# 5.3: Determining the Sensitivity

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To standardize an analytical method we also must determine the value of \(k_A\) in Equations \ref{5.1} or \ref{5.2}.

\[S_\textrm{total}=k_\textrm An_\textrm A+S_\textrm{reag}\label{5.1}\]

\[S_\textrm{total}=k_\textrm AC_\textrm A+S_\textrm{reag}\label{5.2}\]

In principle, it should be possible to derive the value of \(k_A\) for any analytical method by considering the chemical and physical processes generating the signal. Unfortunately, such calculations are not feasible when we lack a sufficiently developed theoretical model of the physical processes, or are not useful because of nonideal chemical behavior. In such situations we must determine the value of \(k_A\) by analyzing one or more standard solutions, each containing a known amount of analyte. In this section we consider several approaches for determining the value of \(k_A\). For simplicity we will assume that \(S_{reag}\) has been accounted for by a proper reagent blank, allowing us to replace \(S_{total}\) in Equations \ref{5.1} and \ref{5.2} with the analyte’s signal, \(S_A\).

\[S_\textrm A=k_\textrm An_\textrm A\label{5.3}\]

\[S_\textrm A=k_\textrm AC_\textrm A\label{5.4}\]

Equations \ref{5.3} and \ref{5.4} are essentially identical, differing only in whether we choose to express the amount of analyte in moles or as a concentration. For the remainder of this chapter we will limit our treatment to Equation \ref{5.4}. You can extend this treatment to Equation \ref{5.3} by replacing \(C_A\) with \(n_A\).

## 5.3.1 Single-Point versus Multiple-Point Standardizations

The simplest way to determine the value of \(k_A\) in Equation \ref{5.4} is by a **single-point standardization** in which we measure the signal for a standard, \(S_{std}\), containing a known concentration of analyte, \(C_{std}\). Substituting these values into Equation \ref{5.4}

\[k_\textrm A=\dfrac{S_\textrm{std}}{C_\textrm{std}}\label{5.5}\]

gives the value for \(k_A\). Having determined the value for *k*_{A}, we can calculate the concentration of analyte in any sample by measuring its signal, *S*_{samp}, and calculating \(C_A\) using Equation \ref{5.6}.

\[C_\textrm A=\dfrac{S_{\textrm{samp}}}{k_{\textrm{A}}}\label{5.6}\]

A single-point standardization is the least desirable method for standardizing a method. There are at least two reasons for this. First, any error in our determination of \(k_A\) carries over into our calculation of \(C_A\). Second, our experimental value for \(k_A\) is for a single concentration of analyte. Extending this value of \(k_A\) to other concentrations of analyte requires us to assume a linear relationship between the signal and the analyte’s concentration, an assumption that often is not true.^{5} Figure \(\PageIndex{1}\) shows how assuming a constant value of \(k_A\) may lead to a determinate error in the analyte’s concentration. Despite these limitations, single-point standardizations find routine use when the expected range for the analyte’s concentrations is small. Under these conditions it is often safe to assume that \(k_A\) is constant (although you should verify this assumption experimentally). This is the case, for example, in clinical labs where many automated analyzers use only a single standard.

**Figure \(\PageIndex{1}\): **Example showing how a single-point standardization leads to a determinate error in an analyte’s reported concentration if we incorrectly assume that the value of \(k_A\) is constant.

The preferred approach to standardizing a method is to prepare a series of standards, each containing the analyte at a different concentration. Standards are chosen such that they bracket the expected range for the analyte’s concentration. A **multiple-point standardization** should include at least three standards, although more are preferable. A plot of \(S_{std}\) versus \(C_{std}\) is known as a **calibration curve**. The exact standardization, or calibration relationship is determined by an appropriate curve-fitting algorithm.

Linear regression, which also is known as the method of least squares, is one such algorithm. Its use is covered in Section 5D.

There are at least two advantages to a multiple-point standardization. First, although a determinate error in one standard introduces a determinate error into the analysis, its effect is minimized by the remaining standards. Second, by measuring the signal for several concentrations of analyte we no longer must assume that the value of \(k_A\) is independent of the analyte’s concentration. Constructing a calibration curve similar to the “actual relationship” in Figure \(\PageIndex{1}\), is possible.

## 5.3.2 External Standards

The most common method of standardization uses one or more **external standards**, each containing a known concentration of analyte. We call them “external” because we prepare and analyze the standards separate from the samples.

