Chemistry of Solutions 202-SN2-RE • CEGEP, Montreal

Revision sheet: measurement, uncertainty and graphical analysis of data (202-SN2-RE)

This is the revision sheet for the experimental side of Chemistry of Solutions, 202-SN2-RE (formerly 202-NYB): measurement, uncertainty and the graphical treatment of data. It does not summarize the course, it takes every place where a paper loses marks, in the lab report as on the exam, and gives the move that avoids it.

The thread fits in one sentence: the constant you are after is almost never a reading, it is a SLOPE. A slope is blind to anything that shifts every point by the same amount, which makes it more solid than an isolated reading, but its uncertainty can be read neither off a point nor off the coefficient of determination $r^{2}$ . The whole chapter consists of linearizing the law, reading the constant in the slope, then bracketing that slope with the extreme lines, which need no statistics.

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The thread of the chapter

In the chemistry of solutions, the constant you are after is almost never a reading, it is a SLOPE. A slope is blind to anything that shifts every point by the same amount, and its uncertainty can be read neither off a single point nor off the coefficient of determination.

This chapter is part of 202-SN2-RE, Chemistry of Solutions

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Relative uncertainty
Propagation of uncertainties
Linear regression
Coefficient of determination
Residual analysis
Extreme-lines method
Linearization of a law
Beer-Lambert law
Calibration curve
Arrhenius equation
Systematic error
Significant figures of a logarithm

The essentials

The constant is a slope, and a slope is blind to a shift

• The four laws of the course are read in a slope, never off an isolated reading: Beer-Lambert $A=\varepsilon \ell C$ , first order $\ln[A]=\ln[A]_{0}-kt$ , second order $\dfrac{1}{[A]}=\dfrac{1}{[A]_{0}}+kt$ , Arrhenius $\ln k=\ln A-\dfrac{E_{a}}{R}\cdot\dfrac{1}{T}$ .
• A slope is a ratio of DIFFERENCES: adding the same quantity to every point does not change it by a digit. A failed blank, a scratched cuvette, a slightly coloured reagent move the y-intercept $b$ and leave the slope $a$ intact.
• A corollary to remember for the exam: the y-intercept is the DETECTOR of systematic error, the slope is its shield. On a well-made calibration curve, $b$ must be compatible with zero.
• What does shift the slope, on the other hand, is a defect that turns with the x-axis: a miscalibrated pipette that always delivers $1\,\%$ too much makes every standard wrong by the same percentage, and the whole line tilts.

Both clouds have the SAME slope

0.191

: the systematic error of the scratched cuvette raised every point by

0.040

and moved only the y-intercept.

Writing “the slope is $\varepsilon\ell$ , so $\varepsilon=\dots$ ” before calculating earns the half mark for physical meaning that almost nobody takes.

Random or systematic: two errors, two remedies

• RANDOM error: it changes sign from one trial to the next, it SCATTERS the points around the line, it widens the extreme lines, and it shrinks with repetition.
• SYSTEMATIC error: it always falls on the same side, it SHIFTS, it leaves the points perfectly aligned and the coefficient of determination superb, and repeating a thousand times changes nothing.
• The three signatures that give it away: a y-intercept that excludes zero, a deviation from a reference value of more than two or three uncertainties although the points are well aligned, and a disagreement between two independent methods.
• The three remedies, and there are no others: calibrate (electrode on two buffers, spectrophotometer blank), run a control sample of known concentration, or redo the measurement by a DIFFERENT procedure.

The ten random measurements spread on both sides of the reference and their mean returns to it; the ten systematic ones are tight and all shifted to the same side.

The exam question is almost always phrased like this: “does this defect distort the slope or the y-intercept?” The answer rests on a single criterion: does it shift all the points by the same amount, or in proportion to the x-axis?

The uncertainty of a slope: what $r^{2}$ does not give, what the extreme lines do

• $r^{2}=\dfrac{S_{xy}^{2}}{S_{xx}S_{yy}}$ measures how well the points LINE UP around the fitted line. It says neither that the model is right, nor by how much the slope would move if the experiment were repeated.
• Extreme lines: draw with a ruler the STEEPEST and the least steep line that still pass through every error bar, then $a=\dfrac{a_{max}+a_{min}}{2}$ and $\Delta a=\dfrac{a_{max}-a_{min}}{2}$ . No statistics, and it is the method expected at CEGEP.
• The steepest line passes through the BOTTOM of the first point's bar and the TOP of the last one's, the least steep one the other way round: it is the end points that give the longest lever arm.
• A free check on any regression: the least-squares line ALWAYS passes through the mean point $(\bar x, \bar y)$ . A line drawn with a ruler that misses it is wrong before any calculation.

The least-squares line passes through the mean point

(1.5, 0.287)

; the line drawn with a ruler misses it, so it is wrong before any calculation.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The laws of the course and the axis that straightens them

You do not pick the plot out of habit: you pick it by the constant you are after. The last column says what the slope IS worth, hence what you write once it is calculated. The red box is the plot that decides nothing.

