Error analysis

The exact calculation of standard deviations of nonlinear function of variables that are subject to error is generally a problem of great mathematical complexity. A linearization based on a Taylor expansion of the nonlinear portion of the expansion allows to obtain approximate estimates of standard deviations [102]; this approximation is quite adequate for most practical applications.

ΔT is a function of several variables, i.e. pX, log KS, log KT all independent of each other. The systematic error present in pX, log KS, and log KT, are propagated to give an overall systematic error in a calculate quantity

$$E\_{\rm sys(\Delta T)} = \left(\frac{\partial \Delta T}{\partial pX}\right) E\_{\rm sys(pX)} + \left(\frac{\partial \Delta T}{\partial \log K\_S}\right) E\_{\rm sys(\log K\_S)} + \left(\frac{\partial \Delta T}{\partial \log K\_T}\right) E\_{\rm sys(\log K\_T)}\tag{79}$$

provided the errors Esys(pX), Esys(log KS), Esys(log KT) are small enough for higher order derivatives to be discarded.

For random error [103], the variance of ΔT can be calculated according to the propagation of variance

$$E\_{\rm ran(AT)}^2 = \left(\frac{\partial \Delta T}{\partial pX}\right)^2 s\_{pX}^2 + \left(\frac{\partial \Delta T}{\partial \log K\_S}\right) s\_{\log K\_S}^2 + \left(\frac{\partial \Delta T}{\partial \log K\_T}\right) s\_{\log K\_T}^2 \tag{80}$$

where spX<sup>2</sup> , slogKs<sup>2</sup> , and slogKT<sup>2</sup> are the variances of the components pX, log KS, and log KT, respectively. The partial derivatives are taken, as before, as values equal or closest to the measured values.

For any measurement the total absolute error Eabs(ΔT) is related to the different types of error present by

$$E\_{\rm abs(\wedge T)} = \Delta T - \Delta \tau = E\_{\rm ran} + E\_{\rm sys} + E\_{\rm bl} \tag{81}$$

where ΔT is the value of the measurement, Δτ the true value, Erand the random error, Esys is the systematic error, and Ebl is the error due to blunders.

In any case, in order to know the proper error, it is necessary to know the standard deviation of the experimentally measured quantities, i.e., Esys(i) and si.

For the sake of convenience Eq. (22) may be put in the form

$$
\Delta T = \frac{A}{1 + [X]^a K\_S} [X]^p \sinh(\mathcal{C} \Delta p X) \tag{82}
$$

where

$$A = 2\sqrt{\frac{K\_S}{K\_T}}\tag{83}$$

$$p = \frac{a - b}{2} \tag{84}$$

$$\mathcal{C} = \ln 10 \frac{a+b}{2} = \ln 10q \tag{85}$$

Make now

$$\mu = \frac{A[\mathbf{X}]^p}{\mathbf{1} + [\mathbf{X}]^a K\_S} \tag{86}$$

and

$$
\sigma = \sinh\left(\mathbf{C} \Delta p \mathbf{X}\right) \tag{87}
$$

in order to may differentiate easily ΔT against d[X]. Thus

$$
\Delta T = \mu v \tag{88}
$$

and we get for the derivative of a product

$$(\Delta T)' = \mathfrak{u}'\mathfrak{v} + \mathfrak{v}'\mathfrak{u} \tag{89}$$

The derivative of u will be given by

$$u' = A \left( \frac{p[X]^{p-1} (1 + [X]^a K\_S) - K\_S a [X]^{a-1} [X]^p}{\left(1 + [X]^a K\_S\right)^2} \right) = A \left( \frac{p[X]^{p-1} (1 + [X]^a K\_S) - K\_S a [X]^a [X]^{p-1}}{\left(1 + [X]^a K\_S\right)^2} \right)$$

$$= \frac{A [X]^{p-1}}{1 + [X]^a K\_S} \left( \frac{p - a K\_S [X]^a}{1 + [X]^a K\_S} \right) \tag{90}$$

On the other hand

$$\nabla v' = \cosh(\mathsf{C}\Delta pX)\mathsf{C}\frac{\partial(\Delta pX)}{\partial[X]} = \cosh(\mathsf{C}\Delta pX)\mathsf{C}\frac{\partial pX}{\partial[X]} = -\cosh(\mathsf{C}\Delta pX)\mathsf{C}\frac{1}{\ln 10[X]}\tag{91}$$

