Category: Advanced Chemistry

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_a and a strong base of concentration C_b with the following equilibria:

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,

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,

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:

Substitute eq16b in eq16a

Substitute eq16b and eq17 in eq16 and rearranging,

Substitute eq17a in eq16b

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

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^st stoichiometric 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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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_a and a strong base of concentration C_b with the following equilibria:

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,

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,

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:

Substituting eq24c into eq24b gives:

Substituting eq25 into eq24a yields:

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

Substituting eq23 into eq24a and eq24b gives:

and

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

where

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:

 

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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,

or

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

 

 

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

where β_w is the buffer capacity of water, β_a is the buffer capacity of a weak acid and β_b is the buffer capacity of a weak base.

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

 

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What is the formula describing the buffer capacity of water?

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

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,

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

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

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

Letting yields

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

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.

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

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^-pH in eq5 to give

The plot of the above equation with β_w as 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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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:

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,

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

Multiplying the term [A^–] on the RHS of the above equation by  and noting that , yields

where [HA]_T =[HA] + [A^–].

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

Substituting eq4a in the above equation gives

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)

Since K_a is a constant

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

Let

A stationary point for the function in eq10 occurs when K_a = [H⁺]. Since

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

Substituting eq13 in the above equation yields

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^-pH in eq10 to give

Plotting the above equation with β_a as 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.

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

Using eq16 to plot a graph of β_s versus 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_s versus 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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The fourth constitutive relation hypothesis states that the properties of a fluid are isotropic, i.e., independent of direction.

For example, an object moving in a fluid encounters the same resistance regardless of the direction of movement. From eq11, the properties of a fluid are described by β_ijkl , which must be isotropic.

As described in the articles on tensors, the general form of a fourth-order isotropic tensor is:

From eq11 of the previous article, τ_ij is symmetric. Therefore, τ_ij =τ_ji and

Substitute eq12 in the above equation, we have

Substitute eq13 in eq12,

Substitute eq14 in eq11,

Substitute eq15 in eq3,

The values of μ and λ can only be determined through experiments. μ is known as the shear viscosity of the fluid while λ is the volume viscosity of the fluid, which is zero for an incompressible fluid. Eq16 becomes:

 

 

 

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The second hypothesis, formulating the constitutive relation for an incompressible and viscous Newtonian fluid, states that the shear stress tensor is a linear function of the deformation tensor.

Consider a shear stress τ that is coplanar to a cross section of a Newtonian fluid ABCD as shown in the diagram below.

The fluid is continuously deformed by the force with different layers of the fluid having different velocities, for example, u(y) at point D and u(y + Δy) at point A. At time t, the body of fluid moves from ABCD to A’B’CD. Let’s define the shear strain in the fluid as  where

If Δy → 0,

If t is small,

We call  the shear strain rate and  the velocity gradient of the fluid. Clearly, the greater the shear stress applied, the higher the value of the shear strain rate, i.e.:

where μ is the constant of proportionality, which is attributed to the resistance of the fluid to the shear stress.

We call this resistance, the viscosity of the fluid. Substituting eq5 in eq6, we have the Newton’s law of viscosity,

When extended to three dimensions (see above diagram), eq7 becomes:

From eq4 of the previous article, shear stress is a second-order tensor with nine components τ_ij . Hence, the general form of eq7 is:

where for a Cartesian system, i = 1, 2, 3, j = 1, 2, 3 and x₁  = x, x₂  = y, x₃  = z.

 can therefore be expressed as a second-order tensor (deformation tensor) of nine components. Since each of the nine components of τ_ij is linearly proportional to each of the nine components of  , there are a total of 9² = 81 proportionality constants. In other words,

where  is a fourth-order tensor of eighty one components, which can be representation by a 3x3x3x3 matrix for a Cartesian system. This satisfies hypothesis 2.

The fourth-order tensor, which is composed of the proportionality constants, characterises the properties of the fluid.

 

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An isotropic Cartesian tensor is one where its components are identical in any orthogonal Cartesian system.

An isotropic property is one that is independent of direction, e.g. thermal expansion of a solid, and therefore has a quantity that is independent of the reference frame. An isotropic property can therefore be described by a Cartesian tensor like in eq15. 

To evaluate the properties of an isotropic tensor, we begin by considering a generic fourth-order tensor , whose transformation is expressed by eq14:

Since j, l, n and p, each of which ranges from 1 to 3, the 81 components of  are classified in four groups as follows:

1. 1. Components with four equal subscripts, i.e. ,  and .
   2. Components with three equal subscripts, e.g. , , etc.
   3. Components with two different pairs of equal subscripts, e.g. , , , etc.
   4. Components with one pair of equal subscripts, e.g. , , etc.