Appending the adjective “external” to the noun “standard” might strike you as odd at this point, as it seems reasonable to assume that standards and samples must be analyzed separately. As you will soon learn, however, we can add standards to our samples and analyze them simultaneously.

### Single External Standard

A quantitative determination using a single external standard was described at the beginning of this section, with \(k_A\) given by Equation \ref{5.5}. After determining the value of \(k_A\), the concentration of analyte, \(C_A\), is calculated using Equation \ref{5.6}.

Example \(\PageIndex{1}\)

A spectrophotometric method for the quantitative analysis of Pb^{2}^{+} in blood yields an *S*_{std} of 0.474 for a single standard whose concentration of lead is 1.75 ppb What is the concentration of Pb^{2}^{+} in a sample of blood for which \(S_{samp}\) is 0.361?

*Solution*

Equation \ref{5.5} allows us to calculate the value of \(k_A\) for this method using the data for the standard.

\[k_\textrm A=\dfrac{S_\textrm{std}}{C_\textrm{std}}=\dfrac{0.474}{1.75\textrm{ ppb}}=0.2709\textrm{ ppb}^{-1} \nonumber\]

Having determined the value of \(k_A\), the concentration of Pb^{2}^{+} in the sample of blood is calculated using Equation \ref{5.6}.

\[C_\textrm A=\dfrac{S_\textrm{samp}}{k_\textrm A}=\dfrac{0.361}{0.2709\textrm{ ppb}^{-1}}=1.33\textrm{ ppb} \nonumber\]

### Multiple External Standards

Figure \(\PageIndex{2}\) shows a typical multiple-point external standardization. The volumetric flask on the left is a reagent blank and the remaining volumetric flasks contain increasing concentrations of Cu^{2}^{+}. Shown below the volumetric flasks is the resulting calibration curve. Because this is the most common method of standardization the resulting relationship is called a **normal calibration curve**.

* Figure \(\PageIndex{2}\):* Shown at the top is a reagent blank (far left) and a set of five external standards for Cu

^{2}

^{+}with concentrations increasing from left to right. Shown below the external standards is the resulting normal calibration curve. The absorbance of each standard, \(S_{std}\), is shown by the filled circles.

When a calibration curve is a straight-line, as it is in Figure \(\PageIndex{2}\), the slope of the line gives the value of \(k_A\). This is the most desirable situation since the method’s sensitivity remains constant throughout the analyte’s concentration range. When the calibration curve is not a straight-line, the method’s sensitivity is a function of the analyte’s concentration. In Figure \(\PageIndex{2}\), for example, the value of \(k_A\) is greatest when the analyte’s concentration is small and decreases continuously for higher concentrations of analyte. The value of \(k_A\) at any point along the calibration curve in Figure \(\PageIndex{2}\) is given by the slope at that point. In either case, the calibration curve provides a means for relating \(S_{samp}\) to the analyte’s concentration.

Example \(\PageIndex{2}\)

A second spectrophotometric method for the quantitative analysis of Pb^{2+} in blood has a normal calibration curve for which

\[S_\textrm{std}=(0.296\textrm{ ppb}^{-1})\times C_\textrm{std}+0.003 \nonumber\]

What is the concentration of Pb^{2}^{+} in a sample of blood if \(S_{samp}\) is 0.397?

*Solution*

To determine the concentration of Pb^{2}^{+} in the sample of blood we replace *S*_{std} in the calibration equation with *S*_{samp} and solve for *C*_{A}.

\[C_\textrm A=\dfrac{S_\textrm{samp}-0.003}{0.296\textrm{ ppb}^{-1}}=\dfrac{0.397-0.003}{0.296\textrm{ ppb}^{-1}}=1.33\textrm{ ppb} \nonumber\]

It is worth noting that the calibration equation in this problem includes an extra term that does not appear in Equation \ref{5.6}. Ideally we expect the calibration curve to have a signal of zero when \(C_A\) is zero. This is the purpose of using a reagent blank to correct the measured signal. The extra term of +0.003 in our calibration equation results from the uncertainty in measuring the signal for the reagent blank and the standards.

