Tugas: Latihan Soal Titrasi A. Soal Latihan Kelompok 6-20 A 0.3367-g sample of primary-standard-grade Na2CO3 required 28.66 mL of a H 2SO4 solution to
reach the end point in the reaction 2-
+
CO3 + 2H H2O + CO 2(g) What is the molarity of the H 2SO4? th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 119)
Solusi: Mr Na2CO3 = 2x23+12+3x16 = 106 g/mol Mol Na2CO3 =
= 3.176 x 10
2-
-3
mol = 3.1764 mmol
+
Reaksi: CO3 + 2H H2O + CO2(g) +
2-
2 mol H ≡ 1 mol CO3 +
Mol H = 2 x 3.1764 mmol = 6.3528 mmol +
2-
H2SO4 2H + SO4
= 0.11 M =
Mol H2SO4 = x 6.3528 mmol = mmol M H2SO4
6-28 The thiourea in a 1.455-g sample of organic material was extracted into a dilute H 2SO4 solution 2+
and titrated with 37.31 mL of 0.009372 M Hg via the reaction 4(NH2)2CS + Hg
2+
[(NH2)2CS]4Hg
Calculate the percent (NH2)2CS (76.12 g/mol) in the sample. th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 120) Solusi :
Mol Hg2+ = 37.31 mL x 0.009372 M = 0.34967 mmol Reaksi:
4(NH2)2CS 1.3987 mmol
+
2+
Hg
0.34967 mmol
2+
[(NH2)2CS]4Hg
0.34967 mmol
Mol (NH2)2CS yang bereaksi = 4 x 0.34967 mmol = 1.3987 mmol
2+
-3
Massa (NH2)2CS dalam campuran = (1.3987 x 10 )mol x 76.12 g/mol = 0.10647 g
x 100% x 100% =
Persentase (NH2)2CS dalam campuran =
= 7.317 % (NH2)2CS
6-30 A solution of Ba(OH)2 was standardized against 0.1016 g of primary-standard-grade benzoic
acid C6H5COOH (122.12 g/mol). An end point was observed after addition of 44.42 mL of base. (a) Calculate the molarity of the base. (b) Calculate the standard deviation of the molarity if the standard deviation for weighing was ± 0.2 mg and that for the volume measurement was ± 0.03 mL. (c) Assuming an error of – 0.3 mg in the weighing, calculate the absolute and relative systematic error in the molarity th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 120) Solusi :
(a) Mol asam benzoat =
= 8.3197 x 10
-4
mol = 0.832 mmol
Reaksi: 2C6H5COOH + Ba(OH)2 (C6H5COO)2Ba + 2H2O 2 mol C6H5COOH ≡ 1 mol Ba(OH) 2
= 9.36 x 10 M M Ba(OH) = = √ () () () ) ( ) = √ (
Mol Ba(OH )2 = x 0.832 mmol = 0.416 mmol -3
2
(b)
= ± 2.0811 x 10
-3
-3
-3
-5
sy = y x (± 0.01252) = 9.36 x 10 M x (± 2.0811 x 10 ) = ± 1.95 x 10 M 3
(c) Error -0.3 mg, massa Ba(OH)2 = (0.1016 x 10 )mg – 0.3 mg = 101.3 mg Mol asam benzoat (error) =
= 8.3197 x 10 mol = 0.8295 mmol -4
Reaksi: 2C6H5COOH + Ba(OH)2 (C6H5COO)2Ba + 2H2O 2 mol C6H5COOH ≡ 1 mol Ba(OH) 2
Mol Ba(OH )2 = x 0.8295 mmol = 0.414756 mmol M Ba(OH)2 (error) =
= 9.337 x 10
-3
-3
-3
M -5
Absolute error = 9.337 x 10 M – 9.36 x 10 M = -2.7652 x 10 M
Relative error =
= -2.95 x 10
-3
= -2.95 ppt
6-32 a. A 0.3147-g sample of primary standard Na 2C2O4 was dissolved in dilute H 2SO4 and titrated with
31,672 g of dilute KMnO4.