Law	What goes on the y-axis and the x-axis	What the slope is worth
Beer-Lambert	$A$ against $C$	$\varepsilon\ell$
Example: Slope $0.191$ L/(mg $\cdot$ cm) in a $1.00$ cm cuvette, so $\varepsilon=0.191$ L/(mg $\cdot$ cm).
Zero order	$[A]$ against $t$	$-k$
Example: A descending line of slope $-2.0\times 10^{-4}$ gives $k=2.0\times 10^{-4}$ mol/(L $\cdot$ s).
First order	$\ln[A]$ against $t$	$-k$
Example: Slope $-0.0116$ min $^{-1}$ , so $k=0.0116$ min $^{-1}$ and $t_{1/2}=\dfrac{\ln 2}{k}=59.8$ min.
Second order	$\dfrac{1}{[A]}$ against $t$	$+k$
Example: Slope $+0.300$ , so $k=0.300$ L/(mol $\cdot$ min) and $t_{1/2}=\dfrac{1}{k[A]_{0}}=66.6$ min.
Arrhenius	$\ln k$ against $\dfrac{1}{T}$	$-\dfrac{E_{a}}{R}$
Example: Slope $-7.218\times 10^{3}$ K, so $E_{a}=7.218\times 10^{3}\times 8.314=60.0$ kJ/mol.
Nernst	$E$ against $\log Q$	$-\dfrac{0.0592}{n}$
Example: Slope $-0.0296$ V at $25\,^{\circ}$ C, so $n=\dfrac{0.0592}{0.0296}=2$ electrons exchanged.
First or second order, raw plot	$[A]$ against $t$	settles nothing plot that decides nothing
Example: On the same six points, $r^{2}=0.983$ for $\ln[A]$ and $0.9995$ for $\dfrac{1}{[A]}$ : the reaction is second order. What to do: Plot BOTH linearizations and keep the one that lines up. The raw plot can only rule out zero order.

The Arrhenius temperature is in KELVINS, always. If the x-axis is $\dfrac{1000}{T}$ to handle numbers around $3$ , the slope is $-\dfrac{E_{a}}{1000R}$ and the factor $1000$ has to be put back by hand.

How an uncertainty goes through a calculation

You first identify the FORM of the operation, and only then do you add. The last column says what adds, hence what you write. The red box is a rule that does not exist and is refuted by two numbers.

Form of the operation	Example from the course	What adds
Sum or difference	$V=V_{1}+V_{2}$	the ABSOLUTE ones
Example: $(25.00\pm 0.05)+(10.00\pm 0.02)=(35.00\pm 0.07)$ mL.
Product or quotient	$C_{1}=C_{0}\dfrac{V_{p}}{V_{f}}$	the RELATIVE ones
Example: Pipette $0.200\,\%$ and flask $0.080\,\%$ give $0.280\,\%$ on the diluted solution.
Power $n$	$K_{sp}=4s^{3}$	relative $\times\,n$
Example: $4.6\,\%$ on $s$ gives $3\times 4.6=13.8\,\%$ on $K_{sp}$ .
Exact factor	$s=\dfrac{[OH^{-}]}{2}$	nothing at all
Example: $4.6\,\%$ on $[OH^{-}]$ is still $4.6\,\%$ on $s$ : the $2$ comes from the equation.
Coming out of a logarithm	$[H_{3}O^{+}]=10^{-\mathrm{pH}}$	relative $=\ln(10)\,\Delta\mathrm{pH}$
Example: $\Delta\mathrm{pH}=0.02$ gives $2.303\times 0.02=4.6\,\%$ on the concentration.
Going into a logarithm	$\mathrm{p}K_{a}=-\log K_{a}$	absolute $=\dfrac{\text{relative}}{\ln 10}$
Example: $9.7\,\%$ on $K_{a}$ gives $\dfrac{0.097}{2.303}=0.042$ on $\mathrm{p}K_{a}$ .
Percentage read off the pH	$\dfrac{\Delta\mathrm{pH}}{\mathrm{pH}}$	makes no sense rule that does not exist
Example: $\dfrac{0.02}{2.87}=0.70\,\%$ , whereas the concentration carries $4.6\,\%$ : a factor of $6.6$ . What to do: A $\Delta\mathrm{pH}$ is an ABSOLUTE uncertainty on an exponent: you go through $\ln(10)\times\Delta\mathrm{pH}$ .

A square root is a power of $\dfrac{1}{2}$ : it HALVES the relative uncertainty. And you never add two uncertainties of different kinds, an absolute one in millilitres and a relative one in percent.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Taking a coefficient of determination close to 1 for a proof of the model

the whole question, and the following ones with it, since

k

and

t_{1/2}

follow from it

What not to write

“ $r^{2}=0.983$ , so the reaction is first order and the law is verified.”