Exercise \(\PageIndex{2}\)

Figure \(\PageIndex{2}\) shows a normal calibration curve for the quantitative analysis of Cu^{2}^{+}. The equation for the calibration curve is

\[S_\textrm{std}=29.59\textrm{ M}^{-1}\times C_\textrm{std}+0.0015 \nonumber\]

What is the concentration of Cu^{2}^{+} in a sample whose absorbance, *S*_{samp}, is 0.114? Compare your answer to a one-point standardization where a standard of 3.16 × 10^{–3} M Cu^{2}^{+} gives a signal of 0.0931.

(The one-point standardization in this exercise uses data from the third volumetric flask in Figure \(\PageIndex{2}\))

Click here to review your answer to this exercise.

An external standardization allows us to analyze a series of samples using a single calibration curve. This is an important advantage when we have many samples to analyze. Not surprisingly, many of the most common quantitative analytical methods use an external standardization.

There is a serious limitation, however, to an external standardization. When we determine the value of \(k_A\) using Equation \ref{5.5}, the analyte is present in the external standard’s matrix, which usually is a much simpler matrix than that of our samples. When using an external standardization we assume that the matrix does not affect the value of \(k_A\). If this is not true, then we introduce a proportional determinate error into our analysis. This is not the case in Figure \(\PageIndex{3}\), for instance, where we show calibration curves for the analyte in the sample’s matrix and in the standard’s matrix. In this example, a calibration curve using external standards results in a negative determinate error. If we expect that matrix effects are important, then we try to match the standard’s matrix to that of the sample. This is known as **matrix matching**. If we are unsure of the sample’s matrix, then we must show that matrix effects are negligible, or use an alternative method of standardization. Both approaches are discussed in the following section.

**Figure \(\PageIndex{3}\):** Calibration curves for an analyte in the standard’s matrix and in the sample’s matrix. If the matrix affects the value of k_{A}, as is the case here, then we introduce a determinate error into our analysis if we use a normal calibration curve.

The matrix for the external standards in Figure \(\PageIndex{3}\), for example, is dilute ammonia, which is added because the Cu(NH_{3})_{4}^{2+} complex absorbs more strongly than Cu^{2}^{+}. If we fail to add the same amount of ammonia to our samples, then we will introduce a proportional determinate error into our analysis.

## 5.3.3 Standard Additions

We can avoid the complication of matching the matrix of the standards to the matrix of the sample by conducting the standardization in the sample. This is known as the **method of standard additions**.

### Single Standard Addition

The simplest version of a standard addition is shown in Figure \(\PageIndex{4}\). First we add a portion of the sample, \(V_o\), to a volumetric flask, dilute it to volume, *V*_{f}, and measure its signal, \(S_{samp}\). Next, we add a second identical portion of sample to an equivalent volumetric flask along with a spike, *V*_{std}, of an external standard whose concentration is \(C_{std}\). After diluting the spiked sample to the same final volume, we measure its signal, \(S_{spike}\). The following two equations relate *S*_{samp} and *S*_{spike} to the concentration of analyte, \(C_A\), in the original sample.

**Figure \(\PageIndex{4}\): **Illustration showing the method of standard additions. The volumetric flask on the left contains a portion of the sample, V_{o}, and the volumetric flask on the right contains an identical portion of the sample and a spike, V_{std}, of a standard solution of the analyte. Both flasks are diluted to the same final volume, V_{f}. The concentration of analyte in each flask is shown at the bottom of the figure where C_{A} is the analyte’s concentration in the original sample and C_{std} is the concentration of analyte in the external standard.

\[S_\textrm{samp}=k_\textrm AC_\textrm A\dfrac{V_\textrm o}{V_\textrm f}\label{5.7}\]

\[S_\textrm{spike}=k_\textrm A\left(C_\textrm A\dfrac{V_\textrm o}{V_\textrm f}+C_\textrm{std}\dfrac{V_\textrm{std}}{V_\textrm f}\right)\label{5.8}\]

The ratios *V*_{o}/*V*_{f} and *V*_{std}/*V*_{f} account for the dilution of the sample and the standard, respectively.

As long as *V*_{std} is small relative to *V*_{o}, the effect of the standard’s matrix on the sample’s matrix is insignificant. Under these conditions the value of \(k_A\) is the same in Equations \ref{5.7} and \ref{5.8}. Solving both equations for \(k_A\) and equating gives

\[\dfrac{S_\textrm{samp}}{C_\textrm A\dfrac{V_\textrm o}{V_\textrm f}}=\dfrac{S_\textrm{spike}}{C_\textrm A\dfrac{V_\textrm o}{V_\textrm f}+C_\textrm{std}\dfrac{V_\textrm{std}}{V_\textrm f}}\label{5.9}\]

which we can solve for the concentration of analyte, *C*_{A}, in the original sample.

