# 3.3: Propagation of Uncertainty

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Suppose we dispense 20 mL of a reagent using the Class A 10-mL pipet whose calibration information is given in Table 4.2.8. If the volume and uncertainty for one use of the pipet is 9.992 ± 0.006 mL, what is the volume and uncertainty if we use the pipet twice?

As a first guess, we might simply add together the volume and the maximum uncertainty for each delivery; thus

(9.992 mL + 9.992 mL) ± (0.006 mL + 0.006 mL) = 19.984 ± 0.012 mL

It is easy to appreciate that combining uncertainties in this way overestimates the total uncertainty. Adding the uncertainty for the first delivery to that of the second delivery assumes that with each use the indeterminate error is in the same direction and is as large as possible. At the other extreme, we might assume that the uncertainty for one delivery is positive and the other is negative. If we subtract the maximum uncertainties for each delivery,

(9.992 mL + 9.992 mL) ± (0.006 mL – 0.006 mL) = 19.984 ± 0.000 mL

we clearly underestimate the total uncertainty.

So what is the total uncertainty? From the discussion above, we reasonably expect that the total uncertainty is greater than ±0.000 mL and that it is less than ±0.012 mL. To estimate the uncertainty we use a mathematical technique known as the propagation of uncertainty. Our treatment of the propagation of uncertainty is based on a few simple rules.

## A Few Symbols

A propagation of uncertainty allows us to estimate the uncertainty in a result from the uncertainties in the measurements used to calculate that result. For the equations in this section we represent the result with the symbol R, and we represent the measurements with the symbols A, B, and C. The corresponding uncertainties are uR, uA, uB, and uC. We can define the uncertainties for A, B, and C using standard deviations, ranges, or tolerances (or any other measure of uncertainty), as long as we use the same form for all measurements.

The requirement that we express each uncertainty in the same way is a critically important point. Suppose you have a range for one measurement, such as a pipet’s tolerance, and standard deviations for the other measurements. All is not lost. There are ways to convert a range to an estimate of the standard deviation. See Appendix 2 for more details.

## Uncertainty When Adding or Subtracting

When we add or subtract measurements we propagate their absolute uncertainties. For example, if the result is given by the equation

$R = A + B - C \nonumber$

the the absolute uncertainty in R is

$u_R = \sqrt{u_A^2 + u_B^2 + u_C^2} \label{4.1}$

Example $$\PageIndex{1}$$

If we dispense 20 mL using a 10-mL Class A pipet, what is the total volume dispensed and what is the uncertainty in this volume? First, complete the calculation using the manufacturer’s tolerance of 10.00 mL±0.02 mL, and then using the calibration data from Table 4.2.8.

Solution

To calculate the total volume we add the volumes for each use of the pipet. When using the manufacturer’s values, the total volume is

$V = 10.00 \text{ mL} + 10.00 \text{ mL} = 20.00 \text{ mL} \nonumber$

and when using the calibration data, the total volume is

$V = 9.992 \text{ mL} + 9.992 \text{ mL} = 19.984 \text{ mL} \nonumber$

Using the pipet’s tolerance as an estimate of its uncertainty gives the uncertainty in the total volume as

$u_R = (0.02)^2 + (0.02)^2 = 0.028 \text{ mL} = 0.028 \text{ mL} \nonumber$

and using the standard deviation for the data in Table 4.2.8 gives an uncertainty of

$u_R = (0.006)^2 + (0.006)^2 = 0.0085 \text{ mL} \nonumber$

Rounding the volumes to four significant figures gives 20.00 mL ± 0.03 mL when we use the tolerance values, and 19.98 ± 0.01 mL when we use the calibration data.

