Chemist: MQL recipes | MentDB Weak

Recipes

Prepare a stock solution and dilute it
pH of a weak acid without the approximation
Amount of gas in a pressurized cylinder
Equivalence point of a weak acid titration
Balancing a redox equation with linear algebra
Isotope pattern of a tetrachloro compound in MS
Lattice parameter, X-ray density and crystallite size from XRD
Method bias on a certified reference material
Two dyes in one drink by least squares (QR)
Phosphate buffer with activity correction (Davies)

executed on MentDB

Prepare a stock solution and dilute it

A lab technician must prepare 250 mL of 0.150 mol/L NaCl, then 100 mL at 0.010 mol/L from it. The script gives the mass to weigh (n = C.V, m = n.M) and the stock volume to pipette (C1V1 = C2V2), with the units checked.

Pipeline

01 · massMass m = C.V.M in gunit calc→02 · dilutionStock volume C2V2/C1 in mLunit calc→03 · reportRounded instructionsmath round

Script

admin

#1. Mass of NaCl to weigh for 250 mL at 0.150 mol/L;
-> "[m]" (unit calc "0.150 mol/L * 250 mL * 58.44 g/mol" "g");
#2. Volume of this stock for 100 mL at 0.010 mol/L, C1V1 = C2V2;
-> "[v1]" (unit calc "0.010 mol/L * 100 mL / (0.150 mol/L)" "mL");
#3. Report;
concat "Weigh " (math round [m] 3) " g of NaCl into a 250 mL flask, then pipette " (math round [v1] 2) " mL of it and make up to 100 mL."

Result

mentdb

Weigh 2.192 g of NaCl into a 250 mL flask, then pipette 6.67 mL of it and make up to 100 mL.

Go further

Propagate the balance and flask tolerances with num uncertainty.
Use unit calc with mg/L or ppm-like ratios for trace standards.

executed on MentDB

pH of a weak acid without the approximation

What is the pH of 0.10 mol/L acetic acid (Ka = 1.8e-5)? The script solves the exact equilibrium x^2/(C - x) = Ka, compares it with the classic sqrt(Ka.C) shortcut and gives the degree of dissociation.

Pipeline

01 · equilibriumSolve x^2 - Ka(C - x) = 0num solve→02 · phpH = -log10[H+]math calc→03 · reportExact vs approximateconcat

Script

admin

#1. Acetic acid 0.10 mol/L, Ka = 1.8e-5;
-> "[c]" 0.10; -> "[ka]" 1.8e-5;
#2. Exact [H+]: x^2 / (c - x) = Ka;
-> "[h]" (num solve (concat "x^2 - " [ka] "*(" [c] " - x)") "x" 0 [c]);
#3. pH exact, pH with the approximation sqrt Ka*c, degree of dissociation;
-> "[ph]" (math calc "-log10([h])");
-> "[ph_approx]" (math calc "-log10(sqrt([ka]*[c]))");
concat "pH = " (math round [ph] 3) " (approx. " (math round [ph_approx] 3) "), alpha = " (math round (math calc "100*[h]/[c]") 2) " %"

Result

mentdb

pH = 2.875 (approx. 2.872), alpha = 1.33 %

Go further

Add the water autoprotolysis for very dilute acids with num solve_system.
Plot the titration curve with num table over the added base volume.

executed on MentDB

Amount of gas in a pressurized cylinder

How much nitrogen is in a 50 L cylinder at 200 bar and 20 degC? The script compares the ideal gas law n = PV/RT with the van der Waals equation (a = 0.137 Pa.m6/mol2, b = 3.87e-5 m3/mol) solved for n, and gives the mass in kg.

Pipeline

01 · idealn = PV/RT with unitsunit calc→02 · realVan der Waals solved for nnum solve→03 · massMass of gasmath calc

Script

admin

#1. Nitrogen cylinder: 50 L at 200 bar and 20 degC;
-> "[n_ideal]" (unit calc "200 bar * 50 L / (_R * 293.15 K)" "mol");
#2. Van der Waals for N2: a = 0.137 Pa.m6/mol2, b = 3.87e-5 m3/mol, SI units;
-> "[n_vdw]" (num solve "(200e5 + 0.137*n^2/0.05^2)*(0.05 - n*3.87e-5) - n*8.314462618*293.15" "n" 1 600);
#3. Mass of gas in kg, M = 28.014 g/mol;
concat "Ideal gas: " (math round [n_ideal] 1) " mol, van der Waals: " (math round [n_vdw] 1) " mol = " (math round (math calc "[n_vdw]*28.014/1000") 2) " kg of N2"

