Advanced Chemistry - Mono Mole

September 15, 2019November 8, 2022

Complete pH titration curve: overview

Titration curves reveal important data of acid-base systems, like the pH of the stoichiometric point, pH at maximum buffer capacity, equilibrium constants, etc. They have different shapes that are governed by the equilibria of the acid, the base and water. In general, titrations are categorized into four groups:

- Strong acid versus strong base
- Weak acid versus strong base
- Strong acid versus weak base
- Weak acid versus weak base

The pH versus volume profile of each group can be expressed explicitly with a unique pH titration curve formula. Knowing the formula for all categories of titration therefore allows us to mathematically explain the shape of the curves, including the reasons for the sharp change of pH near stoichiometric points. In the following derivation of pH curve equations of the four categories of titration, we shall assume that the activity of every chemical species is equivalent to its concentration.

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September 15, 2019September 26, 2024

Monoprotic weak acid versus monoprotic weak base

What is the formula of the titration curve of a monoprotic weak acid versus a monoprotic weak base?

With reference to previous articles, substitute eq6, eq10, K_w= [H⁺][OH^–] and [H⁺] = 10^-pHinto eq1 (where C_aand C_bare now the concentration of a weak acid and the concentration of a weak base respectively), we have,

$10^{-pH}+\frac{K_bC_bV_b10^{-pH}}{(V_a+V_b)(K_w+K_b10^{-pH})}=\frac{K_w}{10^{-pH}}+\frac{K_aC_aV_a}{(V_a+V_b)(10^{-pH}+K_a)}\; \; \; \; \; \; \; \; (12)$

Eq12 is the complete pH titration curve for a monoprotic weak acid versus monoprotic weak base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of CH₃COOH (K_a= 1.75 x 10^-5) with 0.100 M of aqueous NH₃ (K_b= 1.8 x 10^-5), we have the following:

To understand the change in pH near the stoichiometric point versus the change in V_b, we assume that one drop of base is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq12. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm³	19.95	20.00	20.05
pH	6.91	7.01	7.10

The data shows that two drops of base cause a change of only 0.19 in pH before and after the stoichiometric point. Since the change in pH at the stoichiometric point occurs within a very narrow range, no indicator is suitable to accurately monitor the stoichiometric point, not even bromothymol blue (see diagram below for bromothymol blue colour at various pH).

Lastly, we can derive the gradient equation for a weak acid to weak base titration and investigate the inflexion point at pH 7.01 by differentiating eq12 implicitly (see this article), resulting in $\frac{dpH}{dV_b}=1.94\, cm^{-3}$ , i.e. a gradient that makes angle of 62.71^o with the horizontal.

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September 15, 2019September 26, 2024

Comparison of strong acid versus strong base curve and weak acid versus strong base curve

How does the titration curve of a strong acid versus a strong base compare with the titration curve of a weak acid versus a strong base?

Superimposing the weak acid versus strong base titration curve on the strong acid versus strong base titration curve, we have

From the graph, the two curves appears to coalesce when pH > 8. This can be rationalised by comparing eq4 and eq8:

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (4)$

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+\frac{K_aC_aV_a}{(V_a+V_b)(10^{-pH}+K_a)}\; \; \; \; \; \; \; \; (8)$

where the two equations are approximately the same when 10^-pH→ 0, i.e. at high pH. Note that the two curves do not actually coalesce and are still two separate curves when pH > 8 (discernible if the axes of the plot are scaled to a very high resolution).

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September 15, 2019September 26, 2024

Polyprotic acid versus monoprotic strong base

What is the formula of the titration curve of a polyprotic acid versus a monoprotic strong base?

Comparing eq8, eq19 and eq30 the general equation for a polyprotic acid versus a monoprotic strong base titration is:

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+\frac{C_aV_a}{V_a+V_b}\left \{ \frac{\sum_{i=1}^{n}[i10^{-(n-i)pH}\prod_{j=1}^{i}K_{aj}]}{10^{-npH}+\sum_{i=1}^{n}[10^{-(n-i)pH}\prod_{j=1}^{i}K_{aj}]} \right \}$

where n is the basicity of the acid, i.e. the number of replaceable hydrogen atoms in one molecule of the acid.

