equation (1): $$\ce{[Na+] +10^{-4.15}=\frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$

$$\ce{[Na+]_0 + [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$ $$0.50 +\ce{ [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$ $$\ce{ [Na+]_{\mathrm{added}}} \approx 0.4902 \mathrm{M} $$

$$n=\ce{ [Na+]_{\mathrm{added}}} \times V_{\mathrm{sol}}= 0.4902\times 50\times 10^{-3} =24.51 \times 10^{-3} \mathrm{mol} $$

The volume of sodium hydroxide is $$V_{\mathrm{added}} = 24.51 \times 10^{-3}\times 1= 24.51 \times 10^{-3} \mathrm{L} $$ I suggest to use a solution of sodium hydroxide more concentrated, so as to neglect the effect of the added volume on the initial concentrations of the buffer system.

  • $\begingroup$ I think I understand what you're getting at, but I'm still really confused. I understand all of the theory individually, but how do I actually use it together solve for the number of moles of NaOH I'll need? By the way, the buffer is 25mL of 1M of each of the components, so there's a total of 2M in the 50mL solution. $\endgroup$
    – deusy
    May 16, 2015 at 8:47
  • $\begingroup$ Hey, just want to say thanks. I did the experiment today and your numbers were spot on. $\endgroup$
    – deusy
    May 20, 2015 at 9:09

You have a phosphate system so compute

$$r_1 = 10^{(pH - pK_1)}$$ $$r_2 = 10^{(pH - pK_2)}$$

$$r_3 = 10^{(pH - pK_3)}$$

$$f_0 = 1/(1 + r_1 + r_1*r_2 + r_1*r_2*r_3)$$

$$f_1 = f_0 r_1$$

$$f_2 = f_1 r_2$$

$$f_3 = f_2 r_3$$

$$Q = (f_1 + 2 f_2 + 3 f_3)$$

Do this for closely spaced values of pH near your buffer design pH. Obviously, use a spreadsheet, MatLab, Igor or some other program.

If the total phosphate in your buffer is 1 mole then $f_0$ is the number of moles of phosphoric acid, $f_1$ the moles of $\ce{H2(PO4)-}$, $f_2$ the moles of $\ce{H2PO4^{-2}}$ and $f_3$ the moles of $\ce{PO4^{-3}}$. $Q$ represents the moles of negative charge on the phosphate species. If you add acid (protons) the balance between the 4 phosphate species shift and so does $Q$. The buffering capacity at any pH is the change in $Q$ corresponding to a small change in pH. If you differentiate $Q$ with respect to pH you will obtain a curve of the buffering capacity in units of $Eq·pH^-1·mol^{-1}$. It will exhibit three peaks at the three pK's where the buffering is maximum. Multiply this curve by the total moles of $\ce(PO4)$ you put into the buffer and you have Eq/pH. The Eq of acid to be added to the buffer for a shift of ${\Delta pH}$, when that is small is ${\Delta pHEq/pH}$.The same amount of base will shift the pH by the same amount but upwards.

For a larger pH shift the amount of acid or base required to go from $pH_1$ to $pH_2 is $Q(pH_2) - Q(pH_1)$$.

If $pH_2 > pH_1$ then $Q(pH_2) > Q(pH_1)$ and $Q(pH_2) - Q(pH_1) > 0$. If,for example, this is the case and you find $Q(pH_2) - Q(pH_1) = .010$ Eq (10 mEq) that means that you would need to add 10 mL of 1 N $\ce{NaOH}$ or 1 mL of 10 N.

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    – andselisk
    Jan 23, 2018 at 0:24

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