Corrected exercises: measurement, uncertainty and graphical analysis of data (202-SN2-RE)

Pipette: $\dfrac{0.02}{10.00}\times 100=0.200\,\%$. Flask: $\dfrac{0.08}{100.00}\times 100=0.080\,\%$. Cylinder: $\dfrac{1}{100}\times 100=1.0\,\%$. The cylinder is five times worse than the pipette although it measures ten times more: a tolerance can only be compared once it is expressed relative to the volume measured.

The diluted concentration is $C_{1}=C_{0}\dfrac{V_{pipette}}{V_{flask}}$: a quotient, so the RELATIVE uncertainties add. $0.200+0.080=0.280\,\%$. Note that the two instruments do not weigh the same: the pipette alone supplies five sevenths of the total, because it measures the smallest volume in the chain.

Each dilution divides by $10$, so $C_{3}=\dfrac{C_{0}}{1000}=1.000\times 10^{-5}$ mol/L. The three steps are three successive quotients: $3\times 0.280=0.840\,\%$, plus the $0.10\,\%$ of the stock solution, giving $0.94\,\%$. Absolute uncertainty: $0.0094\times 1.000\times 10^{-5}=9.4\times 10^{-8}$ mol/L, so $C_{3}=(1.000\pm 0.009)\times 10^{-5}$ mol/L. The lesson of the calculation: dilution factors MULTIPLY but uncertainties ADD, so one more dilution always costs something, even when it is carried out perfectly.

One step rises to $\dfrac{0.1}{10.0}\times 100+0.080=1.08\,\%$, the three steps to $3.24\,\%$, and the total to $3.34\,\%$ with the stock solution. The factor is $\dfrac{3.34}{0.94}=3.6$: the uncertainty on every standard is multiplied by more than three, and with it the uncertainty on the slope that will be drawn from them. This is no saving of time, it is a net loss of precision on the final result, about half a significant figure.

Because the two defects do not do the same damage to a line. An IMPRECISE pipette delivers $9.98$ mL one time and $10.02$ mL the next: the errors change sign from one standard to the next, the points scatter around the line, and that scatter is visible, it widens the extreme lines and so it shows up in the published uncertainty. A miscalibrated pipette always delivers $9.90$ mL: all five standards are $1\,\%$ too dilute, in the same direction, the points stay perfectly aligned, the coefficient of determination stays superb, and the whole line tilts by $1\,\%$. The systematic error therefore hides exactly where nobody looks, in the SLOPE, while every quality indicator stays green. Only an independent check, weighing the water delivered by the pipette, reveals it.

$\bar C=\dfrac{0.500+1.000+1.500+2.000+2.500}{5}=1.500$ mg/L and $\bar A=\dfrac{1.436}{5}=0.2872$. This point has a property that costs nothing and serves as a permanent check: the least-squares line ALWAYS passes through the mean point, however scattered the data. If your line drawn with a ruler does not pass through $(1.500, 0.2872)$, it is wrong before any calculation.

The deviations in $C$ are $-1.00$, $-0.50$, $0$, $+0.50$, $+1.00$, so $S_{CC}=1+0.25+0+0.25+1=2.50$. For $S_{CA}$: $(-1.00)(-0.1912)+(-0.50)(-0.0972)+0+(0.50)(0.0948)+(1.00)(0.1908)=0.1912+0.0486+0+0.0474+0.1908=0.4780$. Hence $a=\dfrac{0.4780}{2.50}=0.1912$ L/(mg$\cdot$cm) for a $1.00$ cm cuvette. Physically, this slope is the product $\varepsilon \ell$ of Beer-Lambert: it, and it alone, holds the chemical information of the calibration.

$b=\bar A-a\bar C=0.2872-0.1912\times 1.500=0.2872-0.2868=0.0004$. It should be zero: at zero concentration the cuvette holds the blank, and the spectrophotometer was set to zero on that very blank. A y-intercept of $0.040$ would have meant that every absorbance is too high by $0.040$, so a failed blank, a scratched cuvette or a reagent that is coloured by itself. It is the cheapest check of the session, and it detects only one thing, but precisely the one that escapes the coefficient of determination: a SHIFT common to all the points.

$r^{2}=\dfrac{(0.4780)^{2}}{2.50\times 0.091405}=\dfrac{0.228484}{0.228513}=0.9999$. The points are therefore practically aligned.

First mistake: $r^{2}$ measures how well the points LINE UP, not whether the model is right. A slightly curved line over a narrow interval also gives $0.999$, and a perfect systematic error gives exactly $1.000$ while being entirely wrong. An $r^{2}$ is never read alone: it is compared with that of the competing model, or backed up by an examination of the residuals. Second mistake: $r^{2}$ contains NO uncertainty. It says the points are aligned, it does not say by how much the slope could move if the experiment were repeated. The four digits the calculator displays for the slope are the result of a division, not a measurement: a separate method is needed, the extreme lines of the next exercise, to know how many of those digits survive.