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September 15, 2019December 9, 2024

Charge balance equations (chemistry)

Charge balance equations are derived using the concept of electroneutrality, where the sum of positive charges equals to the sum of negative charges in a solution. Such equations are useful for analysing acid-base equilibria and formulating complex acid-base titration equations.

Consider a solution containing water, a strong acid of concentration C_a, and a strong base of concentration C_b, with the following equilibria:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

$HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)$

$BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)$

To maintain electroneutrality, the sum of the number of moles of cations H⁺ and B⁺ must equal to that of anions A^– and OH^–. As the volume of the solution is common to all ions,

$[H^+]+[B^+]=[OH^-]+[A^-]$

When formulating charge balance equations for aqueous compounds with multiple equilibria, we need to account for every charged species on the LHS of a particular equilibrium, which can be complicated. To avoid mistakes, we select equilibrium expressions where species on the LHS are neutral. For example, the equilibrium equations for a diprotic acid can be presented in the following ways:

$H_2A(aq)\rightleftharpoons H^+(aq)+HA^-(aq)$

$HA^-(aq)\rightleftharpoons H^+(aq)+A^{2-}(aq)$

$H_2A(aq)\rightleftharpoons 2H^+(aq)+A^{2-}(aq)$

Select the first and third equilibria, i.e., we can imagine part of the initial number of moles of H₂A dissociating into H_a⁺ and HA^–, with the remaining part of the initial number of moles of H₂A dissociating into 2H_b⁺ and A^2-. For the first equilibrium, the charge balance equation is

$[H_a^{\, +}]=[HA^-]\; \; \; \; \; \; \; \; (31)$

For the third equilibrium, the charge balance equation is

$[H_b^{\, +}]=2[A^{2-}]\; \; \; \; \; \; \; \; (32)$

Combining eq31 and eq32,

$[H_a^{\, +}]+[H_b^{\, +}]=[HA^-]+2[A^{2-}]$

$[H^+]=[HA^-]+2[A^{2-}]$

Question

Write the charge balance equation for a triprotic acid.

Answer

Since,

$H_3A(aq)\rightleftharpoons H^+(aq)+H_2A^-(aq)$

$H_3A(aq)\rightleftharpoons 2H^+(aq)+HA^{2-}(aq)$

$H_3A(aq)\rightleftharpoons 3H^+(aq)+A^{3-}(aq)$

we have,

$[H^+]=[H_2A^-]+2[HA^{2-}]+3[A^{3-}]$

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September 15, 2019June 12, 2025

Monoprotic strong acid versus monoprotic strong base

What is the formula of the titration curve of a monoprotic strong acid versus a monoprotic strong base?

Consider a solution containing water, a strong acid (analyte) of concentration C_aand a strong base of concentration C_bwith the following equilibria:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

$HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)$

$BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)$

With reference to charge balance of the above equilibria, the sum of the number of moles of cations H⁺ and B⁺ must equal to that of anions A^– and OH^–. As the volume is of the solution is common to all ions,

$[H^+]+[B^+]=[OH^-]+[A^-]\; \; \; \; \; \; \; \; (1)$

Assume that the strong acid is fully ionised in water, at any point of the titration, the number of moles of A^– must equal to the initial number of moles of HA. As the change in volume of the solution is common to all ions,

$[A^-]=\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (2)$

where V_a and V_b are the volume of acid in the solution and the volume of base in the solution respectively.

Similarly, assuming that the strong base is fully ionised in water, at any point of the titration, the sum of the number of moles of B⁺ must equal to the number of moles of BOH added to the solution. As the change in volume of the solution is common to all ions,

$[B^+]=\frac{C_bV_b}{V_a+V_b}\; \; \; \; \; \; \; \; (3)$

Substitute eq2, eq3, K_w= [H⁺][OH^–] and [H⁺] = 10^-pH in eq1

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (4)$

Eq4 is the complete pH titration curve for a monoprotic strong acid versus monoprotic strong base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of HCl with 0.100 M of NaOH, we have the following:

To understand the sharp change in pH near the stoichiometric point versus the change in V_b, we assume that one drop of base is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq4. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm³	19.95	20.00	20.05
pH	3.78	7.00	10.22