What to write

“ $r^{2}=0.983$ for first order against $0.9995$ for second order: the gap from $1$ is thirty times smaller for second order, and the residuals of first order draw an arc, $+\,-\,-\,-\,+\,+$ . The reaction is second order.”

The six residuals of first order are positive at both ends and negative in the middle: this arc pattern is the signature of a line fitted to curved points, and

r^{2}=0.983

does not see it.

Why: $r^{2}$ adds up squares, so it LOSES the sign of the deviations: it cannot tell six random deviations from six deviations arranged in an arc. An $r^{2}$ is never read alone, it is compared with that of the competing model, and it is the gap from $1$ that is compared.

2. Looking for the uncertainty of the slope in $r^{2}$ or in an isolated point

2 marks on the lab report, and the final result becomes unverifiable

What not to write

“The calculator displays $a=0.1912$ with $r^{2}=0.9999$ , so the slope is known to the fourth decimal place.”

What to write

“Extreme lines: $a_{max}=0.199$ and $a_{min}=0.183$ , so $a=(0.191\pm 0.008)$ L/(mg $\cdot$ cm), that is $4.2\,\%$ . Two digits survive, not four.”

The two lines still pass through every error bar: their half-gap,

0.008

, IS the uncertainty on the slope, and it already falls on the third digit of

0.191

Why: The four digits displayed by the calculator are the result of a division, not a measurement. $r^{2}$ contains NO uncertainty: a separate method is needed, and at CEGEP it is the extreme lines.

3. Propagating an uncertainty through a logarithm as through a product

3 marks, and the whole compatibility discussion that follows becomes wrong by a factor of 6

What not to write

“ $\mathrm{pH}=2.87\pm 0.02$ , so $\dfrac{0.02}{2.87}=0.70\,\%$ on $[H_{3}O^{+}]$ .”

What to write

“ $\dfrac{\Delta[H_{3}O^{+}]}{[H_{3}O^{+}]}=\ln(10)\times\Delta\mathrm{pH}=2.303\times 0.02=4.6\,\%$ , so $[H_{3}O^{+}]=(1.35\pm 0.06)\times 10^{-3}$ mol/L.”

A band of

\pm 0.02

in pH gives a band of

\pm 4.6\,\%

[H_{3}O^{+}]

: that is

\ln(10)\times 0.02

, and not

\dfrac{0.02}{2.87}

Why: A $\Delta\mathrm{pH}$ is an ABSOLUTE uncertainty on an EXPONENT, hence a RELATIVE uncertainty on the concentration. The percentage read off the pH has no chemical meaning: it depends on the scale, not on the solution.

4. Counting the significant figures of a pH like those of a mass

1 mark per result, and the fault is repeated on every line of a report

What not to write

“ $\mathrm{pH}=2.87$ : three digits written, so $[H_{3}O^{+}]=1.35\times 10^{-3}$ mol/L.”

What to write

“In a logarithm, only the MANTISSA carries significant figures: $87$ has two, so $[H_{3}O^{+}]=1.3\times 10^{-3}$ mol/L.”

Why: The characteristic, here the $2$ , merely places the power of ten, exactly like the exponent of a scientific notation. The uncertainty confirms it separately: $4.6\,\%$ already puts the second digit in question.

5. Forgetting that the exponent multiplies the relative uncertainty

2 marks, and a compatibility conclusion reversed in one case out of two

What not to write

“ $K_{sp}=4s^{3}=5.61\times 10^{-6}$ and $s$ is known to $4.6\,\%$ , so $K_{sp}$ is too.”

What to write

“ $K_{sp}$ goes as $s^{3}$ : a power MULTIPLIES the relative uncertainty by its exponent, so $3\times 4.6=13.8\,\%$ and $K_{sp}=(5.6\pm 0.8)\times 10^{-6}$ .”

Why: The $4$ of $4s^{3}$ is an exact number and contributes nothing, whereas the cube triples everything. Two hundredths of a pH unit become fourteen percent on the constant through nothing more than the composition of a logarithm and a cube.

6. Believing that a systematic error shrinks when the measurement is repeated

the whole critical discussion, 3 to 4 marks on the lab report

What not to write

“The electrode is rated at $\pm 0.02$ pH units, but I made ten measurements, so my uncertainty is divided by $\sqrt{10}$ .”

What to write

“The $\pm 0.02$ is the uncertainty of the electrode's CALIBRATION: it shifts every reading the same way. Repeating improves precision and corrects nothing. Only a recalibration, a control sample or a second method would.”

The more you repeat, the tighter the bar around the mean; but the mean stays shifted by the calibration, and the true value stays outside (schematic).

Why: Repeating averages only what changes sign. A poorly rinsed burette, an electrode calibrated on an expired buffer, a scratched cuvette give a thousand almost identical values, all wrong, with a reassuring scatter thrown in.