## Uncertainty When Multiplying or Dividing

When we multiple or divide measurements we propagate their relative uncertainties. For example, if the result is given by the equation

$R = \frac {A \times B} {C} \nonumber$

then the relative uncertainty in R is

$\frac {u_R} {R} \sqrt{\left( \frac {u_A} {A} \right)^2 + \left( \frac {u_B} {B} \right)^2 + \left( \frac {u_C} {C} \right)^2} \label{4.2}$

Example $$\PageIndex{2}$$

The quantity of charge, Q, in coulombs that passes through an electrical circuit is

$Q = i \times t \nonumber$

where i is the current in amperes and t is the time in seconds. When a current of 0.15 A ± 0.01 A passes through the circuit for 120 s ± 1 s, what is the total charge and its uncertainty?

Solution

The total charge is

$Q = (0.15 \text{ A}) \times (120 \text{ s}) = 18 \text{ C} \nonumber$

Since charge is the product of current and time, the relative uncertainty in the charge is

$u_R = \sqrt{\left( \frac {0.01} {0.15} \right)^2 + \left( \frac {1} {120} \right)^2} = 0.0672 \nonumber$

and the charge’s absolute uncertainty is

$u_R = R \times 0.0672 = (18 \text{ C}) \times (0.0672) = 1.2 \text{ C} \nonumber$

Thus, we report the total charge as 18 C ± 1 C.

## Uncertainty for Mixed Operations

Many chemical calculations involve a combination of adding and subtracting, and of multiply and dividing. As shown in the following example, we can calculate the uncertainty by separately treating each operation using Equation \ref{4.1} and Equation \ref{4.2} as needed.

Example $$\PageIndex{3}$$

For a concentration technique, the relationship between the signal and the an analyte’s concentration is

$S_{total} = k_A C_A + S_{mb} \nonumber$

What is the analyte’s concentration, CA, and its uncertainty if Stotal is 24.37 ± 0.02, Smb is 0.96 ± 0.02, and kA is $$0.186 \pm 0.003 \text{ ppm}^{-1}$$?

Solution

Rearranging the equation and solving for CA

$C_A = \frac {S_{total} - S_{mb}} {k_A} = \frac {24.37 - 0.96} {0.186 \text{ ppm}^{-1}} = \frac {23.41} {0.186 \text{ ppm}^{-1}} = 125.9 \text{ ppm} \nonumber$

gives the analyte’s concentration as 126 ppm. To estimate the uncertainty in CA, we first use Equation \ref{4.1} to determine the uncertainty for the numerator.

$u_R = \sqrt{(0.02)^2 + (0.02)^2} = 0.028 \nonumber$

The numerator, therefore, is 23.41 ± 0.028. To complete the calculation we use Equation \ref{4.2} to estimate the relative uncertainty in CA.

$\frac {u_R} {R} = \sqrt{\left( \frac {0.028} {23.41} \right)^2 + \left( \frac {0.003} {0.186} \right)^2} = 0.0162 \nonumber$

The absolute uncertainty in the analyte’s concentration is

$u_R = (125.9 \text{ ppm}) \times (0.0162) = 2.0 \text{ ppm} \nonumber$

Thus, we report the analyte’s concentration as 126 ppm ± 2 ppm.

Exercise $$\PageIndex{1}$$

To prepare a standard solution of Cu2+ you obtain a piece of copper from a spool of wire. The spool’s initial weight is 74.2991 g and its final weight is 73.3216 g. You place the sample of wire in a 500-mL volumetric flask, dissolve it in 10 mL of HNO3, and dilute to volume. Next, you pipet a 1 mL portion to a 250-mL volumetric flask and dilute to volume. What is the final concentration of Cu2+ in mg/L, and its uncertainty? Assume that the uncertainty in the balance is ±0.1 mg and that you are using Class A glassware.