The data shows that just two drops of base cause a drastic change of 6.44 in pH before and after the stoichiometric point. We can therefore use many different indicators that work within the range of the change in pH at the stoichiometric point to monitor the titration. Furthermore, we can derive the gradient equation for a monoprotic strong acid versus monoprotic strong base titration and investigate the inflexion point at pH 7 by rearranging eq4 as:

$10^{-2pH}+\frac{C_bV_b}{V_a+V_b}10^{-pH}-\frac{C_aV_a}{V_a+V_b}10^{-pH}-K_w=0$

and finding $\frac{dpH}{dV_b}$ by implicit differentiation to give:

$\frac{dpH}{dV_b}=\frac{(C_bV_a+C_aV_a)}{ln10(V_a+V_b)[10^{-pH}2(V_a+V_b)+C_bV_b-C_aV_a]}$

At the stoichiometric point, C_bV_b – C_aV_a= 0 for a monoprotic acid-base reaction and the above equation becomes

$\frac{dpH}{dV_b}=\frac{(C_bV_a+C_aV_a)}{10^{log[H^+]}ln100(V_a+V_b)^2}$

Since [H⁺] = 10^-7 M at the stoichiometric point for a strong acid versus strong base titration, the denominator of the above equation is a very small number and hence the gradient of the pH versus V_bcurve, or $\frac{dpH}{dV_b}$ , is a very large value (computed to be 7238.2 cm^-3), making at angle of 89.99^o with the horizontal, i.e. a vertical line.

Question

How do we differentiate 10^-2pH implicitly?

Answer

Let u = 10^-2pHand use the chain rule $\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}$ where y = pH and x = V_b. The result is $-10^{-2pH}2ln10\frac{dpH}{dV_b}$ .

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September 15, 2019September 26, 2024

Monoprotic strong acid versus monoprotic weak base

What is the formula of the titration curve of a monoprotic strong acid versus a monoprotic weak base?

We make use of eq1 from a previous article except that C_bis now the concentration of a weak base. Since the weak base is partially ionised in water, at any point of the titration, the sum of the number of moles of BOH and B⁺ must equal to the number of moles of BOH if it were undissociated. As the change in volume of the solution is common to all ions,

$[BOH]+[B^+]=\frac{C_bV_b}{V_a+V_b}\; \; \; \; \; \; \; \; (9)$

where V_a and V_b are the volume of strong acid in the solution and the volume of weak base in the solution respectively. Substitute $[BOH]=\frac{[B^+][OH^-]}{K_b}$ in eq9 where K_bis the dissociation constant of BOH and rearranging, we have,

$[B^+]=\frac{K_bC_bV_b}{(V_a+V_b)([OH^-]+K_b)}\; \; \; \; \; \; \; \; (10)$

Substitute eq10, eq2, K_w= [H⁺][OH^–] and [H⁺] = 10^-pH in eq1,

$10^{-pH}+\frac{K_bC_bV_b10^{-pH}}{(V_a+V_b)(K_w+K_b10^{-pH})}=\frac{K_w}{10^{-pH}}+\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (11)$

Eq11 is the complete pH titration curve for a monoprotic strong acid versus monoprotic weak base system. We can input it in a mathematical software to generate a curve of pH against V_a. For example, if we titrate 10 cm³ of 0.200 M of aqueous NH₃ (K_b= 1.8 x 10^-5) with 0.100 M of HCl, we have the following:

To understand the change in pH near the stoichiometric point versus the change in V_a, we assume that one drop of acid is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq11. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm³	19.95	20.00	20.05
pH	6.65	5.22	3.78

The data shows that two drops of acid cause a change of 2.87 in pH before and after the stoichiometric point. Since the change in pH at the stoichiometric point is smaller than that of a strong acid versus strong base titration, we can use fewer indicators that work within the range to monitor the titration. Lastly, we can derive the gradient equation for a strong acid to weak base titration and investigate the inflexion point at pH 5.22 by differentiating eq11 implicitly (see previous article), resulting in $\frac{dpH}{dV_a}=-118.9\, cm^{-3}$ , i.e. a gradient that makes angle of –89.52^o with the horizontal.