Taking into account Eqs. (87)–(91), we get

$$\frac{\partial \Delta T}{\partial [X]} = \frac{A[X]^{p-1}}{1 + [X]^q K\_S} \left( \frac{p - aK\_S[X]^q}{1 + [X]^q K\_S} \right) \sinh(\mathsf{C.d} pX) - \frac{A[X]^p}{1 + [X]^q K\_S} \cosh(\mathsf{C.d} pX) \frac{\mathsf{C.d}}{\ln 10 [X]} \qquad (92)$$

and then

On the Titration Curves and Titration Errors in Donor Acceptor Titrations of Displacement and Electronic… http://dx.doi.org/10.5772/intechopen.68750 143

$$\frac{\partial \Delta T}{\partial [X]} = \frac{A[X]^{p-1}}{1 + [X]^d K\_S} \left( \left( \frac{p - aK\_S[X]^d}{1 + [X]^d K\_S} \right) \sinh(\mathsf{C} \Delta pX) - q \cosh(\mathsf{C} \Delta pX) \right) \tag{93}$$

By multiplying through [X] and taking into account Eq. (31) we get

$$\mathbb{E}\left[\mathbf{X}\right]\frac{\partial\Delta T}{\partial\left[\mathbf{X}\right]} = \frac{\Delta T}{\sinh(\mathbf{C}\Delta p\mathbf{X})} \left( \left(\frac{p - aK\_{\mathbb{S}}[\mathbf{X}]^{\mathbb{A}}}{1 + [\mathbf{X}]^{\mathbb{A}}K\_{\mathbb{S}}}\right)\sinh(\mathbf{C}\Delta p\mathbf{X}) - q\cosh(\mathbf{C}\Delta p\mathbf{X})\right) \tag{94}$$

and so

ð83Þ

ð86Þ

<sup>2</sup> <sup>ð</sup>84<sup>Þ</sup>

<sup>2</sup> <sup>¼</sup> ln 10<sup>q</sup> <sup>ð</sup>85<sup>Þ</sup>

v ¼ sinh ðCΔpXÞ ð87Þ

ΔT ¼ uv ð88Þ

<sup>¼</sup> <sup>A</sup> p X½ �<sup>p</sup>�<sup>1</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup> ð Þ� KS KSa X½ �<sup>a</sup>

<sup>∂</sup>½ � <sup>X</sup> ¼ �coshð Þ <sup>C</sup>ΔpX <sup>C</sup> <sup>1</sup>

coshð Þ <sup>C</sup>ΔpX <sup>C</sup>

u ð89Þ

2

<sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup> ð Þ KS

!

½ � <sup>X</sup> <sup>p</sup>�<sup>1</sup>

ln 10½ � <sup>X</sup> <sup>ð</sup>91<sup>Þ</sup>

ln 10½ � <sup>X</sup> <sup>ð</sup>92<sup>Þ</sup>

ð90Þ

where

142 Redox - Principles and Advanced Applications

Make now

and

A ¼ 2

<sup>C</sup> <sup>¼</sup> ln 10 <sup>a</sup> <sup>þ</sup> <sup>b</sup>

ðΔTÞ

<sup>0</sup> ¼ u<sup>0</sup>

½ � <sup>X</sup> <sup>p</sup>

<sup>∂</sup>½ � <sup>X</sup> <sup>¼</sup> coshð Þ <sup>C</sup>ΔpX <sup>C</sup> <sup>∂</sup>pX

sinhð Þ� <sup>C</sup>ΔpX A X½ �<sup>p</sup>

<sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

KS

v þ v<sup>0</sup>

in order to may differentiate easily ΔT against d[X]. Thus

and we get for the derivative of a product

<sup>u</sup><sup>0</sup> <sup>¼</sup> <sup>A</sup> p X½ �<sup>p</sup>�<sup>1</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup> ð Þ� KS KSa X½ �<sup>a</sup>�<sup>1</sup>

<sup>v</sup><sup>0</sup> <sup>¼</sup> coshð Þ <sup>C</sup>ΔpX <sup>C</sup> <sup>∂</sup>ð Þ <sup>Δ</sup>pX

KS

Taking into account Eqs. (87)–(91), we get

<sup>∂</sup>½ � <sup>X</sup> <sup>¼</sup> A X½ �<sup>p</sup>�<sup>1</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

<sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup> ð Þ KS

KS

<sup>p</sup> � aKS½ � <sup>X</sup> <sup>a</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

� �

KS

<sup>p</sup> � aKS½ � <sup>X</sup> <sup>a</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

� �

!