Example \(\PageIndex{3}\)

A third spectrophotometric method for the quantitative analysis of Pb^{2+} in blood yields an *S*_{samp} of 0.193 when a 1.00 mL sample of blood is diluted to 5.00 mL. A second 1.00 mL sample of blood is spiked with 1.00 mL of a 1560-ppb Pb^{2}^{+} external standard and diluted to 5.00 mL, yielding an *S*_{spike} of 0.419. What is the concentration of Pb^{2}^{+} in the original sample

of blood?

*Solution*

We begin by making appropriate substitutions into Equation \ref{5.9} and solving for *C*_{A}. Note that all volumes must be in the same units; thus, we first covert *V*_{std} from 1.00 mL to 1.00 × 10^{–3} mL.

\[ \begin{align*} \dfrac{0.193}{C_\textrm A\mathrm{\dfrac{1.00\;mL}{5.00\;mL}}} &=\dfrac{0.419}{C_\textrm A\mathrm{\dfrac{1.00\;mL}{5.00\;mL}+1560\;ppb\;\dfrac{1.00\times10^{-3}\;mL}{5.00\;mL}}} \\[4pt] \dfrac{0.193}{2.00C_\textrm A}&=\dfrac{0.419}{200C_\textrm A+0.3120\textrm{ ppb}}\\[4pt] 0.0386C_\textrm A+0.0602\textrm{ ppb} &=0.0838C_\textrm A \\[4pt] 0.0452C_\textrm A&=0.0602\textrm{ ppb} \\[4pt] C_\textrm A&=1.33\textrm{ ppb} \end{align*}\]

The concentration of Pb^{2}^{+} in the original sample of blood is 1.33 ppb.

It also is possible to make a standard addition directly to the sample, measuring the signal both before and after the spike (Figure \(\PageIndex{3}\)). In this case the final volume after the standard addition is *V*_{o} + *V*_{std} and Equations \ref{5.7}, \ref{5.8}, \ref{5.9} become

\[S_\textrm{samp}=k_\textrm AC_\textrm A\]

\[S_\textrm{spike}=k_\textrm A\left(C_\textrm A\dfrac{V_\textrm o}{V_\textrm o+V_\textrm{std}}+C_\textrm{std}\dfrac{V_\textrm{std}}{V_\textrm o+V_\textrm{std}}\right)\label{5.10}\]

\[\dfrac{S_\textrm{samp}}{C_\textrm A}=\dfrac{S_\textrm{spike}}{C_\textrm A\dfrac{V_\textrm o}{V_\textrm o+V_\textrm{std}}+C_\textrm{std}\dfrac{V_\textrm{std}}{V_\textrm o+V_\textrm{std}}}\label{5.11}\]

**Figure \(\PageIndex{3}\): **Illustration showing an alternative form of the method of standard additions. In this case we add a spike of the external standard directly to the sample without any further adjust in the volume.

Example \(\PageIndex{4}\)

A fourth spectrophotometric method for the quantitative analysis of Pb^{2}^{+} in blood yields an *S*_{samp} of 0.712 for a 5.00 mL sample of blood. After spiking the blood sample with 5.00 mL of a 1560-ppb Pb^{2}^{+} external standard, an *S*_{spike} of 1.546 is measured. What is the concentration of Pb^{2}^{+} in the original sample of blood?

*Solution*

To determine the concentration of Pb^{2}^{+} in the original sample of blood, we make appropriate substitutions into Equation \ref{5.11} and solve for \(C_A\).

\[\mathrm{\dfrac{0.712}{\mathit{C}_{A}} = \dfrac{1.546}{\mathit{C}_A\:\dfrac{5.00\: mL}{5.005\: mL} + 1560\: ppb \:\dfrac{5.00×10^{-3}\: mL}{5.005\: mL}}} \nonumber\]

(Note: \(V_\textrm o+V_\textrm{std}=\mathrm{5.00\;mL+5.00\times10^{-3}\;mL=5.005\;mL}\))

\[ \begin{align*} \dfrac{0.712}{C_\textrm A} &=\dfrac{1.546}{0.9990C_\textrm A+1.558\textrm{ ppb}} \\[4pt] 0.7113C_\textrm A+1.109\textrm{ ppb}&=1.546C_\textrm A \\[4pt] C_\textrm A&=1.33\textrm{ ppb} \end{align*}\]

The concentration of Pb^{2}^{+} in the original sample of blood is 1.33 ppb.