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Monoprotic weak acid versus monoprotic strong base

What is the formula of the titration curve of a monoprotic weak acid versus a monoprotic strong base?

We make use of eq1 from the previous article except that C_ais now the concentration of a weak acid. Since the weak acid is partially ionised in water, at any point of the titration, the sum of the number of moles of HA and A^– must equal to the number of moles of HA if it were undissociated. As the change in volume of the solution is common to all ions,

$[HA]+[A^-]=\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (5)$

where V_a and V_b are the volume of weak acid in the solution and the volume of strong base in the solution respectively. Substitute $[HA]=\frac{[H^+][A^-]}{K_a}$ in eq5 where K_ais the dissociation constant of HA and rearranging, we have,

$[A^-]=\frac{K_aC_aV_a}{(V_a+V_b)([H^+]+K_a)}\; \; \; \; \; \; \; \; (6)$

Assuming that the strong base is fully ionised in water, at any point of the titration, the sum of the number of moles of B⁺ must equal to the number of moles of BOH added to the solution. As the change in volume of the solution is common to all ions,

$[B^+]=\frac{C_bV_b}{(V_a+V_b)}\; \; \; \; \; \; \; \; (7)$

Substitute eq6, eq7, K_w= [H⁺][OH^–] and [H⁺] = 10^-pH in eq1,

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+\frac{K_aC_aV_a}{(V_a+V_b)(10^{-pH}+K_a)}\; \; \; \; \; \; \; \; (8)$

Eq8 is the complete pH titration curve for a monoprotic weak acid versus monoprotic strong base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of CH₃COOH (K_a= 1.75 x 10^-5) with 0.100 M of NaOH, we have the following:

To understand the change in pH near the stoichiometric point versus the change in V_b, we assume that one drop of base is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq8. The respective pH just before and just after the stoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm³	19.95	20.00	20.05
pH	7.36	8.79	10.22

The data shows that two drops of base cause a change of 2.86 in pH before and after the stoichiometric point. Since the change in pH at the stoichiometric point is smaller than that of a strong acid versus strong base titration, we can use fewer indicators that work within the range to monitor the titration. Lastly, we can derive the gradient equation for a weak acid versus strong base titration and investigate the inflexion point at pH 8.79 by differentiating eq8 implicitly (see previous article), resulting in $\frac{dpH}{dV_b}=117.3\, cm^{-3}$ , i.e. a gradient that makes angle of 89.51^o with the horizontal.

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Diprotic acid versus monoprotic strong base

What is the formula of the titration curve of a diprotic acid versus a monoprotic strong base?

Consider a solution containing water, a diprotic acid of concentration C_aand a strong base of concentration C_bwith the following equilibria:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

$H_2A(aq)\rightleftharpoons H^+(aq)+HA^-(aq)\; \; \; \; \; \; \; \; (13)$

$HA^-(aq)\rightleftharpoons H^+(aq)+A^{2-}(aq)\; \; \; \; \; \; \; \; (14)$

$BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)$

With reference to charge balance of the above equilibria, the sum of the number of moles of cations H⁺ and B⁺ must equal to that of anions HA^–, A^2- and OH^–. As the volume is of the solution is common to all ions,

$[H^+]+[B^+]=[OH^-]+[HA^-]+2[A^{2-}]\; \; \; \; \; \; \; \; (15)$

Read this article to understand how to arrive at eq15.

With reference to eq13 and eq14, at any point of the titration, the sum of the number of moles of H₂A, HA^– and A^2– must equal to the number of moles of H₂A if it were undissociated. As the change in volume of the solution is common to all ions,

$[H_2A]+[HA^-]+[A^{2-}]=\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (16)$

where V_a and V_b are the volume of diprotic acid in the solution and the volume of strong monoprotic base in the solution respectively. The equilibrium constant equations of eq13 and that of eq14 are:

$[H_2A]=\frac{[H^+][HA^-]}{K_{a1}}\; \; \; \; \; \; \; \; (16a)$

$[HA^-]=\frac{[H^+][A^{2-}]}{K_{a2}}\; \; \; \; \; \; \; \; (16b)$

Substitute eq16b in eq16a

$[H_2A]=\frac{[H^+]^2[A^{2-}]}{K_{a1}K_{a2}}\; \; \; \; \; \; \; \; (17)$

Substitute eq16b and eq17 in eq16 and rearranging,

$[A^{2-}]=\frac{C_aV_aK_{a1}K_{a2}}{(V_a+V_b)([H^+]^2+[H^+]K_{a1}+K_{a1}K_{a2})}\; \; \; \; \; \; \; \; (17a)$

