Review of Fundamental Concepts (Gary Christian, hal 65)
Atomic, and Molecular Weights
th e weight weight of a specified specif ied Atomic weight weight (BA) (BA)for any elements is the number of atoms of that tha t element, and that number is the same from one to another. Examples: Ca 40.08, S 32.06, O 16.00 Molecular weight (BM) is the sum of formula for mula weight 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 40.08+32.06+11x16.00+14x1.09 = 262.25
Atomic, and Molecular Weights
th e weight weight of a specified specif ied Atomic weight weight (BA) (BA)for any elements is the number of atoms of that tha t element, and that number is the same from one to another. Examples: Ca 40.08, S 32.06, O 16.00 Molecular weight (BM) is the sum of formula for mula weight 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 40.08+32.06+11x16.00+14x1.09 = 262.25
Moles, Molarity, Normality
Mole = grams/formula grams/formul a weight weight Molarity = mole/liter 1 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 molarity of the silver nitrate solution. How many millimoles AgNO 3 were dissolved?
Moles, Molarity, Normality
Answer …………………………………………………………… …………………………………………………………… ……………………………………………………………
Millimoles millimole = 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 Calculations
Parts per million (ppm)
= (berat terlarut/berat sampel) x 106 = mg/kg = mg/L = L/L = g/g = g/mL = nL/mL Example. 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 ug berat sampel 2,6 gram konsentrasi 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
10 3
ppb (part per billion) = (berat zat terlarut/berat sampel) x 109
= mg/g
= ng/g
= g/kg
= ug/kg
Contoh soal
Berapa mg sampel yang harus ditimbang jika ingin membuat 50 ppm dalam 100 ml. 50 ppm = (x gram/100 ml) x 10 6 x = 50 x 100 106 = 5 x 10-3 gram = 5 mg
Atau
ppm = ug/ml 50 ppm = ..... g/100 ml 50 ppm x 100 ml = 5000 g = 5 mg
Contoh soal
Hitung berapa Molar jika BM = 100 dengan konsentrasi 1 ppm ppm = mg/L = 10-3 gram/L M = gram/BM/L M = 10-3/100/L = 10-5 mol/L = 10-5 M
Soal Hitung : 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 Molarity
Consider the general reaction aA+tT
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 t mmolA = MT x mlT x a t mgA = mmolT x BMA x a t mgA = 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
Example A 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 = ½) % = MT x mlT x a/t x BM A x 100% mg sample
Normality Many 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 = meq L mL (eq) = Number of equivalent
eq = mol x no. Of reacting units per molecule
Normalitas
Banyaknya ekivalen (ek) zat terlarut tiap liter larutan, atau N = ek/V ek = gram/BE N = (gram x n) / (BM x V)
BE = BM/n
Berat Ekivalen (BE) BE = berat molekul dibagi dengan valensi = BM/n Cara penentuan valensi bergantung pada reaksi yang terjadi
Cara penentuan valensi a.
b.
Reaksi asam basa, valensi ditentukan berdasarkan banyaknya H+ atau OH- yang dihasilkan tiap satu mol asam atau basa Reaksi redoks, valensinya ditentukan o/ banyaknya elektron yang hilang atau timbul pada reaksi oksidasi reduksi
Contoh : a. b. c.
H3PO4 : 1 mol ekivalen dengan 3 mol ion H+ Ca(OH)2 : valensinya 2 I2 + 2e ----- 2Imaka valensinya = 2 sebab 1 mol ekivalen dengan 2 elektron
Contoh soal
Hitung berat ekivalen Na2C2O4 dan KMnO4 dlm reaksi redoks suasana asam
The Equivalent Weight
Berat ekivalen sama dengan berat molekul dibagi dengan valensi
For 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+)
Solution (a) eq. wt. = NH3/1 =17.03/1 (b) Eq. wt. = 90.04/2 = 45.02 (c) The Mn goes a five electron change, from valence +7 to +2: MnO4- + 8 H+ + 5e- = Mn2+ + 4H2O Eq. 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 K2Cr2O7
Solution (a) CO32- reacts with 2H+ to H2CO3 N=5,3/105.99/2 =0.1000 eq/L (a) Each Cr4+ is reduced to Cr3+, a total change of 6e/molecule K2Cr2O7 Cr2O72- + 14H+ + 6e- = 2Cr3+ + 7H2O N=5,267/294.19/6 =0.1074 eq/L
Keuntungan menggunakan satuan Normalitas The 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. meq A = meqT meq A = mg A = NT x mL BE A mg A = NT x mLT x BE A
How about the equation for calculating the percent of a constituent in the sample????
Example 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) calculation
Density is required for a calculation of molarity Density is the weight per unit volume at the specified temperature, usually g/mL at 20 C. °
ρ =
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?
Penyelesaian
Consider 1 cm3 = 1 mL. From density For 1 L solution = 1843 g solution From percent g H2SO4 = 94% x 1843 = 1732.42 Mole H2SO4 = g/BM = 1732.42/98 = 17.68 mol in 1 L solution M= 17.68 Mole 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 analyses For small sample, smaller unit are employed mg = 10-3 gram ug = 10-6 gram ng = 10-9 gram pico = 10-12 gram femto = 10-15 gram
Problems
Calculate 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 CuSO4 How many milliliters of concentrated hydrochloric acid, 38% (wt/wt), specific gravity 1.19, are required to prepare 1L of a 0.100 M solution You 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
Problems
How 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
Problems
A preparation of soda ash is known to contain 98.6% Na 2CO3. 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