Result

mentdb

Ideal gas: 410.3 mol, van der Waals: 408.9 mol = 11.45 kg of N2

Go further

Compute the compressibility factor Z = PV/nRT from both results.
Get R, Vm or NA with their CODATA metadata using const get.

executed on MentDB

Equivalence point of a weak acid titration

Acetic acid titrated by 0.100 mol/L NaOH: the equivalence volume is the steepest point of the pH curve. The script differentiates a monotone interpolation on a fine grid, takes the maximum of the slope, reads the pKa at half equivalence and gives the acid concentration.

Pipeline

01 · gridFine grid of volumesnum table→02 · slopedpH/dV of the interpolantnum interpolate_derivative→03 · maxSteepest pointdsp peaks→04 · pKapH at half equivalencenum interpolate

Script

admin

#1. 25.0 mL of acetic acid titrated with NaOH 0.100 mol/L: volume in mL and pH;
-> "[V]" "[0, 2, 4, 6, 8, 9, 9.5, 9.8, 10, 10.2, 10.5, 11, 12, 14]";
-> "[pH]" "[2.88, 4.15, 4.57, 4.93, 5.35, 5.70, 6.03, 6.45, 8.72, 10.98, 11.38, 11.68, 11.96, 12.22]";
#2. Slope dpH/dV of the monotone interpolation on a fine grid from 9 to 11 mL;
-> "[g]" (matrix col (num table "v" "v" 9 11 0.01) 1);
-> "[d]" (num interpolate_derivative [V] [pH] [g] "pchip");
#3. Equivalence volume at the steepest point;
json load "rx_eq" (dsp peaks [d] "{\"distance\": 1000}");
-> "[Veq]" (math calc (concat "9+0.01*" (json select "rx_eq" /indices[0])));
#4. pKa = pH at half equivalence, and the acid concentration;
-> "[pKa]" (num interpolate [V] [pH] (math calc "[Veq]/2") "pchip");
concat "Veq = " (math round [Veq] 2) " mL, max slope = " (math round (json select "rx_eq" /values[0]) 1) " pH/mL, pKa = " (math round [pKa] 2) ", c acid = " (math round (math calc "0.100*[Veq]/25.0") 4) " mol/L"

Result

mentdb

Veq = 9.93 mL, max slope = 14.3 pH/mL, pKa = 4.74, c acid = 0.0397 mol/L

Go further

Draw the curve and its interpolation with plot interpolate [V] [pH] "pchip".
The second derivative (order 2 of num interpolate_derivative) crosses zero at the equivalence point.

executed on MentDB

Balancing a redox equation with linear algebra

Balance KMnO4 + HCl -> KCl + MnCl2 + Cl2 + H2O. Each element gives one equation; the reduced row echelon form of the composition matrix gives the null space, and the lcm of the denominators gives the smallest integer coefficients.

Pipeline

01 · matrixElement rows, species columnsmatrix rref→02 · fractionsCoefficients as exact fractionsarith to_fraction→03 · integerslcm of the denominatorsarith lcm→04 · checkRank = species - 1matrix rank

Script

admin

#1. Permanganate oxidation of hydrochloric acid: a KMnO4 + b HCl -> c KCl + d MnCl2 + e Cl2 + f H2O;
#Rows K, Mn, O, H, Cl, one column per species, products negative;
-> "[A]" "[[1, 0, -1, 0, 0, 0], [1, 0, 0, -1, 0, 0], [4, 0, 0, 0, 0, -1], [0, 1, 0, 0, 0, -2], [0, 1, -1, -2, -2, 0]]";
json load "rx_tu1_sp" (string split "KMnO4 HCl KCl MnCl2 Cl2" " " -1);
#2. Reduced row echelon form: rank 5, one-dimensional null space, coefficients relative to f = 1;
-> "[R]" (matrix rref [A]);
-> "[co]" ""; -> "[l]" 1;
for (-> "[i]" 1) (<= [i] 5) (++ "[i]") {
	-> "[q]" (arith to_fraction (math calc (concat "-1*" (matrix get [R] [i] 6))));
	-> "[co]" (concat [co] [q] " ");
	#3. Least common multiple of the denominators;
	-> "[l]" (arith lcm [l] (string regexp_extract (concat [q] "/1") "/([0-9]+)" 1));
};
#4. Smallest integer coefficients;
-> "[eq]" "";
for (-> "[i]" 1) (<= [i] 5) (++ "[i]") {
	-> "[eq]" (concat [eq] (arith frac_mul (arith to_fraction (math calc (concat "-1*" (matrix get [R] [i] 6)))) [l]) " " (json select "rx_tu1_sp" (concat "/[" (- [i] 1) "]")));
	if (== [i] 2) {-> "[eq]" (concat [eq] " -> ");} {-> "[eq]" (concat [eq] " + ");};
};
concat "rank " (matrix rank [A]) ", ratios to H2O: " [co] "=> " [eq] [l] " H2O"