2

The derivative of u will be given by

KS

<sup>¼</sup> A X½ �<sup>p</sup>�<sup>1</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

On the other hand

∂ΔT

and then

<sup>p</sup> <sup>¼</sup> <sup>a</sup> � <sup>b</sup>

<sup>u</sup> <sup>¼</sup> A X½ �<sup>p</sup> <sup>1</sup> <sup>þ</sup> ½ � <sup>X</sup> <sup>a</sup>

KS

ffiffiffiffiffiffi KS KT r

$$\frac{\partial \Delta T}{\partial pX} = -\ln 10[X] \frac{\partial \Delta T}{\partial [X]} = \ln 10 \Delta T \left( q \coth(\mathsf{C} \Delta p X) - \left( \frac{p - aK\_S[X]^a}{1 + [X]^a K\_S} \right) \right) \tag{95}$$

Eq. (22) may be presented in the form

$$
\Delta T = \frac{2[X]^p}{\sqrt{K\_T}} \frac{K\_S^{1/2}}{1 + [X]^d K\_S} \sinh(\text{CAp}X) \tag{96}
$$

Differentiation of Eq. (96) with respect to KS gives

$$\frac{\partial \Delta T}{\partial \mathbf{K}\_S} = \frac{2[\mathbf{X}]^p}{\sqrt{K\_T}} \left( \left( \frac{\frac{1}{2} K\_S^{\frac{1}{2} - 1} (1 + [X]^a K\_S) - K\_S^{\frac{1}{2}} [X]^a}{(1 + [X]^a K\_S)} \right) \text{sech}(\mathbf{C} \Delta p \mathbf{X}) + \frac{K\_S^{\frac{1}{2}}}{(1 + [X]^a K\_S)} \cosh(\mathbf{C} \Delta p \mathbf{X}) \mathbf{C} \frac{\partial \Delta p \mathbf{X}}{\partial \mathbf{K}\_S} \right) \tag{97}$$

On the other hand

$$\frac{\partial \Delta pX}{\partial K\_S} = \frac{\partial \left(pX\_{\text{end}} - \frac{\log K\_T + \log K\_S}{a+b}\right)}{\partial K\_S} = \frac{-1}{a+b} \frac{\partial \log K\_S}{\partial K\_S} = \frac{-1}{(a+b)\ln 10 K\_S} \tag{98}$$

$$\mathcal{C}\frac{\partial \Delta pX}{\partial K\_S} = \ln 10 \left(\frac{a+b}{2}\right) \frac{-1}{(a+b)\ln 10 K\_S} = \frac{-1}{2K\_S} \tag{99}$$

By combining Eqs. (97) and (99) we obtain

$$\frac{\partial \Delta T}{\partial \mathbf{K}s} = \frac{2[\mathbf{X}]^p}{\sqrt{K\_T}} \left( \left( \frac{\frac{1}{2} \mathbf{K}\_S^{-\frac{1}{2}} - \frac{1}{2} \mathbf{K}\_S^{\frac{1}{2}} [\mathbf{X}]^a}{\left(1 + [\mathbf{X}]^a \mathbf{K}\_S\right)^2} \right) \sinh(\mathbf{C} \Delta p \mathbf{X}) - \frac{\frac{1}{2} \mathbf{K}\_S^{-\frac{1}{2}}}{\left(1 + [\mathbf{X}]^a \mathbf{K}\_S\right)} \cosh(\mathbf{C} \Delta p \mathbf{X}) \right) \tag{100}$$

which on rearranging gives

$$\frac{\partial \Delta T}{\partial \mathbf{K}\_S} = \frac{\frac{K\_s^{-\frac{1}{2}}}{\sqrt{K\_I}}}{\mathbf{1} + [\mathbf{X}]^d K\_S} [\mathbf{X}]^p \left( \left( \frac{\mathbf{1} - \mathbf{K}\_S [\mathbf{X}]^d}{\mathbf{1} + K\_S [\mathbf{X}]^d} \right) \sinh(\mathbf{C} \Delta p \mathbf{X}) - \cosh(\mathbf{C} \Delta p \mathbf{X}) \right) \tag{101}$$

Differentiating now ΔT against log KS leads to

$$\frac{\partial \Delta T}{\partial \log K\_S} = \ln 10K\_S \frac{\partial \Delta T}{\partial K\_S} = \frac{\Delta T}{2 \sinh(\mathsf{C} \Delta p X)} \left( \left( \frac{1 - K\_S[X]^d}{1 + K\_S[X]^d} \right) \sinh(\mathsf{C} \Delta p X) - \cosh(\mathsf{C} \Delta p X) \right)$$

$$= \frac{\ln 10}{2} \Delta T \left( \left( \frac{1 - K\_S[X]^d}{1 + K\_S[X]^d} \right) - \coth(\mathsf{C} \Delta p X) \right) \tag{102}$$

