how to find standard deviation chemistry

Chemistry 223

Problem Set 2 Answer Key

This problem set deals with statistics.

1. An enterprising student carried out 12 replica titrations of 25.00 mL of 0.1000 M potassium hydrogen phthalate (KHP) with NaOH of unknown molarity. Part way through the series of titrations, the student went to talk to the TA. Spotting this opportunity for mischief, those villains from"Bullwinkle", Boris and Natasha, ran into the lab and exhaled into the NaOH, adding CO₂ to the base and thus reducing its concentration. They disappeared before the student returned. During one of the titrations, the student sneezed just as the end point was being reached, and overshot by quite a bit. Given below are the volumes of NaOH used to titrate the KHP. Your mission: on a statistically sound basis, determine after which titration the CO₂ was added, discard the titration that was ruined by the sneeze, and determine the molarity of the NaOH, both before and after the tampering incident. Report both means and standard deviations, and use the Q and t tests to establish confidence in your answers

Titration number	Volume NaOH

1	27.63
2	27.60
3	27.64
4	27.64
5	27.77
6	27.63
7	27.60
8	27.64
9	27.95
10	27.93
11	27.91
12	27.91

First, find the mean and standard deviation for the entire data set. Sum all data and divide by 12 gives mean = 332.85/12 = 27.7375 ml (ignoring significant figures). We'll round this off once we get the standard deviation. If you just plug into your spreadsheet, you'll find TWO standard deviation functions: @std and @stds. @std gives which is a reasonable approximation to the standard deviation of the population for large N. But 12 is not "large;" we should use the standard deviation of the sample, @stds = . Either by calculator or by spreadsheet, we find the standard deviation (to too many significant figures) is 0.1453 ml. So the correct mean and standard deviation are 27.74

0.15 ml. NaOH molarity would appear to be 25 mL * 0.1000 M/27.74 mL = 0.0901 M

0.0005 M.

But visual inspection shows there's a change in behavior for samples 9-12 vs. 1-8 (with 5 being the likely sneeze-perturbed example). Looking at just the last 4 samples gives mean volume = 27.925

0.019 mL. NaOH = 25*0.1000/27.925 = 0.08952 M + 0.00006 M. For the first 8 samples:

Samples	Mean volume	Standard Deviation	Range	Concentration
1-8	27.644 mL	0.054 mL	27.60 - 27.77	0.09043 M + 0.00018 M
1-8 excluding 5	27.626 mL	0.018 mL	27.60 - 27.64	0.09049 M + 0.00006 M

With all these choices, we are now ready to use the Q and t tests to see what's really going on. Let's do the easy one first -- the Q test to see if the sneeze gave a statistically significantly different answer. We can use volumes rather than concentrations (since they're proportional to each other). Recall Q = gap/range. The "gap" between the ostensible sneeze-altered end point (27.77 mL) and the nearest value (27.64 mL) is 27.77-27.64 = 0.13 mL. The range is from low (27.60 mL) to high (27.77 mL) or 27.77-27.60 = 0.17 mL Q = 0.13/0.17 = 0.765. Recall that if this ratio exceeds Q in Table 4-5 in Harris, then at greater than 90% confidence we can exclude it as an aberration which, in this case, is of known cause. For 8 observations, Q₉₀ = 0.47. Our computed ratio is nearly twice this, so with much greater than 90% confidence, observation #5 is an outlier.

So for the rest of our calculations, we will ignore measurement #5 (but for consistency, I won't change the labeling of each measurement). We can see by eyeball that the CO₂ was added between measurements 8 and 9. To prove this, we do two things: a) show that we get minimum standard deviation if we break the data set between [8 and 9], not between [7 and 8] or [9 and 10]. Looking at volumes, we get:

Break between measurements	Mean and standard deviation of measurements before break (excluding #5)	Mean and standard deviation of measurements after break
7 and 8	27.623 + 0.019 mL	27.868 + 0.129 mL
8 and 9 (already computed)	27.626 + 0.018 mL	27.925 + 0.019 mL
9 and 10	27.666 + 0.116 mL	27.917 + 0.012mL

Note that for BOTH "before" and "after" data subsets to have standard deviations below 0.1 or even 0.02 mL, the break point must be between 8 and 9. Now -- can we prove that the two titration volumes are statistically different? We have sample means and standard deviations for both data sets.