The first step is to determine the concentration of Cu2+ in the final solution. The mass of copper is

$74.2991 \text{ g} - 73.3216 \text{ g} = 0.9775 \text{ g Cu} \nonumber$

The 10 mL of HNO3 used to dissolve the copper does not factor into our calculation. The concentration of Cu2+ is

$\frac {0.9775 \text{ g Cu}} {0.5000 \text{ L}} \times \frac {1.000 \text{ mL}} {250.0 \text{ mL}} \times \frac {1000 \text{ mg}} {\text{g}} = 7.820 \text{ mg } \ce{Cu^{2+}} \text{/L} \nonumber$

Having found the concentration of Cu2+, we continue with the propagation of uncertainty. The absolute uncertainty in the mass of Cu wire is

$u_\text{g Cu} = \sqrt{(0.0001)^2 + (0.0001)^2} = 0.00014 \text{ g} \nonumber$

The relative uncertainty in the concentration of Cu2+ is

$\frac {u_\text{mg/L}} {7.820 \text{ mg/L}} = \sqrt{\left( \frac {0.00014} {0.9775} \right)^2 + \left( \frac {0.20} {500.0} \right)^2 + \left( \frac {0.006} {1.000} \right)^2 + \left( \frac {0.12} {250.0} \right)^2} = 0.00603 \nonumber$

Solving for umg/L gives the uncertainty as 0.0472. The concentration and uncertainty for Cu2+ is 7.820 mg/L ± 0.047 mg/L.

## Uncertainty for Other Mathematical Functions

Many other mathematical operations are common in analytical chemistry, including the use of powers, roots, and logarithms. Table $$\PageIndex{1}$$ provides equations for propagating uncertainty for some of these function where A and B are independent measurements and where k is a constant whose value has no uncertainty.

Table $$\PageIndex{1}$$: Propagation of Uncertainty for Selected Mathematical Functions
Function $$u_R$$ Function $$u_R$$
$$R = kA$$ $$u_R = ku_A$$ $$R = \ln (A)$$ $$u_R = \frac {u_A} {A}$$
$$R = A + B$$ $$u_R = \sqrt{u_A^2 + u_B^2}$$ $$R = \log (A)$$ $$u_R = 0.4343 \times \frac {u_A} {A}$$
$$R = A - B$$ $$u_R = \sqrt{u_A^2 + u_B^2}$$ $$R = e^A$$ $$\frac {u_R} {R} = u_A$$
$$R = A \times B$$ $$u_R = \sqrt{\left( \frac {u_A} {A} \right)^2 +\left( \frac {u_B} {B} \right)^2}$$ $$R = 10^A$$ $$\frac {u_R} {R} = 2.303 \times u_A$$
$$R = \frac {A} {B}$$ $$u_R = \sqrt{\left( \frac {u_A} {A} \right)^2 +\left( \frac {u_B} {B} \right)^2}$$ $$R = A^k$$ $$\frac {u_R} {R} = k \times \frac {u_A} {A}$$

Example $$\PageIndex{4}$$

If the pH of a solution is 3.72 with an absolute uncertainty of ±0.03, what is the [H+] and its uncertainty?

Solution

The concentration of H+ is

$[\ce{H+}] = 10^{-\text{pH}} = 10^{-3.72} = 1.91 \times 10^{-4} \text{ M} \nonumber$

or $$1.9 \times 10^{-4}$$ M to two significant figures. From Table $$\PageIndex{1}$$ the relative uncertainty in [H+] is

$\frac {u_R} {R} = 2.303 \times u_A = 2.303 \times 0.03 = 0.069 \nonumber$

The uncertainty in the concentration, therefore, is

$(1.91 \times 10^{-4} \text{ M}) \times (0.069) = 1.3 \times 10^{-5} \text{ M} \nonumber$

We report the [H+] as $$1.9 (\pm 0.1) \times 10^{-4}$$ M, which is equivalent to $$1.9 \times 10^{-4} \text{ M } \pm 0.1 \times 10^{-4} \text{ M}$$.