Through the BOTTOM of the bar of the first point and the TOP of the last: $(0.500, 0.088)$ and $(2.500, 0.486)$. Two reasons. First, a slope is a ratio of differences: to make it as large as possible, you lower the left-hand point and raise the right-hand one, and it is the end points that give the longest lever arm. Second, the three middle points constrain nothing here, since their bars are crossed by the line obtained; they would have to be checked, and the line moved if one of them were missed, which happens as soon as a point is an outlier.

$a_{max}=\dfrac{0.486-0.088}{2.500-0.500}=\dfrac{0.398}{2.000}=0.199$ L/(mg$\cdot$cm).

Symmetrically, through the top of the first point and the bottom of the last: $a_{min}=\dfrac{0.470-0.104}{2.000}=\dfrac{0.366}{2.000}=0.183$ L/(mg$\cdot$cm).

$a=\dfrac{0.199+0.183}{2}=0.191$ and $\Delta a=\dfrac{0.199-0.183}{2}=0.008$, that is $a=(0.191\pm 0.008)$ L/(mg$\cdot$cm) and $\dfrac{0.008}{0.191}\times 100=4.2\,\%$. The midpoint of the two extreme lines lands on $0.191$, within a thousandth of the regression: the two methods agree on the VALUE. They do not agree on the number of digits. Of the four digits of $0.1912$, TWO survive: the uncertainty already falls on the third. Writing $0.1912$ in a report claims a precision about eighty times better than that of the instrument.

Upper line: $b_{max}=0.088-0.199\times 0.500=-0.0115$. Lower line: $b_{min}=0.104-0.183\times 0.500=+0.0125$. The interval $[-0.012, +0.013]$ contains zero: the blank is compatible with zero, nothing contradicts it, and that is all you can assert. An interval $[0.030, 0.055]$, on the other hand, would exclude zero: there would then be a residual absorbance of a few hundredths that the scatter of the measurements cannot explain, hence a SYSTEMATIC error, a scratched cuvette, a blank prepared without the reagent or a slightly coloured reagent. And that error, it must be stressed, would leave the slope untouched: it raises every point by the same amount, which moves $b$ without touching $a$. This is the deep reason why a constant is sought in a slope rather than in an isolated reading.

Invert the line: $C=\dfrac{A-b}{a}=\dfrac{0.305-0.000}{0.191}=1.596$ mg/L, which the calculator will give to as many digits as you like. It is the easiest calculation of the chapter, and also the one that is spoiled by stopping there.

The quotient $\dfrac{A}{a}$ makes the RELATIVE uncertainties add: $\dfrac{0.008}{0.305}\times 100=2.6\,\%$ for the absorbance, $4.2\,\%$ for the slope, total $6.8\,\%$. Absolute uncertainty: $0.068\times 1.596=0.11$ mg/L, rounded to one significant figure, so $0.1$ mg/L. Publishable result: $C=(1.6\pm 0.1)$ mg/L. Two digits, not four. Note which of the two contributions dominates: the SLOPE, for nearly two thirds. Improving the absorbance reading would gain almost nothing as long as the calibration curve stays what it is, which is the opposite of intuition, because the reading is what we have just taken and the curve dates from an hour ago.

The dilution multiplied by $\dfrac{100.00}{5.00}=20.0$, so $C_{river}=20.0\times 1.596=31.9$ mg/L. It adds its own relative uncertainties: $\dfrac{0.015}{5.00}\times 100=0.30\,\%$ and $\dfrac{0.08}{100.00}\times 100=0.08\,\%$, giving $6.8+0.38=7.2\,\%$ and $\Delta C=0.072\times 31.9=2.3$ mg/L. We publish $C=(32\pm 2)$ mg/L. The glassware added only $0.4$ percentage points to $7.2$: it is not the culprit, and that is a useful result, because it is the glassware that we fuss over by instinct.

Because the Beer-Lambert law stops being linear well before $A=1.20$. Above $A\approx 1$, stray light in the instrument, the association of coloured molecules with one another and the change in refractive index of the solution bend the relationship downward: the extended line UNDERESTIMATES the concentration, and all the more the further you go. So it is not only that we have left the calibrated interval, it is that the law itself has changed shape, and adding more concentrated standards would save nothing. The correct move: dilute the sample by a known factor, read again, and multiply. A one-in-five dilution would bring $A$ to about $0.24$, right in the calibrated zone.