Calculate the weight molarity of the KMnO 4 solution. th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 121) Solution :
moles of Na 2C2O4 =
= 0.00235 moles
moles of dilute KMnO 4 = 0.00235 moles x moles of dilute KMnO4 = 0.00094 moles Weight molarity =
= = =
b. The iron in a 0.6656-g ore sample was reduced quantitatively to the + 2 state and then titrated
with 26.753 g of the KMnO 4 solution. Calculate the percent Fe 2O3 in the sample. Solusi : 2+
-
+
3+
2+
5Fe + MnO 4 + 8H 5Fe + Mn + 4H2O Mol KMnO4 = 2+
= 0.1693 mol
Mol Fe = 5 x mol KMnO 4 =0.8466139 mol 2+
5 mol Fe2O3 ≡10 Fe ≡ 2KMnO4 mol Fe2O3 : mol KMnO 4 = 5 : 2
Mol Fe2O3 = x 0.8466139 mol = 2.11653475 mol Mr Fe2O3 = 2x56 + 3x16 = 160 g/mol Massa Fe2O3 = mol Fe 2O3 x Mr Fe2O3 = 2.11653475 mol x 160 g/mol = 338.64556 g Persentase Fe 2O3 =
x 100% = 50878.24%
14-18 A 1.509-g sample of a Pb/Cd alloy was disolved in acid and diluted to exactly 250.0 mL in a
volumetric flask. A 50.00-mL aliquot of the diluted solution was brought to a pH of 10.0 with an +
NH4 /NH3 buffer. The subsequent titration involved both cations and required 28.89 mL of 0.06950 M EDTA. A second 50.00-mL aliquot was brought to a pH of 10.0 with an HCN/NaCN buffer, which 2+
2+
also served to mask the Cd ; 11.56 mL of the EDTA solution were needed to titrate the Pb . Calculate the percent Pb and Cd in the sample. th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 301)
Solusi : 2+
2+
Mol total = mol (Cd + Pb ) = 28.89 mL x 0.06950 M = 2.00786 mmol Mol EDTA = 0.06950 M x 11.56 mL = 0.80342 mmol EDTA 2+
2+
1 mol Pb ≡ 1 mol EDTA mol Pb = 0.80342 mmol 2+
Mol Cd = mol total – mol Pb
2+
= 2.00786 mmol – 0.80342mmol = 1.20444 mmol Cd
2+
2+
-3
2+
-3
Massa Pb = 0.80342 mmol x 10 mol/mmol x 207.2 g/mol = 0.1665 g Massa Cd = 1.20444 mmol x 10 mol/mmol x 112.4 g/mol = 0.1354 g
= 0.3018 g x 100% = 55.16 % Pb di sampel = x 100% = 44.86 % Cd di sampel =
Massa sampel = 1.509 g x Persentase Pb
2+
2+
Persentase Cd
2+
2+
B. Soal Individu (Titrasi Pengendapan dan Titrasi Kompleksometri) Hansel (NPM: 2012620044) 6-24 Titration of the I 2 produced from 0.1238 g of primary standard KIO3 required 41.27 mL of
sodium thiosulfate
th
Calculate the molarity of the Na 2S2O3 solution. (Skoog, Fundamentals of Analytical Chemistry 7 ed, hlm. 277) Solusi :
Mr KIO3 = (39+127+3x16)g/mol = 214 g/mol Mol of KIO3 =
= 5.785x10
-4
mol -4
-3
Mol I2 = 3 x mol KIO3 = 3 x 5.785x10 mol = 1.7355x10 mol -3
-3
Mol Na2S2O3 = 2 x mol I2 = 2 x 1.7355x10 mol = 3.47x10 mol
Konsentrasi Na2S2O3 =
= 0.0841 M
14-15 An EDTA solution was prepared by dissolving approximately 4 g of the disodium salt in