Result

mentdb

rank 5, ratios to H2O: 1/4 2 1/4 1/4 5/8 => 2 KMnO4 + 16 HCl -> 2 KCl + 2 MnCl2 + 5 Cl2 + 8 H2O

Go further

Check the electron balance: Mn +7 -> +2 gains 5, Cl -1 -> 0 loses 1, so 2 Mn for 10 Cl
Try an equation with two independent reactions: the rank drops by 2 and the null space has 2 vectors

executed on MentDB

Isotope pattern of a tetrachloro compound in MS

Predict the relative intensities of the M, M+1, M+2, M+4 and M+6 peaks of C6H2Cl4 from the binomial law of the chlorine isotopes and the 13C abundance, to recognise the Cl4 signature in a mass spectrum.

Pipeline

01 · binomialC n,k from Gammamath gamma→02 · peaksM+2k intensitiesmath calc→03 · carbon13C contribution by log-Gammamath lngamma→04 · reportIsotope patternconcat

Script

admin

#1. Molecular ion of a tetrachlorobenzene C6H2Cl4: 4 chlorine atoms, abundance of 37Cl = 0.2424;
-> "[n]" 4; -> "[p]" 0.2424;
#2. Binomial coefficients from the Gamma function: C n,k = n! / k!.n-k!;
-> "[c1]" (math calc (concat (math gamma 5) "/(" (math gamma 2) "*" (math gamma 4) ")"));
-> "[c2]" (math calc (concat (math gamma 5) "/(" (math gamma 3) "*" (math gamma 3) ")"));
#3. Peaks M, M+2, M+4 relative to M: C n,k.p^k.1-p^-k;
-> "[m2]" (math calc "100*[c1]*([p]/(1-[p]))");
-> "[m4]" (math calc "100*[c2]*([p]/(1-[p]))^2");
-> "[m6]" (math calc "100*4*([p]/(1-[p]))^3");
#4. 13C contribution to M+1 for 6 carbons, by log-Gamma to avoid big factorials;
-> "[m1]" (math calc (concat "100*exp(" (math lngamma 7) "-" (math lngamma 6) "-" (math lngamma 2) ")*0.0107/0.9893"));
concat "M = 100, M+1 = " (math round [m1] 1) ", M+2 = " (math round [m2] 1) ", M+4 = " (math round [m4] 1) ", M+6 = " (math round [m6] 1) " : the 100 / 128 / 61 pattern is the signature of Cl4"

Result

mentdb

M = 100, M+1 = 6.5, M+2 = 128, M+4 = 61.4, M+6 = 13.1 : the 100 / 128 / 61 pattern is the signature of Cl4

Go further

Combine Cl and Br patterns by convolution with num poly_mul.
For a big molecule like a peptide, use math lngamma for all the binomial terms.

executed on MentDB

Lattice parameter, X-ray density and crystallite size from XRD

From the 200 reflection of NaCl, get the d-spacing (Bragg), the cubic cell edge, the X-ray density, the cell edge implied by a pycnometer density, and the crystallite size by the Scherrer equation.