In order to differentiate ΔT against KT we put (from Eq. (22))

$$\frac{\partial \Delta T}{\partial \mathbf{K}\_T} = \frac{2\sqrt{\mathbf{K}\_S}}{1 + [\mathbf{X}]^d \mathbf{K}\_S} [\mathbf{X}]^p \left( \frac{\partial}{\partial \mathbf{K}\_T} \left( \frac{1}{\sqrt{\mathbf{K}\_T}} \right) \cdot \sinh(\mathbf{C} \Delta p \mathbf{X}) + \frac{1}{\sqrt{\mathbf{K}\_T}} \cosh(\mathbf{C} \Delta p \mathbf{X}) \mathbf{C} \frac{\partial \Delta p \mathbf{X}}{\partial \mathbf{K}\_T} \right) \tag{103}$$

From Eqs. (18), (20), and (34), we get

$$\mathbb{C}\frac{\partial \Delta pX}{\partial K\_T} = \frac{-1}{2K\_T} \tag{104}$$

By differentiating 1/√KT against KT in Eq. (52) and combining the resulting expression with Eq. (53) we get

$$\frac{\partial \Delta T}{\partial \mathbf{K}\_T} = \frac{2\sqrt{K\_S}}{1 + [\mathbf{X}]^d K\_S} [\mathbf{X}]^p \left( \frac{-1}{2} K\_T^{-\frac{3}{2}} \sinh(\mathbf{C} \Delta p \mathbf{X}) - \frac{-1}{2} K\_T^{-\frac{3}{2}} \cosh(\mathbf{C} \Delta p \mathbf{X}) \right)$$

$$= \frac{-\sqrt{\frac{K\_S}{K\_T}}}{K\_T (1 + [\mathbf{X}]^d K\_S)} [\mathbf{X}]^p (\sinh(\mathbf{C} \Delta p \mathbf{X}) + \cosh(\mathbf{C} \Delta p \mathbf{X})) \tag{105}$$

As before with the case of [X] and KS, we may express the partial derivative of ΔT against log KT as a function of the derivative against KT and then

$$\frac{\partial \Delta T}{\partial \log K\_T} = \ln 10K\_T \frac{\partial \Delta T}{K\_T} = -\ln 10 \frac{\sqrt{\frac{K\_S}{K\_T}}}{1 + [X]^\theta K\_S} [X]^\theta (\sinh(\mathcal{C} \Delta pX) + \cosh(\mathcal{C} \Delta pX))$$

$$= -\frac{\ln 10}{2} \frac{\Delta T}{\sinh(\mathcal{C} \Delta pX)} (\sinh(\mathcal{C} \Delta pX) + \cosh(\mathcal{C} \Delta pX)) = -\frac{\ln 10}{2} \Delta T (1 + \coth(\mathcal{C} \Delta pX)) \tag{106}$$

By combining Eqs. (28), (44), (51), and (55), we get for the standard deviation of systematic error

$$E\_{\rm sys(\triangle T)} = \ln 10\Delta T \left( \left( \frac{p - aK\_S[X]^d}{1 + [X]^d K\_S} - q \coth(\mathsf{C} \Delta p \mathbf{X}) \right) \right) E\_{\rm sys(pX)}$$

$$+ \frac{1}{2} \left( \frac{1 - K\_S[X]^d}{1 + K\_S[X]^d} - \coth(\mathsf{C} \Delta p \mathbf{X}) \right) E\_{\rm sys(\log K\_S)} - \frac{1}{2} (1 + \coth(\mathsf{C} \Delta p \mathbf{X})) E\_{\rm sys(\log K\_T)} \right) \qquad (107)$$

In the same way, from Eqs. (29), (44), (51), and (55), we may obtain the variance of the random error

$$E\_{\rm rand}^2(\Delta T) = \ln^2 10 \Delta T^2 \left( \left( \frac{p - aK\_S[X]^a}{1 + [X]^a K\_S} - q \coth(\text{CAp}X) \right)^2 s\_{pX}^2 \right.$$

$$+ \frac{1}{4} \left( \frac{1 - K\_S[X]^a}{1 + K\_S[X]^a} - \coth(\text{CAp}X) \right)^2 s\_{\log K\_S}^2 + \frac{1}{4} (1 + \coth(\text{CAp}X))^2 s\_{\log K\_T}^2 \right) \tag{108}$$

Note that in spite that relative complex expressions are involved in the required differentiations carried out with the purpose to propagate the systematic and random errors implied in the donor/acceptor titration, the algebra involved is simple, and the final expressions obtained are compact.