7. Plotting concentration against time to find the order

the whole question, 8 to 10 marks on a kinetics problem

What not to write

“I plot $[A]$ against $t$ : the curve descends and flattens, so it is first order.”

What to write

“The raw plot rules out only zero order. I plot $\ln[A]$ against $t$ AND $\dfrac{1}{[A]}$ against $t$ , and it is the one that LINES UP that names the order.”

The six points of a second-order reaction and the best first-order model: two curved lines almost on top of each other. The raw plot does not settle it, the linearizations do.

Why: First order and second order both produce a curved decay that flattens, indistinguishable by eye. You do not compare two curves, you compare two ALIGNMENTS, which a human eye can judge and a ruler can check.

8. Adding absolute uncertainties in a product

2 marks, and the result does not even have the right unit

What not to write

“ $m=C\times V$ with $\Delta C=0.1$ mg/L and $\Delta V=0.15$ mL, so $\Delta m=0.25$ .”

What to write

“A product makes the RELATIVE uncertainties add: $\dfrac{\Delta m}{m}=\dfrac{\Delta C}{C}+\dfrac{\Delta V}{V}$ , and you then multiply by $m$ to get back to an absolute one.”

Why: Adding absolute uncertainties in a product makes no dimensional sense: you would be adding millilitres to milligrams per litre. Absolute uncertainties add only in a sum or a difference, where everything is already in the same unit.

9. Forgetting that a serial dilution accumulates uncertainties

1.5 marks, and the error bar of the standards is underestimated by a factor of 3

What not to write

“Three one-in-ten dilutions, so $C_{3}=\dfrac{C_{0}}{1000}$ with the same relative uncertainty as at the start.”

What to write

“Dilution factors MULTIPLY but relative uncertainties ADD: $3\times(0.200+0.080)+0.10=0.94\,\%$ , so $C_{3}=(1.000\pm 0.009)\times 10^{-5}$ mol/L.”

Each one-in-ten dilution adds the pipette and the flask,

0.280\,\%

: starting from

0.10\,\%

, the uncertainty reaches

0.94\,\%

after three steps.

Why: Each step is one more quotient, hence two more relative uncertainties. One more dilution always costs something, even when carried out perfectly, and that is the argument for preparing standards with a volumetric pipette rather than a cylinder.

10. Reading a concentration off the calibration line outside its range

the whole question, and the concentration announced is underestimated

What not to write

“The sample gives $A=1.20$ , so $C=\dfrac{1.20}{0.191}=6.28$ mg/L.”

What to write

“ $A=1.20$ is outside the calibrated range $[0.500, 2.500]$ mg/L and above the limit of linearity $A\approx 1$ . I dilute the sample by a known factor, read again around $A=0.3$ to $0.7$ , then multiply.”

The solid line is the only measured stretch, from

0.500

2.500

mg/L; the

6.3

mg/L read at

A=1.20

come from the dotted extension, where the law has already stopped being linear.

Why: Above $A\approx 1$ , stray light and the association of coloured molecules bend the relationship DOWNWARD: it is not only that you have left the interval, the law has changed shape, and adding more concentrated standards would save nothing. At the other end, below $A\approx 0.1$ , it is the relative uncertainty that explodes, $\dfrac{0.008}{0.012}=67\,\%$ .

Which method to choose

Which plot to make, depending on the constant you are after

You look at what the problem ASKS for, not at what the table of measurements contains. The plot is chosen so that the constant becomes a slope.

The SAME six measurements, linearized two ways: on the left the points drift away from the line in an arc, on the right they line up, and it is this second plot that names second order.

If you are after $\varepsilon$ , or an unknown concentration by calibration → plot $A$ against $C$ , slope $\varepsilon\ell$ , y-intercept zero

Example: slope $0.191$ L/(mg $\cdot$ cm), so $C=\dfrac{A}{0.191}$

the y-intercept serves as a check on the blank, it must be compatible with zero
If the problem asks for the ORDER of a reaction → plot $\ln[A]$ against $t$ AND $\dfrac{1}{[A]}$ against $t$ , keep the one that lines up

Example: $r^{2}=0.983$ against $0.9995$ : second order

the raw plot of $[A]$ against $t$ only serves to rule out zero order, a descending straight line
If the order is already given and you are after $k$ → do the ONE linearization of that order and read $k$ off the slope, up to the sign

Example: second order, slope $+0.300$ , so $k=0.300$ L/(mol $\cdot$ min)

first and zero order give a slope $-k$ , second order a slope $+k$
If the problem gives several TEMPERATURES → Arrhenius: $\ln k$ against $\dfrac{1}{T}$ in kelvins, slope $-\dfrac{E_{a}}{R}$

Example: slope $-7.218\times 10^{3}$ K, so $E_{a}=60.0$ kJ/mol

the slope is necessarily negative, since heating speeds things up
If the problem gives a potential measured at several concentrations → Nernst: $E$ against $\log Q$ , slope $-\dfrac{0.0592}{n}$ at $25\,^{\circ}$ C

Example: slope $-0.0296$ V, so $n=2$

If no branch applies, the quantity you are after is read directly off an instrument: the uncertainty is then that of the instrument, not that of a slope. The rest of the sheet deals with the other case, which is the one in the course.