Pipeline

01 · braggd = lambda / 2.sin thetamath sin→02 · cella = d.sqrt h2+k2+l2math hypot→03 · densityrho = Z.M / NA.a^3unit calc→04 · inversea from a measured densitymath cbrt→05 · scherrerK.lambda.sec theta / betamath sec

Script

admin

#1. Powder XRD of NaCl with Cu K-alpha 1.5406 angstrom: 200 reflection at 2theta = 31.69 deg, FWHM 0.18 deg;
-> "[lam]" 1.5406; -> "[th]" (math calc "31.69/2*pi/180"); -> "[beta]" (math calc "0.18*pi/180");
#2. Bragg law d = lambda/2.sin theta, cubic cell a = d.sqrt of h2+k2+l2;
-> "[d]" (math calc (concat "[lam]/(2*" (math sin [th]) ")"));
-> "[a]" (math calc (concat "[d]*" (math hypot (math hypot 2 0) 0)));
#3. X-ray density: Z = 4 NaCl per cell, M = 58.44 g/mol;
-> "[rho]" (unit calc (concat "4 * 58.44 g/mol / (_NA * (" [a] " angstrom)^3)") "g/cm^3");
#4. Cell edge that a pycnometer density of 2.160 g/cm3 would imply, by the cube root;
-> "[a2]" (math cbrt (unit calc "4 * 58.44 g/mol / (_NA * 2.160 g/cm^3)" "angstrom^3"));
#5. Scherrer crystallite size K.lambda.sec theta / beta, K = 0.9;
-> "[size]" (math calc (concat "0.9*[lam]*" (math sec [th]) "/[beta]/10"));
concat "d200 = " (math round [d] 4) " A, a = " (math round [a] 4) " A, X-ray density = " (math round [rho] 3) " g/cm3, a from the pycnometer = " (math round [a2] 4) " A, crystallite size = " (math round [size] 1) " nm"

Result

mentdb

d200 = 2.8212 A, a = 5.6425 A, X-ray density = 2.161 g/cm3, a from the pycnometer = 5.6432 A, crystallite size = 45.9 nm

Go further

Refine a on several reflections with a Nelson-Riley extrapolation and stat fit.
Subtract the instrumental broadening before Scherrer.

executed on MentDB

Method bias on a certified reference material

Check an ICP-MS method on a certified soil: Dixon Q test on a suspect replicate, then z-test of the mean against the certified value with the known method sigma, and recovery.

Pipeline

01 · sortSorted replicatesmath order_double→02 · dixonQ = gap / rangestat range→03 · biasz-test vs certified valuestat ztest→04 · recoveryMean / certifiedstat mean

Script

admin

#1. Lead in a certified soil, ICP-MS, 7 replicates in mg/kg, certified 25.3 and method sigma 0.6;
-> "[x]" "[25.9, 26.1, 25.6, 27.9, 26.0, 25.8, 26.3]";
#2. Dixon Q test on the suspect high value, on the sorted series;
-> "[s]" (math order_double [x]);
-> "[q]" (math calc (concat "(" (matrix get (concat "[" [s] "]") 1 7) "-" (matrix get (concat "[" [s] "]") 1 6) ")/(" (stat range [x]) ")"));
#3. Q critical at 95 % for n = 7 is 0.568: drop the outlier, then z-test of the bias with the known sigma;
-> "[x6]" "[25.9, 26.1, 25.6, 26.0, 25.8, 26.3]";
json load "rx_p4c_crm" (stat ztest [x6] 25.3 0.6);
#4. Recovery;
concat "Q = " (math round [q] 3) " > 0.568: 27.9 rejected; mean = " (math round (stat mean [x6]) 2) " mg/kg, recovery = " (math round (math calc (concat "100*" (stat mean [x6]) "/25.3")) 1) " %, z = " (math round (json select "rx_p4c_crm" /z) 2) ", p = " (math round (json select "rx_p4c_crm" /p_value) 4) " : significant bias"

Result

mentdb

Q = 0.696 > 0.568: 27.9 rejected; mean = 25.95 mg/kg, recovery = 102.6 %, z = 2.65, p = 0.008 : significant bias

Go further

Use the En number with the uncertainty of the certificate instead of a z-test.
Use stat shapiro to check the normality before the test.

executed on MentDB

Two dyes in one drink by least squares (QR)

Determine two food dyes from absorbances at four wavelengths: solve the overdetermined Beer-Lambert system by the QR decomposition and read the residual norm of the fit directly in Q^T.A.