2. Later this semester, we'll discuss spectrophotometry. For now, it's enough to know that the absorbance (how much light is absorbed in passing through a solution) is proportional to the concentration of whatever is absorbing the light. There is ALWAYS a blank (even if it's tiny) in absorbance spectrophotometry, so for practical determinations of single species, the absorption and the concentration of the species of interest are related by (cut and paste from Lecture 3):

For all practical purposes, the standard deviations are the same for both data sets, so the left column applies. Pre-CO₂ addition, we're using 7 measurements, while afterwards, we're using 4. s_pooled = ((0.018²(7-1)+0.019²(4-1))/(7+4-2))^1/2 = ((0.018²*6+0.019²*3)/9)^1/2 =0.018. t_calc=((27.925-27.626)/0.018)(7*4/(7+4))^1/2 = 0.299/0.018*(28/11)^1/2. 16.6*1.595 = 26.5. So the difference in means corresponds to a t_calc of 26.5. We have 9 degrees of freedom (using 11 data, have computed 2 independent means, leaving 11-2 = 9 unused degrees of freedom). From Table 4-2, we find that even at the 99.9% confidence level, for 9 degrees of freedom t = 4.781. Thus, with greater than 99.9o% confidence, we can say that the addition of CO₂ changed the concentration. The concentrations, then are (from the tables above) 0.09049 M

0.00006 M before the CO₂ addition, and 0.08952 M + 0.00006 M thereafter.

where A is absorption (the dependent variable you measure), C is concentration (known for the standards, not known for your unknown), and b is the blank. We'll talk about what k is when we talk about spectrometry; for now it's OK to simply regard it as a fitting constant. Here are some standardization data and some data on an unknown. Find the concentration of the unknown and provide an uncertainty for that determination.

Standard Concentration	Absorption
(micromolar)

blank	0.0589
20	0.2581
40	0.3732
60	0.4753
80	0.5856
100	0.6972
120	0.8126

Unknown absorption
0.3012
0.2997
0.2981
0.3023

This is a least squares analysis and propagation of error problem. First, we fit the data for the calibration curve. There are 7 data (blank through 120 micromolar). The variable is concentration; the dependent variable is absorbance. So the needed sums are:

So the fitted line is A = 5.985*10^-3+0.1067. Let's do a sanity check and plot raw data, the fitted line, and the differences between fitted and observed data:

Note that if we simply took the data point at nominal 0 micromolar as the intercept, we'd drastically mis-fit the rest of the line. One may ask if there's a systematic error at low concentration, but the available data give no clue on this. We need to know the uncertainty in both slope and intercept. From the lecture notes, , . The d_i values are exactly those shown by the blue line in the graph above. Let's show a table:

	Standard
	Concentration	Absorption						Rounded
	(micromolar)		C^2	C*A	Fitted A	d	d^2	fit

	0	0.0589	0	0	0.1067	-0.0478	0.002285	0.11
	20	0.2581	400	5.162	0.2264	0.0317	0.00100489	0.23
	40	0.3732	1600	14.928	0.3461	0.0271	0.00073441	0.35
	60	0.4753	3600	28.518	0.4658	0.00950	9.0250E-05	0.47
	80	0.5856	6400	46.848	0.5855	1.0E-04	1.0E-08	0.59
	100	0.6972	10000	69.72	0.7052	-0.0080	6.4E-05	0.71
	120	0.8126	14400	97.512	0.8249	-0.0123	0.00015129	0.83

Sum	420	3.2609	36400	262.688	3.2606		0.00432969

	D	78400		rounded	sigma
	b	0.1067		0.11	0.02
	k	0.005985		0.006	0.0003

	sigma b	0.0201
	sigma k	0.000278

So now we can figure out what concentration the unknown absorbance corresponds to and evaluate the uncertainty in that measured concentration. For the four measurements of the unknown absorption, mean and standard deviation are 0.3003 + 0.0018. So while we can measure absorbance to 1 part in 165 roughly (0.3003/0.0018), we only know slope and intercept to 2 and 1 significant figures each! That means we're not going to be able to know the unknown's concentration as well as the absorbance precision might lead us to believe. IF you solve this flat-footedly, 0.3003 = 0.005985C + 0.1067, and C =32.3+0.2 micromolar. Doing a proper propagation of error analysis,

The uncertainties in slope and intercept dominate the uncertainties. This is an absolutely CRITICAL understanding -- the precision of the concentration measurement is NOT limited by the precision of the absorption measurement in this case -- it's by the fit between the line and the data! This is not always the case, but it sure is here. Specifically

Std. dev. = ((0.0018²+0.02²)/0.0060² +0.0003² 32.3²/0.006²)^1/2. Uncertainty in concentration is 3.7 micromolar. So the correct statement is 32 + 4 micromolar. That's an uncertainty of 1 part in 8. At the upper end of the working curve, the uncertainty is ((0.0018²+0.02²)/0.0060² +0.0003² 120²/0.006²)^1/2 = 6.9 micromolar, an uncertainty of only 1 part in 17.