Advanced Chemistry - Mono Mole

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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September 15, 2019September 26, 2024

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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September 15, 2019September 26, 2024

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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September 15, 2019August 30, 2025

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:

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

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

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

Substituting eq24c into eq24b gives:

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

Substituting eq25 into eq24a yields:

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

Substituting eq26, eq25 and eq24c into eq24 and rearranging results in:

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

Substituting eq23 into eq24a and eq24b gives:

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

and

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

Substituting eq7, eq27, eq28, eq29, K_w= [H⁺][OH^–] and [H⁺] = 10^-pH into eq23 yields:

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

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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September 15, 2019October 11, 2024

Buffer capacity (solution chemistry)

The buffer capacity β of a solution is a measure of the solution’s resistance to changes in pH. It is defined as the number of moles of a strong acid or a strong base needed to change one unit of pH of 1.00 L of the solution.

Mathematically,

$\beta=\frac{dc_b}{dpH}\; \; \; \; \; \; \; \; (1)$

$\beta=-\frac{dc_a}{dpH}\; \; \; \; \; \; \; \; (2)$

where c_b is the number of moles per litre of a strong base, and c_ais the number of moles per litre of a strong acid.

Since buffer capacity is defined as a positive value, eq2 has a negative sign because the change in pH is negative when a strong acid is added to the buffer solution.

Question

Is a strong base versus strong acid system a buffer?

Answer

According to the above definition, a strong base versus strong acid system can also be considered a buffer, but with negligible buffer capacity. This is because a strong acid is fully dissociated in water, and any addition of a strong base reduces the concentration of H⁺; whereas an aqueous weak acid is partially dissociated, and any H⁺ removed by the base is partially replenished by further dissociation of the weak acid, thereby minimising the increase in the solution’s pH.

When an acid is added to the strong acid solution, any molecular acid formed by the combination of H⁺ from the added acid and the conjugate base of the strong acid immediately dissociates into the component ions. However, when H⁺from an acid is added to a weak acid solution, it combines with the conjugate base of the weak acid, partially shifts the equilibrium of the weak acid’s dissociation to the left, thus minimising the decrease in the solution’s pH.

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September 15, 2019September 26, 2024

Buffer capacity of an aqueous solution

The buffer capacity of an aqueous solution, β_s, consists of the buffer capacities of the components of the solution:

$\beta_s=\beta_w+\beta_a+\beta_b+...\; \; \; \; \; \; \; \; (7)$

where β_w is the buffer capacity of water, β_ais the buffer capacity of a weak acid and β_bis the buffer capacity of a weak base.

To prove that buffer capacities are additive, we substitute eq1 in eq7

$\frac{dc_{b,s}}{dpH}=\frac{dc_{b,w}}{dpH}+\frac{dc_{b,a}}{dpH}+\frac{dc_{b,b}}{dpH}+...$

$\frac{dc_{b,s}}{dpH}=\frac{d\left (c_{b,w}+c_{b,a}+c_{b,b}+... \right )}{dpH}$

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September 15, 2019January 21, 2025

Buffer capacity of water

What is the formula describing the buffer capacity of water?

Consider a solution containing water and a strong base with the following equilibria:

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

$NaOH(aq)\rightleftharpoons Na^+(aq)+OH^-(aq)$

With reference to the above equilibria, the sum of the number of moles of H⁺ and Na⁺ must equal to that of OH^–(see this article for details). As the volume of the solution is common to all ions,

$[H^+]+[Na^+]=[OH^-]\; \; \; \; \; \; \; \; (3)$

Substituting c_b = [Na⁺] and K_w= [H⁺][OH^–] in eq3 yields

$c_b=\frac{K_w}{[H^+]}-[H^+]$

$\frac{dc_b}{dpH}=\frac{dK_w[H^+]^{-1}}{dpH}-\frac{d[H^+]}{dpH}\; \; \; \; \; \; \; \; (4)$

Substituting eq1 and pH = –log[H⁺] in eq4 gives

$\beta_w=-\frac{dK_w[H^+]^{-1}}{dlog[H^+]}+\frac{d[H^+]}{dlog[H^+]}\; \; \; \; \; \; \; \; (4a)$