Pipeline

01 · qrQ.R = matrix of absorptivitiesmatrix qr→02 · projectQ^T.Amatrix mul→03 · backSolve the triangular 2x2 blockmatrix solve→04 · residualNorm of the last componentsvector norm

Script

admin

#1. Absorbance of a drink with two dyes at 430, 480, 580, 630 nm, 1 cm cell;
-> "[e]" "[[27, 1], [15, 2.5], [0.3, 40], [0.05, 120]]";
-> "[b]" "[[0.545], [0.314], [0.205], [0.600]]";
#2. Overdetermined Beer-Lambert system E.c = A by QR: Q.R = E, molar absorptivities in L/mmol/cm;
json load "rx_p4c_qr" (matrix qr [e]);
-> "[y]" (matrix mul (matrix transpose (json select "rx_p4c_qr" /Q)) [b]);
#3. Back substitution with the upper 2x2 block of R;
-> "[r]" (concat "[[" (json select "rx_p4c_qr" /R/[0]/[0]) ", " (json select "rx_p4c_qr" /R/[0]/[1]) "], [0, " (json select "rx_p4c_qr" /R/[1]/[1]) "]]");
-> "[c]" (matrix solve [r] (concat "[" (matrix get [y] 1 1) ", " (matrix get [y] 2 1) "]"));
#4. The last 2 components of Q^T.A give the residual norm of the fit;
-> "[res]" (vector norm (concat "[" (matrix get [y] 3 1) ", " (matrix get [y] 4 1) "]"));
concat "tartrazine = " (math round (math calc (concat "1000*" (matrix get (concat "[" [c] "]") 1 1))) 2) " umol/L, brilliant blue = " (math round (math calc (concat "1000*" (matrix get (concat "[" [c] "]") 1 2))) 2) " umol/L, residual norm = " (math round [res] 4) " AU"

Result

mentdb

tartrazine = 20.02 umol/L, brilliant blue = 4.99 umol/L, residual norm = 0.0015 AU

Go further

Use the whole spectrum (100 wavelengths) for a classical least squares calibration.
Compare with matrix pinv and with the normal equations.

executed on MentDB

Phosphate buffer with activity correction (Davies)

Prepare 1 L of 0.100 M phosphate buffer at pH 7.40: solve the HPO4/H2PO4 ratio with the Davies activity coefficients (the ionic strength depends on the ratio) and give the masses to weigh.

Pipeline

01 · modelDavies log gamma in the pH equationstring replace→02 · solveRatio r with I = f of rnum solve→03 · idealRatio without activitymath power→04 · massesMasses of the two saltsmath calc

Script

admin

#1. Phosphate buffer 0.100 M total at pH 7.40 and 25 degC, pKa2 = 7.198 at zero ionic strength;
-> "[ct]" 0.100; -> "[pka]" 7.198;
#2. Davies equation log gamma = -0.509.z^2.sqrt I / 1+sqrt I - 0.3 I, NaH2PO4 + Na2HPO4: I = ct + 2.base;
-> "[lg]" "(-0.509*(sqrt(I)/(1+sqrt(I))-0.3*I))";
-> "[ph]" (concat "7.198 + log10(r) + 4*LG - LG - 7.40");
#3. Solve the base/acid ratio r with I = 0.1 + 2.0.1.r/1+r;
-> "[eq]" (string replace (string replace [ph] "LG" [lg]) "I" "(0.1+0.2*r/(1+r))");
-> "[r]" (num solve [eq] "r" 0.1 10);
#4. Masses for 1 L: NaH2PO4.H2O 137.99 g/mol, Na2HPO4.7H2O 268.07 g/mol;
-> "[base]" (math calc "[ct]*[r]/(1+[r])");
concat "ratio HPO4/H2PO4 = " (math round [r] 3) " instead of " (math round (math power 10 (math calc "7.40-[pka]")) 3) " without activity; I = " (math round (math calc "0.1+2*[base]") 3) " M; weigh " (math round (math calc "([ct]-[base])*137.99") 2) " g NaH2PO4.H2O and " (math round (math calc "[base]*268.07") 2) " g Na2HPO4.7H2O"

Result

mentdb

ratio HPO4/H2PO4 = 3.968 instead of 1.592 without activity; I = 0.26 M; weigh 2.78 g NaH2PO4.H2O and 21.41 g Na2HPO4.7H2O

Go further

Add the temperature dependence of pKa2 (dpKa/dT = -0.0028 /K).
Check the buffer capacity beta = 2.303.C.Ka.H / Ka+H ^2.

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