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I've been learning how to simulate a 1M phosphoric acid titration curve using numerical methods in R.

So far has this has been the best curve:

enter image description here

Notice how it flattens out (artificially) towards the left.

I am not certain if this is a programming and / or a "chemistry" issue.


The system is defined by charge balance, mass balance, and equilibrium equations:

$$[\ce{H3A}] + [\ce{H2A-}] + [\ce{HA^{2-}}] + [\ce{A^{3-}}] = P_{CA}$$

$$[\ce{H+}] + [\ce{Na+}] = [\ce{H2A-}] + 2[\ce{HA^{2-}}] + 3[\ce{A^{3-}}] + Kw/[\ce{H+}]$$

$$K_{a, 1} = [\ce{H2A-}] [\ce{H+}] / [\ce{H3A}]$$

$$K_{a, 2} = [\ce{HA^{2-}}] [\ce{H+}] / [\ce{H2A^-}]$$

$$K_{a, 3} = [\ce{A^{3-}}] [\ce{H+}] / [\ce{HA^{2-}}]$$

I've posted a very similar question at SO and shared code for the simulation over there, using R's nleqslv::nleqslv (go have a look if you'd like to).

I thought this is also a good place for this issue. For instance, this answer (and the ones it links to) do seem to be useful, but I'm not sure how to solve that equation for $[\ce{H3O+}]$ (called $x$ over there).

I would like to know which is the best way to simulate a titration curve.

And also: are the equations I used correct? do you know what am I missing?

Thanks!

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  • $\begingroup$ You may compare the visual result with other's solutions in the field, e.g. CurTipot, especially to avoid the sharp transition above $[\ce{Na+}] > 2$. And, for future reference, learn how to use mhchem in the body of a question / answer / comment on ChemSE (not in the title), e.g. here. $\endgroup$
    – Buttonwood
    Mar 31 at 13:46
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Much easier is to calculate the inverse function $[\ce{Na+}]=f([\ce{H+}],K_\mathrm{a1},K_\mathrm{a2},K_\mathrm{a3})$.

Calculate fractions of respective phosphate forms as the function of $\ce{[H+]}$ and acidity constants.

First, calculate the common denominator $CD$:

$$a_1 = [\ce{H+}]^3$$ $$a_2 = [\ce{H+}]^2 \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$ $$a_3 = [\ce{H+}] \cdot K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$ $$a_4 = K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot K_\mathrm{a3} = a_3 \cdot \frac {K_\mathrm{a3}}{[\ce{H+}]}$$

$$CD = a_1 + a_2 + a_3 + a_4$$

.. and then reuse it and its additive terms(*):

$$[\ce{H3A}]=c_0 \cdot \frac {a_1}{CD}$$

$$[\ce{H2A-}]=c_0 \cdot \frac {a_2}{CD}$$

$$[\ce{HA^2-}]=c_0 \cdot \frac {a_3}{CD}$$

$$[\ce{A^3-}]=c_0 \cdot \frac {a_4}{CD}$$

$c_0$ is the total phosphate concentration. From the fractions, you can get $\ce{[Na+]}$:

$$\ce{[Na+]} = \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]}$$


There is one important thing to consider: Titration dilutes the acid.

If there is the volume $V_0$ of $\ce{H3PO4}$ with molar concentration $c_\mathrm{0,init}$, and if we spend the volume $V_1$ of $\ce{NaOH}$ solution of concentration $c_1$, then:

$$[\ce{Na+}]=c_1 \cdot \frac {V_1}{V_0 + V_1 }$$ $$c_0=c_\mathrm{0,init} \cdot \frac {V_0}{V_0 + V_1}$$

So perhaps the best approach can be:

  • There are given concentrations of respective solutions
  • For given volumes of respective solutions calculate $c_0$ and $\ce{[Na+]}$
  • For given $c_0$ and $\ce{[Na+]}$, calculate $\ce{[H+]}$ by finding the solution of the equation below ( $f([\ce{H+}]) = 0$ ) by numerical methods, like Regular falsi, or Newton one.

$$ \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]} - \ce{[Na+]} = 0$$

For $y_1=f(x_1)$, $y_2=f(x_2)$, $y_1 \cdot y_2 \lt 0$, there is the Regula falsi iteration:

$$x_3 = \frac{x_2 \cdot y_1 - x_1 \cdot y_2 }{ y_1 - y_2}$$


(*) Note that the formulas are quite common, I have just rearrange them. They should be able to be found on internet. I have once derived them myself, but not this time, because I remember them.

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  • $\begingroup$ Excelent! Thanks :) I'll post the code in another answer and accept yours. $\endgroup$
    – Naiky
    Mar 31 at 22:54
  • $\begingroup$ Hi again! I've been through this again and thought I'd try getting an analytical solution too. Do you think this equation is right? Thanks! $\endgroup$
    – Naiky
    May 22 at 4:48
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The R code equivalent of Poutnik's answer, for refrence:

# Sumamente facil

Ka.1 <- 7.1 * 10^-3
Ka.2 <- 6.3 * 10^-8
Ka.3 <- 4.5 * 10^-13
Kw <- 10^-14

P_ca <- 1

pH.seq <- seq(from=0,to=14,length.out = 1000)

a1 <- function(H) H^3
a2 <- function(H) H^2 * Ka.1
a3 <- function(H) H * Ka.1 * Ka.2
a4 <- function(H) Ka.1 * Ka.2 * Ka.3
cd <- function(H) a1(H) + a2(H) + a3(H) + a4(H)

H3A <- function(H) P_ca * a1(H) / cd(H)
H2A <- function(H) P_ca * a2(H) / cd(H)
HA <- function(H) P_ca * a3(H) / cd(H)
A <- function(H) P_ca * a4(H) / cd(H)

OH <- function(H) Kw/H

Na <- function(H) H2A(H) + 2*HA(H) + 3*A(H) + OH(H) - H

Na.seq <- Na(H = 10^-pH.seq)

plot(Na.seq, pH.seq)

Also as suggested by Poutnik, a volume correction method was implemented here (it's not Newton nor regula falsi, I got lazy).

Plots

Get your plots here!

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  • $\begingroup$ Note that [H+] giving [Na+]=0 means [H+] of H3PO4 solution at c0. Negative [Na+] have no chemical meaning, it is just mathematical extrapolation. There will be need of iteration as c0=f([Na+]). You dilute the solution by titration. $\endgroup$
    – Poutnik
    Apr 1 at 4:44
  • $\begingroup$ See the A update, addressing dilution. $\endgroup$
    – Poutnik
    Apr 1 at 6:50
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    $\begingroup$ R code syntax note: They could not use boring =, that had to use <- . I would hate to type it 1000 times. $\endgroup$
    – Poutnik
    Apr 1 at 7:04
  • $\begingroup$ Sorry about the arrows "<-" indeed assignment is not the same as equality :P (Alt + - is handy). I wrote an approximation method for the correction, calculating [Na] as before, but using it as an estimate for updating solution volume, then repeat until convergence. Seems to not-crash, but I'm not sure it converges where it's supposed to. $\endgroup$
    – Naiky
    Apr 1 at 14:59

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