Exercise $$\PageIndex{2}$$

A solution of copper ions is blue because it absorbs yellow and orange light. Absorbance, A, is defined as

$A = - \log T = - \log \left( \frac {P} {P_\text{o}} \right) \nonumber$

where, T is the transmittance, Po is the power of radiation as emitted from the light source and P is its power after it passes through the solution. What is the absorbance if Po is $$3.80 \times 10^2$$ and P is $$1.50 \times 10^2$$? If the uncertainty in measuring Po and P is 15, what is the uncertainty in the absorbance?

The first step is to calculate the absorbance, which is

$A = - \log T = -\log \frac {P} {P_\text{o}} = - \log \frac {1.50 \times 10^2} {3.80 \times 10^2} = 0.4037 \approx 0.404 \nonumber$

Having found the absorbance, we continue with the propagation of uncertainty. First, we find the uncertainty for the ratio P/Po, which is the transmittance, T.

$\frac {u_{T}} {T} = \sqrt{\left( \frac {15} {3.80 \times 10^2} \right)^2 + \left( \frac {15} {1.50 \times 10^2} \right)^2 } = 0.1075 \nonumber$

Finally, from Table $$\PageIndex{1}$$ the uncertainty in the absorbance is

$u_A = 0.4343 \times \frac {u_{T}} {T} = (0.4343) \times (0.1075) = 4.669 \times 10^{-2} \nonumber$

The absorbance and uncertainty is 0.40 ± 0.05 absorbance units.

## The Basis Behind the Equations for the Propagation of Error and Extension to other Calculated Results

Now let’s look at a general case of x = fn(p,q,r,…) and we assume p, q, r,… can fluctuate randomly and be treated as independent variables.

Then for any individual measurement of x, say xthen

$$d x_{i}=f n\left(\partial p_{i}, \partial q_{i}, \partial r_{i}, \ldots\right)$$

Now from calculus we know that the total variation in x as a function of the variations in p,q,r,… comes from the partial derivative of x with respect to each variable, p,q,r,… so

$$d x=\left(\frac{\partial x}{\partial p}\right)_{v} d p+\left(\frac{\partial x}{\partial q}\right)_{v} d q+\left(\frac{\partial x}{\partial r}\right)_{v} d r+\ldots$$

To develop a relationship between the standard deviation of x; (sx) and the standard deviations of p,q,r,…; (sp,sq,sr,…) we square the above expression and sum over i = 1 to n values

$$\sum_{i=1}^{n}\left(d x_{i}\right)^{2}=\sum_{i=1}^{n}\left(\left(\frac{\partial x}{\partial p}\right)_{v} d p_{i}+\left(\frac{\partial x}{\partial q}\right)_{v} d q_{i}+\left(\frac{\partial x}{\partial r}\right)_{v} d r_{i}+\ldots\right)^{2}$$

Expanding the right hand side of this last equation we will get

square terms are always positive and always sum to a positive value, such as

$$\left(\frac{\partial x}{\partial p}\right)^{2}(dp)^{2} ;\left(\frac{\partial x}{\partial q}\right)^{2}(dq)^{2} ; \text { etc }$$

and cross terms, such as the one shown below, can be positive or negative and that will sum to 0 as n get large and as long as the variables are independent

$$\left(\frac{\partial x}{\partial p}\right)\left(\frac{\partial x}{\partial q}\right) dp dq$$

Focusing only on the square terms and the sum of those terms for i = 1 to n

$$\sum_{i=1}^{n}\left(d x_{i}\right)^{2}=\left(\frac{\partial x}{\partial p}\right)^{2} \sum_{i=1}^{n}\left(d p_{i}\right)^{2}+\left(\frac{\partial x}{\partial q}\right)^{2} \sum_{i=1}^{n}\left(d q_{i}\right)^{2}+\left(\frac{\partial x}{\partial r}\right)^{2} \sum_{i=1}^{n}\left(d r_{i}\right)^{2}+\ldots$$