My result does not land on the reference value

You first quantify the deviation as a NUMBER of uncertainties, $\dfrac{|x_{meas}-x_{ref}|}{\Delta x}$ , then you look at the shape of the cloud. The percentage alone settles nothing.

The same error bar, three deviations: below one uncertainty we conclude agreement, between one and two a cautious agreement, beyond three a systematic error.

If the deviation is less than 1 uncertainty → the measurement is compatible, and that is all you can assert

Example: $K_{sp}$ at $2.0\,\%$ from the reference with an error bar of $13.8\,\%$ : $0.14$ uncertainty

do not congratulate yourself on the small percentage, a wide error bar makes agreement easy
If the deviation is 1 to 2 uncertainties → acceptable agreement, but state that a bias remains possible and name the candidates

Example: $E_{a}=(60.0\pm 1.0)$ against $58.8$ kJ/mol: $1.2$ uncertainty
If the deviation exceeds 2 or 3 uncertainties AND the points are well aligned → conclude to a SYSTEMATIC error and look for the faulty calibration

Example: a y-intercept in $[0.030, 0.055]$ , which excludes zero

repeating the experiment would be pointless, you must recalibrate or change method
If the points are SCATTERED and the extreme lines are far apart → the error is random: repeat, increase the number of points, improve your technique

Example: $\Delta a/a$ of $12\,\%$ instead of $4\,\%$ on the same calibration
If the residuals draw a PATTERN, an arc or a wave → this is not a question of uncertainty, it is the MODEL that is wrong: change the linearization

Example: residuals $+\,-\,-\,-\,+\,+$ on first order, whereas second order makes them flat

A compatibility sentence is always worth a mark and costs nothing: “the deviation of $1.2$ uncertainties does not allow us to reject agreement with the reference value”. A conclusion without a number of uncertainties earns nothing.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Publishing a constant taken from a slope

When to use it: As soon as a lab report or a problem asks for a constant from a table of measurements: $\varepsilon$ , $k$ , $E_{a}$ , $K_{a}$ , $K_{sp}$ .

1 Name the linearization BEFORE calculating: “the law $k=Ae^{-E_{a}/RT}$ linearizes to $\ln k=\ln A-\dfrac{E_{a}}{R}\cdot\dfrac{1}{T}$ , so I plot $\ln k$ on the y-axis and $\dfrac{1}{T}$ in kelvins on the x-axis.”
2 Calculate the slope, $a=\dfrac{S_{xy}}{S_{xx}}$ , and check its SIGN against the physics: negative for Arrhenius and for first order, positive for second order.
3 Convert the slope into a constant with its units written out: “ $E_{a}=-a\times 1000R=7.218\times 10^{3}\times 8.314=6.00\times 10^{4}$ J/mol”.
4 Bracket the slope with the extreme lines and take $\Delta a$ from them, then propagate to the constant: here $E_{a,max}=61.0$ and $E_{a,min}=59.0$ kJ/mol.
5 Round the uncertainty to ONE significant figure, then line the result up on it: $E_{a}=(60.0\pm 1.0)$ kJ/mol.
6 Compare with the reference value as a NUMBER of uncertainties, and conclude in one sentence.
7 If a defect is reported in the problem, say whether it affects the slope or the y-intercept, and therefore whether it distorts the constant.

Concluding sentence

“The regression of $\ln k$ against $\dfrac{1000}{T}$ gives a slope of $-7.218$ , so $E_{a}=(60.0\pm 1.0)$ kJ/mol, that is $1.7\,\%$ . The deviation from the reference value $58.8$ kJ/mol is $1.2$ uncertainties: the measurement is compatible with the literature.”

The trap: The step everybody skips is the fifth: you publish the five digits of the calculator because no uncertainty was calculated. Without $\Delta a$ , the constant is not a measurement result.

Marking: Typically 1 mark for the named linearization, 2 for the slope and units, 2 for the uncertainty by the extreme lines, 1 for consistent rounding, 1 for the compatibility sentence.

The uncertainty budget of an assay

When to use it: When the problem asks for “the uncertainty on the result” of a chain of measurements: aliquots, dilutions, instrument reading, calibration curve.

1 Write the COMPLETE formula of the result, from the volume taken to the mass you are after: $m=\dfrac{A}{a}\times\dfrac{V_{f}}{V_{p}}\times V_{tot}$ .
2 Note that everything is a product or a quotient, and so announce: “I am working in relative uncertainties”.
3 List the items one by one, each with its calculation: absorbance $\dfrac{0.008}{0.412}=1.9\,\%$ , slope $4.2\,\%$ , pipette $0.30\,\%$ , flasks $0.060\,\%$ each.
4 Add the relative uncertainties, go back to an absolute one by multiplying by the result, round the uncertainty to one figure.
5 Name the DOMINANT item and say what you would gain by improving it, with figures.