Substitute eq17a in eq16b

$[HA^-] =\frac{C_aV_aK_{a1}[H^+]}{(V_a+V_b)([H^+]^2+[H^+]K_{a1}+K_{a1}K_{a2})}\; \; \; \; \; \; \; \; (18)$

Substitute eq7, eq17a, eq18, K_w= [H⁺][OH^–] and [H⁺] = 10^-pH in eq15,

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+\frac{C_aV_aK_{a1}10^{-pH}}{(V_a+V_b)(10^{-2pH}+10^{-pH}K_{a1}+K_{a1}K_{a2})}+\frac{2C_aV_aK_{a1}K_{a2}}{(V_a+V_b)(10^{-2pH}+10^{-pH}K_{a1}+K_{a1}K_{a2})}\; \; \; \; \; \; \; \; (19)$

Eq19 is the complete pH titration curve for a diprotic acid (either strong or weak) versus monoprotic strong base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of H₂SO₄(K_a1= 1000, K_a2= 1.02 x 10^-2) with 0.100 M of NaOH, we have the following:

To understand the change in pH near the 1^st stoichiometric point versus the change in V_b, we assume that one drop of base is about 0.05 cm³ and substitute 19.95 cm³, 20.00 cm³ and 20.05 cm³ into eq19. The respective pH just before and just after the 1^ststoichiometric point (SP) are:

	One drop before SP	SP	One drop after SP
Volume, cm³	19.95	20.00	20.05
pH	1.665	1.668	1.670

The data shows that two drops of base cause a change of only 0.005 in pH before and after the 1^st stoichiometric point, which means we cannot monitor it during the titration process. The 1^st stoichiometric point is not discernable as the 1^st dissociation constant is a very large value, i.e. the first ionisation is complete. As for the 2^nd stoichiometric point,

	One drop before SP	SP	One drop after SP
Volume, cm³	39.95	40.00	40.05
pH	4.69	7.35	10.00

The data shows that just two drops of titrant cause a relatively big change of 5.31 in pH before and after the 2^nd stoichiometric point. We can therefore use many different indicators that work within the range of the change in pH at the 2^nd stoichiometric point to monitor the titration.

Another example is the titration of 10 cm³ of 0.200 M of H₂CO₃ (K_a1= 4.5 x 10^-7, K_a2 = 4.8 x 10^-11) with 0.100M of NaOH. Using eq19, we have:

The changes of pH at both stoichiometric points are not sharp, with the 2^nd one almost non-discernable. Finally, for the titration of 10 cm³ of 0.200 M of H₂SO₃ (K_a1= 1.5 x 10^-2, K_a2 = 6.2 x 10^-8) with 0.100M of NaOH, we have:

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Triprotic acid versus monoprotic strong base

What is the formula of the titration curve of a triprotic acid versus a strong base?

Consider a solution containing water, a triprotic acid of concentration C_aand a strong base of concentration C_bwith the following equilibria:

$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$

$H_3A(aq)\rightleftharpoons H^+(aq)+H_2A^-(aq)\; \; \; \; \; \; \; \; (20)$

$H_2A^-(aq)\rightleftharpoons H^+(aq)+HA^{2-}(aq)\; \; \; \; \; \; \; \; (21)$

$HA^{2-}(aq)\rightleftharpoons H^+(aq)+A^{3-}(aq)\; \; \; \; \; \; \; \; (22)$

$BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)$

With reference to charge balance of the above equilibria, the sum of the number of moles of cations H⁺ and B⁺ must equal to that of anions H₂A^–, HA^2-, A^3- and OH^–. As the volume is of the solution is common to all ions,

$[H^+]+[B^+]=[OH^-]+[H_2A^-]+2[HA^{2-}]+3[A^{3-}]\; \; \; \; \; \; \; \; (23)$

Read this article to understand how to arrive at eq23.