$\beta_w=ln10\left (\frac{K_w}{[H^+]}+[H^+] \right )\; \; \; \; \; \; \; \; (5)$

To determine the maximum or minimum buffer capacity of water, we take the derivative of eq5 with respect to [H⁺] to give:

$\frac{d\beta_w}{d[H^+]}=ln10\left (\frac{dK_w[H^+]^{-1}}{d[H^+]}+\frac{d[H^+]}{d[H^+]} \right )$

$\frac{d\beta_w}{d[H^+]}=ln10\left ( -\frac{K_w}{[H^+]^2}+1 \right )\; \; \; \; \; \; \; \; (6)$

Letting $\frac{d\beta_w}{d[H^+]}=0$ yields

$\frac{K_w}{[H^+]^2}=1$

Substituting K_w= [H⁺][OH^–] in the above equation results in

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

This means that a stationary point for the function in eq5 occurs when [OH^–] = [H⁺] , i.e. at pH = 7. Let’s investigate the nature of this stationary point by differentiating eq6 again.

$\frac{d^{\, 2}\beta_w}{d[H^+]^2}=2ln10\frac{K_w}{[H^+]^3}$

Substituting K_w= [H⁺][OH^–] in the above equation, and noting that [OH^–] = [H⁺], gives

$\frac{d^{\, 2}\beta_w}{d[H^+]^2}=\frac{2ln10}{[H^+]}$

$\frac{d^{\, 2}\beta_w}{d[H^+]^2}> 0$

If we repeat the above steps and consider the case of the addition of a strong acid to water, we end up with the same conclusion. Hence, water has minimum buffer capacity at pH = 7.

The buffer capacity of water over the entire range of pH can be found by substituting [H⁺] = 10^-pHin eq5 to give

$\beta_w=ln10\left ( \frac{K_w}{10^{-pH}}+10^{-pH} \right )\; \; \; \; \; \; \; \; (6a)$

The plot of the above equation with β_was the vertical axis and pH as the horizontal axis is shown in the diagram below.

From the graph, the buffering capacity of water increases infinitesimally from a minimum of 0 at pH 7, to 0.023 at pH 2 on the left and pH 12 on the right, and then drastically to 2.303 at pH 0 on the left and pH 14 on the right. This means that water is only relatively effective as a buffer at extreme pH levels.

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September 15, 2019March 27, 2025

Buffer capacity (monoprotic weak acid and its conjugate base)

What is the formula describing the buffer capacity of an aqueous monoprotic weak acid and its conjugate base?

Consider a solution containing water, a strong base and a weak acid, HA, with the following equilibria:

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

$NaOH(aq)\rightleftharpoons Na^+(aq)+OH^-(aq)$

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

With reference to the above equilibria, the sum of the number of moles of H⁺ and Na⁺ must equal to that of OH^– and A^–(see this article for details). As the volume is of the solution is common to all ions,

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

Substituting c_b = [Na⁺] and K_w= [H⁺][OH^–] in the above equation gives

$c_b=\frac{K_w}{[H^+]}-[H^+]+[A^-]$

Multiplying the term [A^–] on the RHS of the above equation by $\left (\frac{[HA]+[A^-]}{[HA]+[A^-]} \right )\left (\frac{[H^+]/[HA]}{[H^+]/[HA]} \right )$ and noting that $K_a = \frac{[H^+][A^-]}{[HA]}$ , yields

$c_b=\frac{K_w}{[H^+]}-[H^+]+\frac{[HA]_TK_a}{K_a+[H^+]}$

where [HA]_T=[HA] + [A^–].

$\frac{dc_b}{dpH}=\frac{dK_w[H^+]^{-1}}{dpH}-\frac{d[H^+]}{dpH}+\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dpH}$

Substituting eq1 and pH = –log[H⁺] in the above equation results in

$\beta_s=-\frac{dK_w[H^+]^{-1}}{dlog[H^+]}+\frac{d[H^+]}{dlog[H^+]}-\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dlog[H^+]}$

Substituting eq4a in the above equation gives

$\beta_s=\beta_w-\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dlog[H^+]}\; \; \; \; \; \; \; \; (8)$

Comparing eq8 with eq7 with β_s= 0 (since we are considering the case where a strong base is added to a solution containing water and a weak acid)