Concluding sentence

“The uncertainty budget gives $6.5\,\%$ , two thirds of it from the slope of the calibration curve, so $m=(27\pm 2)$ mg: the tablet complies with the label, and this assay could not tell $27$ mg from $28.5$ mg.”

The trap: Taking care over the glassware gains almost nothing here, $6\,\%$ , whereas tightening the calibration gains $32\,\%$ . The visible gesture is not the useful gesture, and that is exactly what question d) expects you to say.

Marking: 1 mark per item correctly quantified, 1 for the total, 1 for consistent rounding, 2 for the discussion of the dominant item.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Does the line pass through the mean point?

Calculate $\bar x$ and $\bar y$ in your head, then check that $a\bar x+b=\bar y$ . The least-squares line ALWAYS passes through it, however scattered the data: it is the cheapest check on the paper and it does not take ten seconds.

$0.1912\times 1.500+0.0004=0.2872=\bar A$ . If the sum does not come out, the slope or the y-intercept is wrong, and there is no point going further.
Is the sign of the slope the one the physics predicts?

Arrhenius and first order give a NEGATIVE slope, second order a POSITIVE one, Beer-Lambert a positive slope through the origin. The opposite sign signals swapped axes or a forgotten sign on the exponent.

A positive Arrhenius slope would give a negative activation energy, hence a reaction that slows down when heated: impossible.
Does the uncertainty carry ONE digit, and does the result line up on it?

Round the uncertainty to one significant figure, then cut the result at the same decimal place. If the result has more digits than the uncertainty allows, it is wrong, even if the calculation is right.

$1.596$ mg/L with an uncertainty of $0.109$ is published $(1.6\pm 0.1)$ mg/L, never $1.596\pm 0.109$ .
Is the absorbance reading inside the useful window?

An absorbance should fall between $0.1$ and $1$ , and preferably between $0.3$ and $0.7$ . Below, the relative uncertainty explodes; above, the law is no longer linear.

A reading at $A=0.012$ carries $\dfrac{0.008}{0.012}=67\,\%$ uncertainty: the result is worthless, and you must concentrate the sample instead of reading it where it happens to be.
Is the deviation from the reference expressed as a number of uncertainties?

Divide the absolute deviation by the absolute uncertainty. A percentage alone settles nothing: $2\,\%$ deviation is excellent with an error bar of $14\,\%$ and catastrophic with one of $0.5\,\%$ .

$K_{sp}=5.61$ against $5.50\times 10^{-6}$ : $2.0\,\%$ , but above all $\dfrac{0.11}{0.78}=0.14$ uncertainty, so compatible.
Are the digits of a result that comes from a pH counted on the mantissa?

Count the decimal places of the pH, not its written digits. Two decimal places give two significant figures to the concentration, three decimal places give three.

$\mathrm{pH}=12.35$ gives $[OH^{-}]=2.2\times 10^{-2}$ mol/L, and not $2.239\times 10^{-2}$ , even if the calculator displays it.

The typical problem, taken apart

A solubility product taken from a single pH, from the measurement to the compatibility sentence

Solid calcium hydroxide is stirred in distilled water until saturation, the mixture is filtered, and the pH of the filtrate is measured: $\mathrm{pH}=12.35\pm 0.02$ at $25\,^{\circ}$ C, the electrode having been calibrated on two buffers.

The dissolution is written $Ca(OH)_{2}(s)\rightleftharpoons Ca^{2+}+2\,OH^{-}$ , so $K_{sp}=4s^{3}$ where $s$ is the molar solubility. We are given $K_{w}=1.0\times 10^{-14}$ and the reference value $K_{sp}=5.5\times 10^{-6}$ .

Give the solubility product with its uncertainty, conclude on compatibility, then say whether repeating the measurement would improve the result.

The measurement in the example: the saturated solution, filtered, and the pH of the filtrate read on the pH meter.

Step 1

$Ca(OH)_{2}(s)\rightleftharpoons Ca^{2+}+2\,OH^{-}$ , so $[OH^{-}]=2s$ and $K_{sp}=[Ca^{2+}][OH^{-}]^{2}=s\times(2s)^{2}=4s^{3}$ .

Why

The equation is written BEFORE any calculation: the factor $2$ comes from no uncertainty, it comes from the stoichiometry, and it is the first fault when it is forgotten.

Step 2

$\mathrm{pOH}=14.00-12.35=1.65$ , so $[OH^{-}]=10^{-1.65}=2.239\times 10^{-2}$ mol/L and $s=\dfrac{[OH^{-}]}{2}=1.119\times 10^{-2}$ mol/L.