$\dfrac{0.008}{0.012}\times 100=67\,\%$: the result is worthless. The absolute uncertainty on the absorbance does not depend on the signal, it is always $0.008$; so it is the RELATIVE uncertainty that explodes as the signal approaches zero. At the other end, above $A=1$, it is linearity that is lost. In between, the relative precision is best near the middle of the range. The working rule that follows, and that is applied without thinking in the lab: bring the sample, by dilution or by concentration, to the MIDDLE of the calibration curve, typically around $A=0.3$ to $0.7$, instead of reading it where it happens to be.

We read a decay that flattens out, so a rate that falls as the reagent becomes scarce, and we can rule out zero order, whose integrated law $[A]=[A]_{0}-kt$ would give a descending STRAIGHT LINE. That is all. First order and second order both produce a curved, flattening decay, and by eye they cannot be told apart. This plot therefore settles nothing, and that is precisely why we linearize: we do not compare two curves, we compare two ALIGNMENTS, which a human eye can judge and a ruler can check.

First order: $r^{2}=\dfrac{(-63.26)^{2}}{7000\times 0.5814}=\dfrac{4001.8}{4069.8}=0.983$. Second order: $r^{2}=\dfrac{(2101.5)^{2}}{7000\times 631.19}=\dfrac{4416300}{4418330}=0.9995$. The second is thirty times closer to $1$ in terms of residual gap: $0.0005$ against $0.017$. The plot of $\dfrac{1}{[A]}$ against $t$ is the one that lines up, so the reaction is SECOND ORDER. The point of method: 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, not the value itself.

For second order, $\dfrac{1}{[A]}=\dfrac{1}{[A]_{0}}+kt$: the slope is $+k$. $k=\dfrac{S_{ty}}{S_{tt}}=\dfrac{2101.5}{7000}=0.300$ L/(mol$\cdot$min). The units: $\dfrac{1}{[A]}$ is in L/mol, divided by minutes, so L$\cdot$mol$^{-1}\cdot$min$^{-1}$, which are indeed those of a second-order rate constant. The y-intercept is $\bar y-k\bar t=35.00-0.300\times 50=19.99$ L/mol, so $[A]_{0}=\dfrac{1}{19.99}=0.0500$ mol/L: we recover the measured starting concentration, which confirms that the line was not forced.

What you read is a PATTERN: the residuals are positive at both ends and negative in the middle, $+\,-\,-\,-\,+\,+$. Purely random scatter would change sign without any order; this one draws a curve. Taking residual as observed minus fitted, the first-order line therefore runs above the data in the middle and below it at the ends, which is the signature of a straight line fitted to curved points. The coefficient of determination sees none of this: it adds up squares and loses the sign, so it cannot tell six random deviations from six deviations arranged in an arc. Looking at the residuals takes thirty seconds and answers the question that $r^{2}$ does not ask, the one about the MODEL.

For second order, $t_{1/2}=\dfrac{1}{k[A]_{0}}=\dfrac{1}{0.300\times 0.0500}=66.6$ min, which is consistent with the table, where $[A]$ is still $0.0262$ at $60$ min. To fall to $0.0100$ mol/L: $t=\dfrac{1}{k}\left(\dfrac{1}{0.0100}-\dfrac{1}{0.0500}\right)=\dfrac{100-20}{0.300}=266$ min. Four times the first half-life does not divide the concentration by sixteen, far from it: a second-order reaction drags on endlessly, its rate falling as the square of what remains.

$[H_{3}O^{+}]=10^{-\mathrm{pH}}=10^{-2.87}=1.349\times 10^{-3}$ mol/L.

An uncertainty does not pass through a logarithm as a percentage. From $\mathrm{pH}=-\log[H_{3}O^{+}]$ we get $\dfrac{\Delta[H_{3}O^{+}]}{[H_{3}O^{+}]}=\ln(10)\times\Delta\mathrm{pH}=2.303\times 0.02=0.046$, that is $4.6\,\%$. Absolute uncertainty: $0.046\times 1.349\times 10^{-3}=6.2\times 10^{-5}$, so $[H_{3}O^{+}]=(1.35\pm 0.06)\times 10^{-3}$ mol/L. Yet the relative uncertainty on the pH is $\dfrac{0.02}{2.87}\times 100=0.70\,\%$: seven tenths of a percent on one side, nearly five percent on the other. This is the central trap of the chapter, and it fits in one sentence: 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.

$K_{a}=\dfrac{(1.349\times 10^{-3})^{2}}{0.100-0.001349}=\dfrac{1.820\times 10^{-6}}{0.09865}=1.845\times 10^{-5}$. The hydronium concentration appears SQUARED, so its relative uncertainty is doubled: $2\times 4.6=9.2\,\%$, plus the $0.5\,\%$ of the prepared solution, giving $9.7\,\%$. Absolute uncertainty: $0.097\times 1.845\times 10^{-5}=1.8\times 10^{-6}$. Publishable result: $K_{a}=(1.8\pm 0.2)\times 10^{-5}$. Two figures on the pH gave two on the constant: the chain gained nothing and lost nothing, but the illusion of the calculator's four digits very nearly had us write $1.845\times 10^{-5}$.