$\beta_a=-\frac{d\frac{[HA]_TK_a}{K_a+[H^+]}}{dlog[H^+]}\; \; \; \; \; \; \; \; (9)$

Since K_a is a constant

$\beta_a=\frac{[HA]_TK_a[H^+]ln10}{{\left ( K_a+[H^+] \right )^2}}\; \; \; \; \; \; \; \; (10)$

To determine the maximum or minimum buffer capacity of the weak monoprotic acid buffer,

$\frac{d\beta_a}{d[H^+]}=K_aln10\frac{d\frac{[HA]_T[H^+]}{\left ( K_a+[H^+] \right )^2}}{d[H^+]}$

$\frac{d\beta_a}{d[H^+]}=K_aln10\left \{\frac{\left (K_a+[H^+] \right )^2 [HA]_T-2[HA]_T[H^+]\left ( K_a+[H^+] \right )}{\left ( K_a+[H^+] \right )^4} \right \}\; \; \; \; \; \; \; \; (11)$

Let $\frac{d\beta_a}{d[H^+]}=0$

$K_a=[H^+]\; \; \; \; \; \; \; \; (12)$

A stationary point for the function in eq10 occurs when K_a= [H⁺]. Since $K_a = \frac{[H^+][A^-]}{[HA]}$

$[HA]=[A^-]\; \; \; \; \; \; \; \; (13)$

Let’s investigate the nature of this stationary point by differentiating eq11.

$\frac{d^2\beta_a}{d[H^+]^2}=K_aln10\frac{d\frac{(K_a+[H^+])[HA]_T-2[HA]_T[H^+]}{(K_a+[H^+])^3}}{d[H^+]}$

$\frac{d^2\beta_a}{d[H^+]^2}=K_aln10\frac{2[HA]_T(K_a+[H^+])^2([H^+]-2K_a)}{(K_a+[H^+])^6}$

$\frac{d^2\beta_a}{d[H^+]^2}=K_aln10\frac{2[HA]_T(K_a+[H^+])^2[H^+]\frac{[HA]-2[A^-]}{[HA]}}{(K_a+[H^+])^6}$

Substituting eq13 in the above equation yields

$\frac{d^2\beta_a}{d[H^+]^2}=-\frac{2K_a(ln10)[HA]_T(K_a+[H^+])^2[H^+]}{(K_a+[H^+])^6}$

$\frac{d^2\beta_a}{d[H^+]^2}< 0$

If we consider the case of the addition of a strong acid to the aqueous solution, we end up with the same conclusion. Hence, a monoprotic weak acid has maximum buffer capacity when [HA] = [A^–]. The buffer capacity of the monoprotic weak acid over the entire range of pH can be found by substituting [H⁺] = 10^-pHin eq10 to give

$\beta_a=\frac{[HA]_TK_a10^{-pH}ln10}{(K_a+10^{-pH})^2}\; \; \; \; \; \; \; \; (14)$

Plotting the above equation with β_aas the vertical axis and pH as the horizontal axis results in:

The total buffer capacity of an aqueous monoprotic weak acid is given by the combined equations of eq8 and eq9, i.e.

$\beta_s=\beta_w+\beta_a\; \; \; \; \; \; \; \; (15)$

The total buffer capacity of an aqueous monoprotic weak acid over the entire range of pH can be found by substituting eq6a and eq14 in eq15 to give

$\beta_s=ln10\left ( \frac{K_w}{10^{-pH}}+10^{-pH} \right )+\frac{[HA]_TK_a10^{-pH}ln10}{(K_a+10^{-pH})^2}\; \; \; \; \; \; \; \; (16)$

Using eq16 to plot a graph of β_sversus pH for 1.0 M aqueous ethanoic acid (K_a = 1.75 x 10^-5) produces:

The graph shows a maximum point at (4.757,0.576) corresponding to the maximum buffer capacity at pH = 4.757, which is attributed to the ethanoic acid/ethanoate buffer pair; while the increased in buffer capacity below pH 2 and above pH 12 is attributed to water. If we scale the graph by multiplying eq16 throughout by a factor of 10 and plotting 10β_sversus pH, we observe another two stationary points (minimum points), each corresponding to the sum of the minimum buffer capacity of the weak acid and the increased buffer capacity of water:

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