### Multiple Standard Additions

We can adapt the single-point standard addition into a multiple-point standard addition by preparing a series of samples containing increasing amounts of the external standard. Figure 5.7 shows two ways to plot a standard addition calibration curve based on Equation \ref{5.8}. In Figure 5.7a we plot *S*_{spike} against the volume of the spikes, *V*_{std}. If \(k_A\) is constant, then the calibration curve is a straight-line. It is easy to show that the *x*-intercept is equivalent to –*C*_{A}*V*_{o}/*C*_{std}.

**Figure \(\PageIndex{6}\):** Shown at the top is a set of six standard additions for the determination of Mn^{2}^{+}. The flask on the left is a 25.00 mL sample diluted to 50.00 mL. The remaining flasks contain 25.00 mL of sample and, from left to right, 1.00, 2.00, 3.00, 4.00, and 5.00 mL of an external standard of 100.6 mg/L Mn^{2+}. Shown below are two ways to plot the standard additions calibration curve. The absorbance for each standard addition, S_{spike}, is shown by the filled circles.

Example \(\PageIndex{4}\)

Beginning with Equation \ref{5.8} show that the equations in Figure 5.7a for the slope, the *y*-intercept, and the *x*-intercept are correct.

*Solution*

We begin by rewriting Equation \ref{5.8} as

\[S_\textrm{spike}=\dfrac{k_\textrm AC_\textrm AV_\textrm o}{V_\textrm f}+\dfrac{k_\textrm AC_\textrm{std}}{V_\textrm f}\times V_\textrm{std} \nonumber\]

which is in the form of the equation for a straight-line

\[Y=y\textrm{-intercept + slope}\times X \nonumber\]

where *Y* is *S*_{spike} and *X* is *V*_{std}. The slope of the line, therefore, is *k*_{A}*C*_{std}/*V*_{f} and the *y*-intercept is *k*_{A}*C*_{A}*V*_{o}/*V*_{f}. The *x*-intercept is the value of *X* when *Y* is zero, or

\[0=\dfrac{k_\textrm AC_\textrm AV_\textrm o}{V_\textrm f}+\dfrac{k_\textrm AC_\textrm{std}}{V_\textrm f}\times x\textrm{-intercept} \nonumber\]

\[x\textrm{-intercept}=-\dfrac{\dfrac{k_\textrm AC_\textrm AV_\textrm o}{V_\textrm f}}{\dfrac{k_\textrm AC_\textrm{std}}{V_\textrm f}}=-\dfrac{C_\textrm AV_\textrm o}{C_\textrm{std}} \nonumber\]

Exercise \(\PageIndex{4}\)

Beginning with Equation \ref{5.8} show that the equations in Figure 5.7b for the slope, the y-intercept, and the x-intercept are correct.

Click here to review your answer to this exercise.

Because we know the volume of the original sample, *V*_{o}, and the concentration of the external standard, *C*_{std}, we can calculate the analyte’s concentrations from the *x*-intercept of a multiple-point standard additions.

Example \(\PageIndex{5}\)

A fifth spectrophotometric method for the quantitative analysis of Pb^{2+} in blood uses a multiple-point standard addition based on Equation \ref{5.8}. The original blood sample has a volume of 1.00 mL and the standard used for spiking the sample has a concentration of 1560 ppb Pb^{2}^{+}. All samples were diluted to 5.00 mL before measuring the signal. A calibration curve of *S*_{spike} versus *V*_{std} has the following equation

\[S_\textrm{spike}=0.266+312\textrm{ mL}^{-1}\times V_\textrm{std} \nonumber\]

What is the concentration of Pb^{2}^{+} in the original sample of blood.

*Solution*

To find the *x*-intercept we set *S*_{spike} equal to zero.

\[0=0.266+312\textrm{ mL}^{-1}\times V_\textrm{std} \nonumber\]

Solving for *V*_{std}, we obtain a value of –8.526 × 10^{–4} mL for the *x*-intercept. Substituting the *x*-interecpt’s value into the equation from Figure 5.7a

\[-8.526\times10^{-4}\textrm{ mL}=-\dfrac{C_\textrm AV_\textrm o}{C_\textrm{std}}=-\dfrac{C_A\times1.00\textrm{ mL}}{1560\textrm{ ppb}} \nonumber\]

and solving for *C*_{A} gives the concentration of Pb^{2}^{+} in the blood sample as 1.33 ppb.

