Review of Fundamental Concepts(Gary Christian, hal 65)

Atomic, and Molecular WeightsAtomic weight (BA)for any elements is the weight of a specified number of atoms of that element, and that number is the same from one to another.Examples: Ca 40.08, S 32.06, O 16.00Molecular weight (BM) is the sum of formula weight of the atoms that make up compound.Example: Calculate the number of grams in one mole of CaSO4.7H2O = Ca + S + 11xO + 14xH = 40.08+32.06+11x16.00+14x1.09= 262.25

Moles, Molarity, NormalityMole = grams/formula weightMolarity = mole/liter1 mol dalam 1 liter larutan

Example: A solution is prepared by dissolving 1,26 g AgNO3 in a 250 ml volumetric flask and diluting to volume. Calculate the molarity of the silver nitrate solution. How many millimoles AgNO3 were dissolved?

Moles, Molarity, NormalityAnswer

Millimolesmillimole= M x milliliter= 0.0297x250= 7.42 mmole

Persen berat/berat% (wt/wt) = (berat zat terlarut) x 100% (berat sampel) = (g zat terlarut/g sampel) x 100 %

Common Units for Expressing Trace CalculationsParts per million (ppm) = (berat terlarut/berat sampel) x 106= mg/kg = mg/L = L/L = g/g = g/mL = nL/mLExample. A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 g zinc. What is the concentration of zinc in the plant in ppm

Answer :berat zat 3,6 ugberat sampel 2,6 gramkonsentrasi zinc dlm sampel = 3.6 g/2.6 g atau = (3.6 x 10-6 /2.6) x 106 = 1.4 ppm

Satuan lainnya.... ppt (part per thousand) = (berat zat terlarut (g)/berat sampel (g)) x 103

= mg/g

= g/kg

ppb (part per billion)

= (berat zat terlarut/berat sampel) x 109

= ng/g

= ug/kg

Contoh soalBerapa mg sampel yang harus ditimbang jika ingin membuat 50 ppm dalam 100 ml.50 ppm = (x gram/100 ml) x 106 x = 50 x 100106= 5 x 10-3 gram= 5 mg

Atau ppm = ug/ml50 ppm = ..... g/100 ml50 ppm x 100 ml = 5000 g = 5 mg

Contoh soalHitung berapa Molar jika BM = 100 dengan konsentrasi 1 ppm

ppm = mg/L = 10-3 gram/LM = gram/BM/LM = 10-3/100/L = 10-5 mol/L = 10-5 M

SoalHitung :a. 1 ppm berapa persen?b. 1 ppb berapa persen? Dan berapa ppm?Jawab :1 ppm = 10-4 %1 ppb = 10-7 % = 10-3 ppm

General calculation with MolarityConsider the general reaction

a A + t T P

Where A is the analyte, T is the titrant, reacting in the ratio a/t to give products P, then

mmolA = mmolT x a tmmolA = MT x mlT x a tmgA = mmolT x BMA x a tmgA = MT x mlT x a x BMA t

Fraction..% A = Fraction analyte x 100 % = mganalyte x 100 % mgsample= mmol titrant x (a/t) x BM x 100 %mg sample= Mt x ml t x a/t x BM x 100 %mg sample

ExampleA 0,2638 gram soda ash sample is analyzed by titrating the sodium carbonate with the standard 0,1288 M hydrochloric solution, requiring 38, 27 ml. The reaction is CO32- + 2H+ ---- H2CO3 + CO

milimoles of sodium carbonate is equal to one-half the milimoles of acid used to titrate it, since they react in a 1:2 ratio (a/t = )

% = MT x mlT x a/t x BMA x 100%mg sample

NormalityMany substance do not react on a 1:1 mole basis.And so solution of equal molar concentration do not react on a 1:1 volume basis.-----consepts of equivalents and normality (N)N = number of equivalents of material per liter solution N = eq = meqL mL (eq) = Number of equivalent eq = mol x no. Of reacting units per molecule

NormalitasBanyaknya ekivalen (ek) zat terlarut tiap liter larutan, atauN = ek/V

ek = gram/BE BE = BM/n

N = (gram x n) / (BM x V)

Berat Ekivalen (BE)BE = berat molekul dibagi dengan valensi= BM/nCara penentuan valensi bergantung pada reaksi yang terjadi

Cara penentuan valensiReaksi asam basa, valensi ditentukan berdasarkan banyaknya H+ atau OH- yang dihasilkan tiap satu mol asam atau basaReaksi redoks, valensinya ditentukan o/ banyaknya elektron yang hilang atau timbul pada reaksi oksidasi reduksi

Contoh :

H3PO4 : 1 mol ekivalen dengan 3 mol ion H+Ca(OH)2 : valensinya 2I2 + 2e ----- 2I- maka valensinya = 2 sebab 1 mol ekivalen dengan 2 elektron

Contoh soalHitung berat ekivalen Na2C2O4 dan KMnO4 dlm reaksi redoks suasana asam

The Equivalent WeightBerat ekivalen sama dengan berat molekul dibagi dengan valensiFor example :HCl, eq. Wt (BE) = the formula weight

H2SO4 it takes only one-half the number of molecules to furnish one mole of H+ , so eq.wt = one half the formula weight

eq.wt H2SO4 = f.wt 2

eq = gram eq. Wt (BE)