If we now divide each term by n-1 then we have

$$\frac{\sum_{i=1}^{n}\left(d x_{i}\right)^{2}}{n-1}=\frac{\left(\frac{\partial x}{\partial p}\right)^{2} \sum_{i=1}^{n}\left(d p_{i}\right)^{2}}{n-1}+\frac{\left(\frac{\partial x}{\partial q}\right)^{2} \sum_{i=1}^{n}\left(d q_{i}\right)^{2}}{n-1}+\frac{\left(\frac{\partial x}{\partial r}\right)^{2} \sum_{i=1}^{n}\left(d r_{i}\right)^{2}}{n-1}+\ldots$$

And if we note that each term $$\left(d x_{i}\right)^{2}=\left(x_{i}-\bar{x}\right)^{2}$$  and likewise $$\left(d q_{i}\right)^{2}=\left(q_{i}-\bar{q}\right)^{2}$$

then we get the important and generally applicable result in terms of the variances (u2)

$$u_{x}^{2}=\left(\frac{\partial x}{\partial p}\right)^{2} u_{p}^{2}+\left(\frac{\partial x}{\partial q}\right)^{2} u_{q}^{2}+\left(\frac{\partial x}{\partial r}\right)^{2} u_{r}^{2}+\ldots$$

### Example $$\PageIndex{5}$$:

Question: What is the uncertainty in the volume of a rectangular solid that has a base = 2.00 +/- 0.05 cm on a side and a height = 5.50 +/- 0.10 cm?

Answer:  In this case V = b2h  and V = (2.00)2 x 5.50 = 22.0 cm3

To get the uncertainty we get the partial derivatives of V with respect ot b and h

$$\left(\frac{\partial V}{\partial b}\right)_{h} = 2bh$$ and $$\left(\frac{\partial V}{\partial h}\right)_{b}= b^{2}$$

so following our general result

$$u_{V}^{2}=\left(\frac{\partial V}{\partial b}\right)^{2} u_{b}^{2}+\left(\frac{\partial V}{\partial h}\right)^{2} u_{b}^{2}$$

and after substitution

$$u_{V}^{2}=$$ [(2 x 2.00 x 5.50)2 x (0.05)2] + [(2.00)2 x (0.10)2] = 1.21 + 0.04 = 1.25  so $$u_{V}=$$ (1.25)0.5 = 1.12 cm3

and the volume of the solid to be 22.0 +/- 1.12 cm3 that would be reported as 22 +/- 1 cm3 .

## Is Calculating Uncertainty Actually Useful?

Given the effort it takes to calculate uncertainty, it is worth asking whether such calculations are useful. The short answer is, yes. Let’s consider three examples of how we can use a propagation of uncertainty to help guide the development of an analytical method.

One reason to complete a propagation of uncertainty is that we can compare our estimate of the uncertainty to that obtained experimentally. For example, to determine the mass of a penny we measure its mass twice—once to tare the balance at 0.000 g and once to measure the penny’s mass. If the uncertainty in each measurement of mass is ±0.001 g, then we estimate the total uncertainty in the penny’s mass as

$u_R = \sqrt{(0.001)^2 + (0.001)^2} = 0.0014 \text{ g} \nonumber$

If we measure a single penny’s mass several times and obtain a standard deviation of ±0.050 g, then we have evidence that the measurement process is out of control. Knowing this, we can identify and correct the problem.

We also can use a propagation of uncertainty to help us decide how to improve an analytical method’s uncertainty. In Example $$\PageIndex{3}$$, for instance, we calculated an analyte’s concentration as 126 ppm ± 2 ppm, which is a percent uncertainty of 1.6%. Suppose we want to decrease the percent uncertainty to no more than 0.8%. How might we accomplish this? Looking back at the calculation, we see that the concentration’s relative uncertainty is determined by the relative uncertainty in the measured signal (corrected for the reagent blank)

$\frac {0.028} {23.41} = 0.0012 \text{ or } 0.12\% \nonumber$

and the relative uncertainty in the method’s sensitivity, kA,

$\frac {0.003 \text{ ppm}^{-1}} {0.186 \text{ ppm}^{-1}} = 0.016 \text{ or } 1.6\% \nonumber$

Of these two terms, the uncertainty in the method’s sensitivity dominates the overall uncertainty. Improving the signal’s uncertainty will not improve the overall uncertainty of the analysis. To achieve an overall uncertainty of 0.8% we must improve the uncertainty in kA to ±0.0015 ppm–1.