Exercise \(\PageIndex{4}\)

Figure 5.7 shows a standard additions calibration curve for the quantitative analysis of Mn^{2}^{+}. Each solution contains 25.00 mL of the original sample and either 0, 1.00, 2.00, 3.00, 4.00, or 5.00 mL of a 100.6 mg/L external standard of Mn^{2+}. All standard addition samples were diluted to 50.00 mL before reading the absorbance. The equation for the calibration curve in Figure 5.7a is

\[S_\textrm{std}=0.0854\times V_\textrm{std}+0.1478 \nonumber\]

What is the concentration of Mn^{2}^{+} in this sample? Compare your answer to the data in Figure 5.7b, for which the calibration curve is

\[S_\textrm{std}=0.0425\times C_\textrm{std}(V_\textrm{std}/V_\textrm f) + 0.1478 \nonumber\]

Click here to review your answer to this exercise.

Since we construct a standard additions calibration curve in the sample, we can not use the calibration equation for other samples. Each sample, therefore, requires its own standard additions calibration curve. This is a serious drawback if you have many samples. For example, suppose you need to analyze 10 samples using a three-point calibration curve. For a normal calibration curve you need to analyze only 13 solutions (three standards and ten samples). If you use the method of standard additions, however, you must analyze 30 solutions (each of the ten samples must be analyzed three times, once before spiking and after each of two spikes).

### Using a Standard Addition to Identify Matrix Effects

We can use the method of standard additions to validate an external standardization when matrix matching is not feasible. First, we prepare a normal calibration curve of *S*_{std} versus *C*_{std} and determine the value of \(k_A\) from its slope. Next, we prepare a standard additions calibration curve using Equation \ref{5.8}, plotting the data as shown in Figure 5.7b. The slope of this standard additions calibration curve provides an independent determination of *k*_{A}. If there is no significant difference between the two values of *k*_{A}, then we can ignore the difference between the sample’s matrix and that of the external standards. When the values of \(k_A\) are significantly different, then using a normal calibration curve introduces a proportional determinate error.

## 5.3.4 Internal Standards

To successfully use an external standardization or the method of standard additions, we must be able to treat identically all samples and standards. When this is not possible, the accuracy and precision of our standardization may suffer. For example, if our analyte is in a volatile solvent, then its concentration increases when we lose solvent to evaporation. Suppose we have a sample and a standard with identical concentrations of analyte and identical signals. If both experience the same proportional loss of solvent then their respective concentrations of analyte and signals continue to be identical. In effect, we can ignore evaporation if the samples and standards experience an equivalent loss of solvent. If an identical standard and sample lose different amounts of solvent, however, then their respective concentrations and signals will no longer be equal. In this case a simple external standardization or standard addition is not possible.

We can still complete a standardization if we reference the analyte’s signal to a signal from another species that we add to all samples and standards. The species, which we call an **internal standard**, must be different than the analyte.

Because the analyte and the internal standard in any sample or standard receive the same treatment, the ratio of their signals is unaffected by any lack of reproducibility in the procedure. If a solution contains an analyte of concentration *C*_{A}, and an internal standard of concentration, *C*_{IS}, then the signals due to the analyte, *S*_{A}, and the internal standard, *S*_{IS}, are

\[S_\textrm A=k_\textrm AC_\textrm A\]

\[S_\textrm{IS}=k_\textrm{IS}C_\textrm{IS}\]

where \(k_A\) and *k*_{IS} are the sensitivities for the analyte and internal standard. Taking the ratio of the two signals gives the fundamental equation for an internal standardization.

\[\dfrac{S_\textrm A}{S_\textrm{IS}}=\dfrac{k_\textrm AC_\textrm A}{k_\textrm{IS}C_\textrm{IS}}=K\times \dfrac{C_\textrm A}{C_\textrm{IS}}\label{5.12}\]

Because *K* is a ratio of the analyte’s sensitivity and the internal standard’s sensitivity, it is not necessary to independently determine values for either \(k_A\) or *k*_{IS}.