Why

This is the only step where you COME OUT of the logarithm, hence the only place where the propagation rule changes. Keep every digit of the calculator, the rounding will come once the uncertainty is known.

Step 3

$K_{sp}=4s^{3}=4\times(1.119\times 10^{-2})^{3}=4\times 1.402\times 10^{-6}=5.61\times 10^{-6}$ .

Why

The raw result is written out in full. Rounding here would force you to start again, since the number of digits to keep is not yet known.

Step 4

Coming out of the logarithm: $\dfrac{\Delta[OH^{-}]}{[OH^{-}]}=\ln(10)\times\Delta\mathrm{pH}=2.303\times 0.02=4.6\,\%$ . Dividing by $2$ , an exact number, does not change this relative uncertainty, so $s$ also carries $4.6\,\%$ .

Why

This is the step that earns the most marks and the one most often missed: writing $\dfrac{0.02}{12.35}=0.16\,\%$ would give an error bar thirty times too small and a reversed conclusion.

Step 5

Power: $K_{sp}$ goes as $s^{3}$ , so $3\times 4.6=13.8\,\%$ . The $4$ is exact and contributes nothing. Absolute: $0.138\times 5.61\times 10^{-6}=7.8\times 10^{-7}$ .

Why

The exponent MULTIPLIES the relative uncertainty, it does not add to it. This is what turns two hundredths of a pH unit into fourteen percent on the constant.

Step 6

Rounding: the uncertainty takes one digit, $0.8\times 10^{-6}$ , and the result lines up on it: $K_{sp}=(5.6\pm 0.8)\times 10^{-6}$ .

Why

Publishing $5.61$ with an error bar of $0.8$ claims a hundredth that the measurement does not know. The uncertainty commands the last digit, never the calculator.

Step 7

Compatibility: relative deviation $\dfrac{|5.61-5.50|}{5.50}\times 100=2.0\,\%$ , and above all $\dfrac{0.11\times 10^{-6}}{0.78\times 10^{-6}}=0.14$ uncertainty.

Why

It is the number of uncertainties that decides, not the percentage. Congratulating yourself on the $2\,\%$ would be a misreading: with an error bar of $14\,\%$ , landing anywhere between $4.8$ and $6.4\times 10^{-6}$ was equally acceptable.

Step 8

Repeat? The $\pm 0.02$ is the uncertainty of the electrode's CALIBRATION, hence systematic: ten measurements would give ten values shifted alike. To aim for $5\,\%$ on $K_{sp}$ you would need $\Delta\mathrm{pH}=\dfrac{0.05}{3\ln 10}=0.007$ , out of reach. The remedy is a TITRATION of the filtrate with hydrochloric acid, which goes through no logarithm.

Why

The question “would repeating be enough?” is a question about the NATURE of the error, never about the number of trials. Naming an independent second method earns the mark for critical discussion.

The conclusion, written out

“The saturated filtrate gives $s=1.12\times 10^{-2}$ mol/L and $K_{sp}=(5.6\pm 0.8)\times 10^{-6}$ , that is $13.8\,\%$ . The deviation from the reference value $5.5\times 10^{-6}$ is $0.14$ uncertainty: the measurement is fully compatible with the literature. Since the uncertainty is dominated by the electrode calibration, which is systematic, repeating the measurement would not improve it; an acid-base titration of the filtrate would divide it by ten.”

The classic mistake on this problem: The classic error on this very exercise is to forget the factor $2$ between $[OH^{-}]$ and $s$ , which multiplies $K_{sp}$ by $8$ and gives $4.5\times 10^{-5}$ : the result is then eight times too large and the compatibility conclusion is reversed, with no uncertainty able to save the paper.

Learn by heart

• The constant is a SLOPE: you linearize first, you read afterward.
• A slope is blind to a shift common to all the points. That shift goes into the y-intercept, which is therefore the detector of systematic error.
• Zero order: $[A]$ against $t$ , slope $-k$ . First order: $\ln[A]$ against $t$ , slope $-k$ . Second order: $\dfrac{1}{[A]}$ against $t$ , slope $+k$ . Arrhenius: $\ln k$ against $\dfrac{1}{T}$ in KELVINS, slope $-\dfrac{E_{a}}{R}$ .
• $r^{2}$ measures ALIGNMENT, never the model, and it supplies NO uncertainty. You compare it with that of the competing model, on the gap from $1$ .
• A pattern in the residuals exposes the model; scatter with no pattern exposes only chance.
• Extreme lines: $a=\dfrac{a_{max}+a_{min}}{2}$ and $\Delta a=\dfrac{a_{max}-a_{min}}{2}$ , the steepest line passing through the BOTTOM of the first point and the TOP of the last.
• Sum or difference: the ABSOLUTE uncertainties add. Product or quotient: the RELATIVE ones. Power $n$ : the relative uncertainty is MULTIPLIED by $n$ .
• Passing through a logarithm: $\dfrac{\Delta[H_{3}O^{+}]}{[H_{3}O^{+}]}=\ln(10)\times\Delta\mathrm{pH}\approx 2.30\,\Delta\mathrm{pH}$ . Never $\dfrac{\Delta\mathrm{pH}}{\mathrm{pH}}$ .
• In a logarithm, only the MANTISSA carries significant figures.
• Repeating reduces RANDOM error and NEVER touches systematic error.
• An absorbance is read between $0.1$ and $1$ , preferably between $0.3$ and $0.7$ .
• The uncertainty is rounded to ONE significant figure and the result lines up on it. Compatibility is quantified as a NUMBER of uncertainties.