With reference to eq20, eq21 and eq22, at any point of the titration, the sum of the number of moles of H₃A, H₂A^–, HA^2-and A^3– must equal to the number of moles of H₃A if it were undissociated. As the change in volume of the solution is common to all ions,

$[H_3A]+[H_2A^-]+[HA^{2-}]+[A^{3-}]=\frac{C_aV_a}{V_a+V_b}\; \; \; \; \; \; \; \; (24)$

where V_a and V_b are the volume of triprotic acid in the solution and the volume of strong monoprotic base in the solution respectively. The equilibrium constant equations of eq20, eq21 and eq22 are:

$[H_3A]=\frac{[H^+][H_2A^-]}{K_{a1}}\; \; \; \; \; \; \; \; (24a)$

$[H_2A^-]=\frac{[H^+][HA^{2-}]}{K_{a2}}\; \; \; \; \; \; \; \; (24b)$

$[HA^{2-}]=\frac{[H^+][A^{3-}]}{K_{a3}}\; \; \; \; \; \; \; \; (24c)$

Substitute eq24c in eq24b

$[H_2A^-]=\frac{[H^+]^2[A^{3-}]}{K_{a2}K_{a3}}\; \; \; \; \; \; \; \; (25)$

Substitute eq25 in eq24a

$[H_3A]=\frac{[H^+]^3[A^{3-}]}{K_{a1}K_{a2}K_{a3}}\; \; \; \; \; \; \; \; (26)$

Substitute eq26, eq25 and eq24c in eq24 and rearranging,

$[A^{3-}]=\frac{C_aV_aK_{a1}K_{a2}K_{a3}}{(V_a+V_b)([H^+]^3+[H^+]^2K_{a1}+[H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}\; \; \; \; \; \; \; \; (27)$

Substitute eq23 in eq24a and eq24b

$[H_2A^-]=\frac{[H^+]^2C_aV_aK_{a1}}{(V_a+V_b)([H^+]^3+[H^+]^2K_{a1}+[H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}\; \; \; \; \; \; \; \; (28)$

$[HA^{2-}]=\frac{[H^+]C_aV_aK_{a1}K_{a2}}{(V_a+V_b)([H^+]^3+[H^+]^2K_{a1}+[H^+]K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}\; \; \; \; \; \; \; \; (29)$

Substitute eq7, eq27, eq28, eq29, K_w= [H⁺][OH^–] and [H⁺] = 10^-pH in eq23,

$10^{-pH}+\frac{C_bV_b}{V_a+V_b}=\frac{K_w}{10^{-pH}}+A+B+C\; \; \; \; \; \; \; \; (30)$

where

$A=\frac{10^{-2pH}C_aV_aK_{a1}}{(V_a+V_b)(10^{-3pH}+10^{-2pH}K_{a1}+10^{-pH}K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}$

$B=\frac{2\times 10^{-pH}C_aV_aK_{a1}K_{a2}}{(V_a+V_b)(10^{-3pH}+10^{-2pH}K_{a1}+10^{-pH}K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}$

$C=\frac{3C_aV_aK_{a1}K_{a2}K_{a3}}{(V_a+V_b)(10^{-3pH}+10^{-2pH}K_{a1}+10^{-pH}K_{a1}K_{a2}+K_{a1}K_{a2}K_{a3})}$

Eq30 is the complete pH titration curve for a triprotic acid (either strong or weak) versus monoprotic strong base system. We can input it in a mathematical software to generate a curve of pH against V_b. For example, if we titrate 10 cm³ of 0.200 M of H₃PO₄(K_a1= 7.11 x 10^-3, K_a2= 6.00 x 10^-8, K_a3= 4.80 x 10^-13) with 0.100 M of NaOH, we have the following:

The 1^st and 2^nd stoichiometric points are titrated using methyl orange and thymolphthalein as indicators respectively, while the 3^rd stoichiometric point is non-discernible. Another way to titrate phosphoric acid is to first react it with excess cation to completely precipitate the phosphate salt, thereby releasing all H⁺, which is then titrated with a strong base:

$2H_3PO_4(aq)+3Ca^{2+}(aq)\rightarrow Ca_3(PO_4)_2(s)+6H^+(aq)$

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