Next, consider the transformation of  to  as a rotation about the x₃-axis of  where x₃ coincides with x’₃ (x’₃ is equal to x₃ and is perpendicular to the plane of the page). From eq12, the transformation matrix is:

Therefore,  and , with the rest of the components of the transformation matrix equal to zero. From eq16, a component of group 2 of the above classification is:

Substituting ,  and  in eq17, we have:

If is an isotropic tensor, (since an isotropic Cartesian tensor is one where its components are identical in any orthogonal Cartesian system). Hence, the only way to satisfy eq18 is for . Otherwise, the value of  varies with the value of , e.g. when , .

Another component of group 2 of the above classification is  where after expanding and substituting with ,  and ,

Eq20 shows that  but does not indicate the value of either component. If the rotation is instead made about the x₂-axis at , the transformation matrix is:

We have , and , giving . If  is an isotropic tensor,

Repeating the above steps and assuming that  is an isotropic tensor, we find that all the components of  in group 2 and group 4 are zero, which means that all the components of  in group 2 and group 4 are zero.

We shall now analyse the components of group 1 by a rotation about the x₃-axis at  with the transformation matrix:

We have .

A rotation about the x₂-axis at  involves the transformation matrix:

and gives .

Comparing the six equations   and , all the components of  in group 1 are equal to one another, which means that all the components of  in group 1 are equal to one another.

Finally, let’s look at the components in group 3. Under a rotation about the x₃-axis at , we have .

A rotation about the x₂-axis at  gives .

Comparing the six equations   and  , components in group 3 of  (which is equal to the components in group 3 of ) with i = k and m = o are equal to one another. Repeating the above steps, we can show that components in group 3 with i = m and k = o are equal to one another and those with i = o and k = m are equal to one another.

Therefore, if  is an isotropic tensor:

i) components of  with i = k = m = o are equal and therefore  with j = l = n = p are equal

ii) components of  with i = k ≠ m = o are equal and therefore  with j = l ≠ n = p are equal

iii) components of  with i = m ≠ k = o are equal and therefore  with j = n ≠ l = p are equal

iv) components of  with i = o ≠ k = m are equal and therefore  with j = p ≠ l = n are equal

v) all other components not mentioned are equal to zero

Note that components of parts ii, iii and iv belong to group3 of the isotropic tensor. Now, consider a case of  where components of  in iii and iv are set to zero ( is still an isotropic tensor but with more components equal to zero). A rotation about the x₃-axis at  gives the transformation matrix:

Since components in part iii and part iv are set to zero and the conditions of part v applies, we have  . Furthermore, part i states that  and part ii states that  . So,

If  is an isotropic tensor,   and therefore,  . This means that all non-zero components of this particular  have the same value, say, λ, where:

where  and  are Kronecker deltas.

Using the same logic by considering the cases of  where components of  in part ii and part iv are set to zero and where components of  in part ii and part iii are set to zero, we have:

and

respectively.

Therefore, we can write the general form of the fourth-order isotropic tensor as:

 

 

 

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A tensor is an array of numbers or functions that can be used to describe properties of a body, such as the scalar for temperature, the vector   for velocity, and the matrix   for stress.

Tensors are categorised by their ranks (also known as order), where

1. 1. A zeroth-order tensor is an array representing a quantity that only has magnitude and no direction, i.e. a scalar.
   2. A first-order tensor is an array representing a quantity that has magnitude and a direction, i.e. a vector.
   3. A second-order tensor is an array representing a quantity that has magnitude and two directions.

The above implies that the number of elements of a tensor is 3ⁿ, where n is the order of the tensor.

Next, we shall investigate how the elements of a tensor, particularly a 2^nd-order tensor, transform between two sets of coordinates. From an earlier article, u_j, which is a set of three quantities, is known as a first order tensor (i.e. a vector). Consider the product of two first order tensors u_j and v_l , whose elements with respect to another set of axes are given by:

Since each term on the RHS of eq12a is a scalar, we can rewrite eq12a as

There are 9 possible products of because i = 1,2,3 and k = 1,2,3. In other words, is a set of 9 elements, each being a sum of nine quantities , e.g.

We can therefore represent the elements of by a 3×3 matrix . Similarly, in eq12b is a multiplication of two first order tensors and can be represented by a 3×3 matrix . This results in:

Eq13 indicates that each element of  in one reference frame is a result of the transformation of  in a different reference frame by  . In general, a set of nine quantities (as is the case of a second-order tensor) with reference to an orthogonal set of axes is transformed to another set of nine quantities with reference to another orthogonal set of axes by the transformation matrix .

Using the same logic, a third order tensor transforms as follows:

and a fourth order tensor:

In general, a tensor of rank n is a quantity that transforms from one set of orthogonal axes to another set of orthogonal axes according to:

 

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