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How can I create a graph of a titration curve (acid-base, strong or weak, not something specific), with an equation or a function. In other words, what is the equation that describes such a curve that could be used in graphing software to recreate the curve without actually needing any experimental values? Also, if they exist, what are the parameters of the function (Molarity? Volume? Etc.)

Mithoron
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MariosA
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    https://doi.org/10.1007/s00897000426a – andselisk Mar 16 '21 at 13:01
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    This has essentially been done in a number of good answers at this site. Among other ways I do this, I use the equations in the spreadsheet screenshots in this answer: https://chemistry.stackexchange.com/a/136203/79678. – Ed V Mar 16 '21 at 13:13
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    Or try Tom O’Haver’s free Excel or OpenOffice titration curve software: https://terpconnect.umd.edu/~toh/models/TitrationDemo.html. There is a lot of stuff at his website, so maybe look around there. And, of course, there are books, other free pH calculation software, etc. – Ed V Mar 16 '21 at 14:06
  • @EdV Thanks, I've visited the answer you mentioned, but there's not a pH variable in any of the equations there. Do I just solve for it with a negative log? – MariosA Mar 16 '21 at 15:45
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    @EdV This is for excel, I'll try it out but I prefer a mathematical function rather than a spreadseet, so that I can use it anywhere. Except if there's a way to extract the function from excel? – MariosA Mar 16 '21 at 15:46
  • The equations are right in the spreadsheet screenshots! They come directly from the equilibrium constant expressions, mass balance, etc. So there is no need to use a spreadsheet. And pH is just negative log of hydrogen ion concentration, ignoring activity coefficients. So you go back and forth with those. – Ed V Mar 16 '21 at 15:50
  • @EdV Alright. So for monoprotic acids I'll be fine with that. I'll try it out and if I find any problems I'll write back. Thanks! – MariosA Mar 16 '21 at 16:02
  • So for monoprotic acids, only the first acid dissociation constant exists. Therefore, just take the equation in the second spreadsheet screenshot, i.e., carbonic acid titrated with NaOH, and let the second and third dissociation constants equal zero, like it says there. This greatly simplifies the equation. – Ed V Mar 16 '21 at 16:07
  • I tried solving for [H+], it's a mess, haven't even tried to take the negative log of it. Any suggestions? – MariosA Mar 16 '21 at 21:03
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    https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.494.1535&rep=rep1&type=pdf – Buck Thorn Mar 17 '21 at 09:03
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    Go here and scroll down past the photos of sapphire tubing. There is no problem solving for [H+], but it is very easy to just take the easy route, as explained in the link. – Ed V Mar 18 '21 at 00:55
  • @EdV I didn't understand how I can graph it for -log([H+]) when it is just a concentration. The formula solved for Va that you described contained the concentration of H+. How do I solve for pH? (This is more of a maths question rather than a chemistry one) – MariosA Mar 18 '21 at 12:48
  • Let us assume, arbitrarily, that you use the easy way. A spreadsheet program is convenient. So, first make a column of pH values you are interested in, e.g., values such as 1.00, 1.10, ..., 13.00. Now in the next column, put in the corresponding [H+] values. These are [H+] = 10^(-pH). So now you have two columns. Now use the Vb equation to compute Vb for each [H+] in the second column. Put these in a third column. This is very easy using an actual spreadsheet program and also easy using any computer language. Then plot pH as the vertical axis and Vb as the horizontal axis. All done! – Ed V Mar 18 '21 at 12:59
  • I updated what was at the link here to show the whole thing: a spreadsheet for titration of acetic acid with MOH. Study the spreadsheet until it makes good sense, OK? – Ed V Mar 18 '21 at 14:11
  • @EdV Sir thank you very much for your work. I will study the spreadsheet and I will let you know. – MariosA Mar 18 '21 at 14:22
  • Good! Please copy what I posted at the link: it is only temporary and I will delete it in a three days. – Ed V Mar 18 '21 at 14:30
  • I updated the spreadsheet screen shot: the volumes Va and Vb are in L, not mL. It does not really matter except to be more realistic: 0.025 L = 25 mL, which is typical in a titration. – Ed V Mar 18 '21 at 15:56

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If $\pu{1 l}$ of $\pu{1 M}$ $\ce{HCl}$ is gradually neutralized by adding $x\,\pu{mol}$ $\ce{NaOH}$ without change in volume, the $\mathrm{pH}$ of the obtained solution is given by

$$\mathrm{pH} = -\log\left(\frac{1-x}{2} + \frac{1}{2}\sqrt{(1 - x)^2 + 4\times10^{-14}}\right).$$

andselisk
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Maurice
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    I want the volume to be a variable here. Suppose we have an unkown molarity weak acid we want to titrate with a strong base of known molarity. We gradually add the base solution to the acid solution. The volume is variable. – MariosA Mar 16 '21 at 21:04
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    @MariosA And that is already done in the equation I referenced in other comments: just zero out the second and third acid dissociation constants and the equation simplifies greatly. Then either use pH as the independent variable, computing base volumes as a function of pH (this is the very easy trick) or use base volume as the independent variable and compute pH as a function of base volume (a little harder, but we have computers now). – Ed V Mar 16 '21 at 21:11