N = eq = gram/BE L L

T

Example: Calculate the equivalent weights f the following substances: (a) NH3 (b) H2C2O4 (c) KMnO4 (MnO4- is reduced to Mn2+)Solutioneq. wt. = NH3/1 =17.03/1Eq. wt. = 90.04/2 = 45.02The Mn goes a five electron change, from valence +7 to +2:MnO4- + 8 H+ + 5e- = Mn2+ + 4H2OEq. wt. = 158.04/5 = 31.608

Example: Calculate the normality of the solutions containing the following (a) 5.30 g/L Na2CO3 (b) 5.267 g/L K2Cr2O7SolutionCO32- reacts with 2H+ to H2CO3N=5,3/105.99/2 =0.1000 eq/LEach Cr4+ is reduced to Cr3+, a total change of 6e-/molecule K2Cr2O7Cr2O72- + 14H+ + 6e- = 2Cr3+ + 7H2ON=5,267/294.19/6 =0.1074 eq/L

Keuntungan menggunakan satuan NormalitasThe advantage of expressing concentrations in normality and quantities as equivalents is that one equivalent of substance A will ALWAYS react with one equivalent of substance B.

NaOH (= 1mol) will react with one equivalent of HCl (=1 mol), or with one equivalent of H2SO4 (1/2 mol)

Cont`n....We can, calculate the weight of analyte from the number of equivalents of titrant, because the latter is equal to equivalents of analyte.meqA = meqTmeqA = mgA = NT x mL BEAmgA = NT x mLT x BEA

How about the equation for calculating the percent of a constituent in the sample????

A 0,467 g sample containing sodium bicarbonate (a monoacidic base) and titrated with a standard solution of HCl, requairing 40,72 ml. The hydrocloric acid was standarized by titrating 0,1876 g sodium carbonate, wich required 37,86 ml. Calculate the percent sodium bicarbonate in the sample.Solution :NHCl = meq Na2CO3 = mg/BE (BM)mL HCl mL

Persen NaHCO3 = ....................

Density (kerapatan) calculationDensity is required for a calculation of molarityDensity is the weight per unit volume at the specified temperature, usually g/mL at 20C.

= gram/mL

Example: How many millimiters of concentrated sulfuric acid 94% (g/100g solution) density 1.843 g/cm3 are required to prepare 1 L of 0,100M solution?

PenyelesaianConsider 1 cm3 = 1 mL. From density For 1 L solution = 1843 g solutionFrom percent g H2SO4 = 94% x 1843 = 1732.42Mole H2SO4 = g/BM = 1732.42/98 = 17.68 mol in 1 L solutionM= 17.68Mole Initial = Mol Final 17.68 x V H2SO4 = 0.1 x 1000 V = 5.66 mL

Gram is basic unit of massa and is the unit employed most often in macro analysesFor small sample, smaller unit are employedmg = 10-3 gramug = 10-6 gramng = 10-9 grampico = 10-12 gramfemto = 10-15 gram

ProblemsCalculate the molar concentration of all the cations and anions in a solution prepared by mixing 10.0 mL each of the following solutions: 0.100 M Mn(NO3)2, 0.100 M KNO3, 0.100 M K2SO4.Calculate the grams of each substance required to prepare the following solution: (a) 250 mL of 0.100 M KOH (b) 1.00L of 0.0275 M K2Cr2O7 (c) 500 mL of 0.500 M CuSO4How many milliliters of concentrated hydrochloric acid, 38% (wt/wt), specific gravity 1.19, are required to prepare 1L of a 0.100 M solutionYou have a 250 ppm solution of K+ as KCl. You wish to prepare from this a 0.00100 M solution of Cl-. How many milliliters must be diluted to 1 L?

Solution

ProblemsHow many milliliters of 0.10 M of H2SO4 must be added to 50 mL of 0.1 M NaOH to give a solution that is 0.050 M in H2SO4? Assumes volumes are additive. A 0.500 g sample is analyzed spectrophotometrically for manganese by dissolving it in acid and transferring to 250 mL flask and diluting to volume. Three aliquots are analyzed by transferring 50 mL portions with a pipet to 500 mL Erlenmeyer flasks and reacting with an oxidizing agent, potassium peroxydisulfate, to convert the manganese to permanganate. After reaction, these are quantitatively transferred to 250 mL volumetric flasks, diluted to volume, and measured spectrometrically. By comparison with standards, the average concentration in the final solution is determined to be 1.25 x 10-5M. What is the percent manganese in the sample?Solution

ProblemsA preparation of soda ash is known to contain 98.6% Na2CO3. If a 0.678 g sample requires 36.8 ml of sulfuric acid to complete neutralization, what is the molarity of the sulfuric acid solution?A sample of USP grade citric acid (H3C6O7, three titratable protons) is analyzed by titrating with 0.1087 M NaOH. If a 0.2678 g sample requires 38.31 ml for titration, what is the purity of the preparation? (USP requires 99.5%)

Solution

ProblemsA solution is prepared by dissolving 7.82 g NaOH and 9.26 g Ba(OH)2 in water and diluting to 500 ml. What is the normality of the solution as a base?What weight of arsenic trioxide, As2O3, is required to prepare 1L of 0.1000 N arsenic (III) solution?The sulfur content of steel sample is determined by converting it to H2S gas, absorbing the H2S in 10 ml of 0.00500 M I2, and then back titrating the excess I2 with 0.00200M Na2S2O3. If 2.6 ml Na2S2O3 is required for titration, how many milligrams of sulfur are contained in the sample?A potassium permanganate solution is prepared by dissolving 4.68 g KMnO4 in water and diluting to 500 ml. How many milliliters of this will react with the iron in 0.500 g of ore containing 35.6% Fe2O3?

Solution