$\mathrm{p}K_{a}=-\log(1.845\times 10^{-5})=4.734$. Going back into the logarithm divides the relative uncertainty by $\ln 10$: $\Delta\mathrm{p}K_{a}=\dfrac{0.097}{2.303}=0.042$, so $\mathrm{p}K_{a}=4.73\pm 0.04$. The reference $1.8\times 10^{-5}$ corresponds to $\mathrm{p}K_{a}=4.74$, a deviation of $0.01$, a quarter of the uncertainty: perfectly compatible. In relative terms on $K_{a}$, the deviation is $2.5\,\%$ for an uncertainty of $9.7\,\%$, which says the same thing. Note in passing that $\Delta\mathrm{p}K_{a}=0.042$ is twice $\Delta\mathrm{pH}=0.02$: the squared numerator comes through intact after the round trip through the logarithms.

TWO. In a logarithm, only the MANTISSA, the part after the decimal point, carries significant figures; the characteristic, here the $2$, merely gives the power of ten, exactly as the exponent of a scientific notation carries none. The mantissa $87$ has two digits, so $[H_{3}O^{+}]=1.3\times 10^{-3}$ mol/L and not one digit more, which the $4.6\,\%$ uncertainty confirms independently. The practical consequence is brutal: to obtain a concentration to three significant figures you would need a pH to three decimal places, $2.870$, that is, a pH meter and a calibration of a different class from that of a teaching laboratory.

$\mathrm{pOH}=14.00-12.35=1.65$, so $[OH^{-}]=10^{-1.65}=2.239\times 10^{-2}$ mol/L. Each formula unit dissolved releases TWO hydroxide ions, so $s=\dfrac{[OH^{-}]}{2}=1.119\times 10^{-2}$ mol/L. The factor $2$ here is the first trap, and it has nothing to do with uncertainties: it comes from the dissolution equation, which is written before any calculation.

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

The solubility is half of $[OH^{-}]$, and dividing by an exact number does not change the relative uncertainty: $\ln(10)\times 0.02=4.6\,\%$, the same as on $[OH^{-}]$. The solubility product, for its part, goes as $s^{3}$: a power MULTIPLIES the relative uncertainty by its exponent, so $3\times 4.6=13.8\,\%$. The $4$ is exact and contributes nothing. Absolute uncertainty: $0.138\times 5.61\times 10^{-6}=7.8\times 10^{-7}$. We publish $K_{sp}=(5.6\pm 0.8)\times 10^{-6}$. Two hundredths of a pH unit have become fourteen percent on the constant, through nothing more than the composition of a logarithm and a cube.

$\dfrac{|5.61-5.50|}{5.50}\times 100=2.0\,\%$. In number of uncertainties: $\dfrac{0.11\times 10^{-6}}{0.78\times 10^{-6}}=0.14$. The deviation fits within a seventh of the uncertainty: the measurement is fully compatible with the reference, and nothing more can be asserted. It would be wrong to congratulate ourselves on the two percent as if it were a performance: with an error bar of fourteen percent, the experiment could just as well have come out at $5.0$ or $6.2\times 10^{-6}$ with nothing to criticize.

We would need $3\times\ln(10)\times\Delta\mathrm{pH}=0.05$, that is $\Delta\mathrm{pH}=\dfrac{0.05}{6.91}=0.007$. No teaching-laboratory pH meter holds seven thousandths of a pH unit, if only because the calibration buffers themselves are given at $\pm 0.01$. And repeating would change nothing: the $\pm 0.02$ is the uncertainty of the electrode's CALIBRATION, hence a systematic error that shifts all the readings the same way. Repeating ten times would give ten almost identical values, all shifted alike, with the reassuring illusion of excellent precision thrown in. The only remedy is a change of method: titrating the filtrate with hydrochloric acid from a burette gives the same solubility with an uncertainty ten times smaller, because the burette passes through no logarithm.

FALSE. The coefficient of determination measures how well the points LINE UP around the fitted line, nothing else. Over a narrow interval, a slightly curved line easily gives $0.998$, and that is exactly what happens in exercise 5, where first order shows $0.983$ while being the wrong model. Two moves are worth more than reading a number: compare the $r^{2}$ with that of the COMPETING model, comparing the gaps from $1$ and not the values, and look at the RESIDUALS, where an arc pattern exposes the curvature that $r^{2}$ cannot see. The correct sentence: a coefficient of determination close to $1$ says the points are aligned, it says neither that the model is right nor by how much the slope can move.