### Single Internal Standard

In a single-point internal standardization, we prepare a single standard containing the analyte and the internal standard, and use it to determine the value of *K* in Equation \ref{5.12}.

\[K=\left(\dfrac{C_\textrm{IS}}{C_\textrm A}\right)_\textrm{std}\times\left(\dfrac{S_\textrm A}{S_\textrm{IS}}\right)_\textrm{std}\label{5.13}\]

Having standardized the method, the analyte’s concentration is given by

\[C_\textrm A=\dfrac{C_\textrm{IS}}{K}\times\left(\dfrac{S_\textrm A}{S_\textrm{IS}}\right)_\textrm{amp}\]

Example \(\PageIndex{5}\)

A sixth spectrophotometric method for the quantitative analysis of Pb^{2+} in blood uses Cu^{2}^{+} as an internal standard. A standard containing 1.75 ppb Pb^{2}^{+} and 2.25 ppb Cu^{2}^{+} yields a ratio of (*S*_{A}/*S*_{IS})_{std} of 2.37. A sample of blood is spiked with the same concentration of Cu^{2}^{+}, giving a signal ratio, (*S*_{A}/*S*_{IS})_{samp}, of 1.80. Determine the concentration of Pb^{2}^{+} in the sample of blood.

*Solution*

Equation 5.13 allows us to calculate the value of *K* using the data for the standard

\[K=\left(\dfrac{C_\textrm{IS}}{C_\textrm A}\right)_\textrm{std}\times\left(\dfrac{S_\textrm A}{S_\textrm{IS}}\right)_\textrm{std}=\mathrm{\dfrac{2.25\;ppb\;Cu}{1.75\;ppb\;Pb^{2+}}\times2.37=3.05\dfrac{ppb\;Cu^{2+}}{ppb\;Pb^{2+}}} \nonumber\]

The concentration of Pb^{2}^{+}, therefore, is

\[C_\textrm A=\dfrac{C_\textrm{IS}}{K}\times\left(\dfrac{S_\textrm A}{S_\textrm{IS}}\right)_\textrm{samp}=\mathrm{\dfrac{2.25\;ppb\;Cu^{2+}}{3.05\dfrac{ppb\;Cu^{2+}}{ppb\;Pb^{2+}}}\times1.80=1.33\;ppb\;Cu^{2+}} \nonumber\]

### Multiple Internal Standards

A single-point internal standardization has the same limitations as a single-point normal calibration. To construct an internal standard calibration curve we prepare a series of standards, each containing the same concentration of internal standard and a different concentrations of analyte. Under these conditions a calibration curve of (*S*_{A}/*S*_{IS})_{std} versus *C*_{A} is linear with a slope of *K*/*C*_{IS}.

Although the usual practice is to prepare the standards so that each contains an identical amount of the internal standard, this is not a requirement.

Example \(\PageIndex{6}\)

A seventh spectrophotometric method for the quantitative analysis of Pb^{2}^{+} in blood gives a linear internal standards calibration curve for which

\[\left(\dfrac{S_\textrm A}{S_\textrm{IS}}\right)_\textrm{std}=(2.11\textrm{ ppb}^{-1})\times C_\textrm A-0.006 \nonumber\]

What is the ppb Pb^{2}^{+} in a sample of blood if (*S*_{A}/*S*_{IS})_{samp} is 2.80?

*Solution*

To determine the concentration of Pb^{2}^{+} in the sample of blood we replace (*S*_{A}/*S*_{IS})_{std} in the calibration equation with (*S*_{A}/*S*_{IS})samp and solve for *C*_{A}.

\[C_\textrm A=\dfrac{\left(\dfrac{S_\textrm A}{S_\textrm{IS}}\right)_\textrm{samp}+0.006}{2.11\textrm{ ppb}^{-1}}=\dfrac{2.80+0.006}{2.11\textrm{ ppb}^{-1}}=1.33\textrm{ ppb} \nonumber\]

The concentration of Pb^{2}^{+} in the sample of blood is 1.33 ppb.

In some circumstances it is not possible to prepare the standards so that each contains the same concentration of internal standard. This is the case, for example, when preparing samples by mass instead of volume. We can still prepare a calibration curve, however, by plotting (*S*_{A}/*S*_{IS})_{std} versus *C*_{A}/*C*_{IS}, giving a linear calibration curve with a slope of *K*.