Frequently asked questions

Does a coefficient of determination of 0.999 prove that the model is right?

No. It only measures how well the points line up around the fitted line. Over a narrow interval, a slightly curved line easily gives 0.999. To settle it, compare it with that of the competing model by looking at the gap from 1, and examine the residuals: an arc pattern exposes a line fitted to curved points, which the coefficient cannot see.

How do I calculate the uncertainty on a slope without statistics?

By the extreme-lines method. Draw with a ruler the steepest and the least steep line that still pass through every error bar, the steepest passing through the bottom of the first point's bar and the top of the last point's. The slope is the mean of the two and the uncertainty is their half-gap. No statistics are involved, and it is the method expected at CEGEP.

How many significant figures does a pH of 2.87 carry?

Two. In a logarithm, only the mantissa, the part after the decimal point, carries significant figures. The characteristic, here the 2, merely places the power of ten, like the exponent of a scientific notation. The hydronium ion concentration is therefore written 1.3 times ten to the minus three moles per litre, and not one digit more.

How do I propagate an uncertainty through a pH?

The relative uncertainty on the concentration equals the natural logarithm of ten times the absolute uncertainty on the pH, that is about 2.30 times delta pH. An uncertainty of 0.02 pH units therefore gives 4.6 percent on the concentration. Never divide delta pH by the pH itself: that percentage depends on the scale chosen and has no chemical meaning.

How do I find the order of a reaction from a table of measurements?

By plotting, not by looking at the raw curve. Concentration against time can only rule out zero order, which alone would give a straight line. Then plot the logarithm of the concentration against time, for first order, and the inverse of the concentration against time, for second order: the one that lines up names the order, and its slope gives the rate constant.

Does repeating a measurement ten times always reduce the uncertainty?

No, only the random part, the one that changes sign from one trial to the next. A systematic error, a poorly rinsed burette, an electrode calibrated on an expired buffer, a scratched cuvette, always falls on the same side: a thousand measurements will give a thousand values shifted alike, with excellent precision thrown in. Only a recalibration, a control sample or a second method corrects it.

Practise it

Corrected exercises: Measurement, uncertainty and graphical analysis, 202-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn2-measurement-uncertainty. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: measurement, uncertainty and graphical analysis of data (202-SN2-RE)

The essentials

The constant is a slope, and a slope is blind to a shift

Random or systematic: two errors, two remedies

The uncertainty of a slope: what r2r^{2}r2 does not give, what the extreme lines do

The rules in table form

The laws of the course and the axis that straightens them

How an uncertainty goes through a calculation

The mistakes that cost marks

1. Taking a coefficient of determination close to 1 for a proof of the model

2. Looking for the uncertainty of the slope in r2r^{2}r2 or in an isolated point

3. Propagating an uncertainty through a logarithm as through a product

4. Counting the significant figures of a pH like those of a mass

5. Forgetting that the exponent multiplies the relative uncertainty

6. Believing that a systematic error shrinks when the measurement is repeated

7. Plotting concentration against time to find the order

8. Adding absolute uncertainties in a product

9. Forgetting that a serial dilution accumulates uncertainties

10. Reading a concentration off the calibration line outside its range

Which method to choose

Which plot to make, depending on the constant you are after

My result does not land on the reference value

How the answer is expected to be written

Publishing a constant taken from a slope

The uncertainty budget of an assay

Check before you hand in

The typical problem, taken apart

A solubility product taken from a single pH, from the measurement to the compatibility sentence

Learn by heart

Frequently asked questions

Does a coefficient of determination of 0.999 prove that the model is right?

How do I calculate the uncertainty on a slope without statistics?

How many significant figures does a pH of 2.87 carry?

How do I propagate an uncertainty through a pH?

How do I find the order of a reaction from a table of measurements?

Does repeating a measurement ten times always reduce the uncertainty?

Corrected exercises: Measurement, uncertainty and graphical analysis, 202-SN2-RE

See also

Looking for a 202-SN2-RE tutor in Montreal?

The uncertainty of a slope: what $r^{2}$ does not give, what the extreme lines do

2. Looking for the uncertainty of the slope in $r^{2}$ or in an isolated point