Exercise $$\PageIndex{3}$$

Verify that an uncertainty of ±0.0015 ppm–1 for kA is the correct result.

An uncertainty of 0.8% is a relative uncertainty in the concentration of 0.008; thus, letting u be the uncertainty in kA

$0.008 = \sqrt{\left( \frac {0.028} {23.41} \right)^2 + \left( \frac {u} {0.186} \right)^2} \nonumber$

Squaring both sides of the equation gives

$6.4 \times 10^{-5} = \left( \frac {0.028} {23.41} \right)^2 + \left( \frac {u} {0.186} \right)^2 \nonumber$

Solving for the uncertainty in kA gives its value as $$1.47 \times 10^{-3}$$ or ±0.0015 ppm–1.

Finally, we can use a propagation of uncertainty to determine which of several procedures provides the smallest uncertainty. When we dilute a stock solution usually there are several combinations of volumetric glassware that will give the same final concentration. For instance, we can dilute a stock solution by a factor of 10 using a 10-mL pipet and a 100-mL volumetric flask, or using a 25-mL pipet and a 250-mL volumetric flask. We also can accomplish the same dilution in two steps using a 50-mL pipet and 100-mL volumetric flask for the first dilution, and a 10-mL pipet and a 50-mL volumetric flask for the second dilution. The overall uncertainty in the final concentration—and, therefore, the best option for the dilution—depends on the uncertainty of the volumetric pipets and volumetric flasks. As shown in the following example, we can use the tolerance values for volumetric glassware to determine the optimum dilution strategy [Lam, R. B.; Isenhour, T. L. Anal. Chem. 1980, 52, 1158–1161].

Example $$\PageIndex{6}$$:

Which of the following methods for preparing a 0.0010 M solution from a 1.0 M stock solution provides the smallest overall uncertainty?

(a) A one-step dilution that uses a 1-mL pipet and a 1000-mL volumetric flask.

(b) A two-step dilution that uses a 20-mL pipet and a 1000-mL volumetric flask for the first dilution, and a 25-mL pipet and a 500-mL volumetric flask for the second dilution.

Solution

The dilution calculations for case (a) and case (b) are

$\text{case (a): 1.0 M } \times \frac {1.000 \text { mL}} {1000.0 \text { mL}} = 0.0010 \text{ M} \nonumber$

$\text{case (b): 1.0 M } \times \frac {20.00 \text { mL}} {1000.0 \text { mL}} \times \frac {25.00 \text{ mL}} {500.0 \text{mL}} = 0.0010 \text{ M} \nonumber$

Using tolerance values from Table 4.2.1, the relative uncertainty for case (a) is

$u_R = \sqrt{\left( \frac {0.006} {1.000} \right)^2 + \left( \frac {0.3} {1000.0} \right)^2} = 0.006 \nonumber$

and for case (b) the relative uncertainty is

$u_R = \sqrt{\left( \frac {0.03} {20.00} \right)^2 + \left( \frac {0.3} {1000} \right)^2 + \left( \frac {0.03} {25.00} \right)^2 + \left( \frac {0.2} {500.0} \right)^2} = 0.002 \nonumber$

Since the relative uncertainty for case (b) is less than that for case (a), the two-step dilution provides the smallest overall uncertainty. Of course we must balance the smaller uncertainty for case (b) against the increased opportunity for introducing a determinate error when making two dilutions instead of just one dilution, as in case (a).

This page titled 3.3: Propagation of Uncertainty is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.