approximately 1L of water. An average of 42.35 mL of this solution was required to titrate 50.00-mL aliquots of a standard that contained 0.7682 g of MgCO3 per liter. Titration of a 25.00-mL sample of mineral water at pH 10 required 18.81 mL of the EDTA solution. A 50.00-mL aliquot of the mineral water was rendered strongly alkaline to precipitate the magnesium at Mg(OH) 2. Titration with a calcium-specific indicator required 31.54 mL of the EDTA solution. Calculate (a) the molarity of the EDTA solution, (b) the concentration of CaCO3 in the mineral water (ppm), (c) the concentration of MgCO3 in the mineral water (ppm). th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 301) Solusi:
Diketahui konsentrasi MgCO 3 = 0.7682 g/L Volume MgCO3 = 50 mL Mr MgCO3 = (24+12+3x16) g/mol = 84 g/mol Volume EDTA= 42.35 mL (a)
mol MgCO3 =
= 9.1452 x 10
-3
mol = 9145.2 mmol
Misal volume larutan = 1L, maka konsentrasi molar MgCO 3 =
= 9.1452 x 10
-3
M
-3
Volume MgCO3 = 50 mL, maka mol MgCO 3 = 9.1452 x 10 M x 50 mL= 0.45726 mmol 2+
Mg + EDTA -> MgEDTA ; sehingga perbandingan EDTA dengan MgCO 3 = 1:1 Mol EDTA = mol MgCO 3 = 0.45726 mmol Jadi, molaritas larutan EDTA = (b)
= 0.0108 M EDTA
Konsentrasi total CaCO3 dan MgCO 3 =
= 8.126 x 10 M
Mol CaCO3 : mol EDTA = 1:1, maka konsentrasi CaCO 3 =
-3
= 6.812 x 10 -3
-3
M
mol CaCO3 pada air mineral = konsentrasi x volume = 6.812 x 10 M x 1mL = 6.812 x 10 mmol -3
massa CaCO3 = 6.812 x 10 mmol x
= 6.812 x 10 mg = = 6.812 x 10 g -1
-4
-3
-4
6
ppm CaCO3 = 6.812 x 10 g x 10 ppm/g = 681.2 ppm (c)
Konsentrasi MgCO3 = konsntrasi total – konsentrasi CaCO 3 -3
-3
= 8.126 x 10 M - 6.812 x 10 M -3
= 1.314 x 10 M -3
-3
Mol MgCO3 = M x V = 1.314 x 10 M x 1mL = 1.314 x 10 mmol -3
massa MgCO 3 = 1.314 x 10 mmol x -3
= 0.11038 mg = = 0.11038 x 10 g -3
6
ppm MgCO3 = 0.11038 x 10 g x 10 ppm/g = 110.38 ppm
Vicky (NPM: 2012620050) 13-11 Titration of a 0.485-g sample by the Mohr method required 36.8 ml of standard 0.1060 M
AgNO3 solution. Calculate the percentage of chloride in the sample. th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 276) Solusi:
Mol AgNO3 yang bereaksi = 36.8 mL x 0.1060 M = 3.9 mmol Mol AgNO3 yang bereaksi = mol klorida yang bereaksi = 3.9 mmol Massa klorida = 3.9 mmol x 35.5 mg/mmol = 138.45 mg Persentase klorida = (138.45 mg/485 mg) x 100% = 28.546%
14-13 The Cr plating on the surface that measured 3 x 4 cm was dissolved in HCl. The pH was suitably
justed, following which 15 ml of 0.01768 M EDTA were introduced. The The excess reagent required a 2+
4.3 ml back titration with 0.008120 M Cu . Calculate the average mass of Cr on each square centimeter of surface. th
(Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 301) Solusi:
Mol EDTA = 15 mL x 0.01768 0.017 68 M= 0.2652 mmol EDTA 2+
Mol Cu = 4.3 mL x 0.008120 M = 0.0349 mmol Cu Cu
2+
+
EDTA
m:
0.0349 mmol
0.2652 mmol
r:
0.0349 mmol
0.0349 mmol
s:
-
2+
Cu(EDTA) 0.0349 mmol
0.2303 mmol 3+
mol EDTA : mol Cr dalam stoikiometri = 1:1 Mol yang bereaksi = 0.2652 – 0.0349 = 0.2303 mmol Massa Cr = 0.2303 mmol x 52 mg/mmol = 11.9756 mg Luas permukaan Cr = 3cm x 4cm = 12 cm 2
Perbandingan massa Cr per cm luas =
2
= = 0.9979 mg/cm
2
Thomas (NPM: 2012620092) 13-25 The action of an alkaline I 2 solution upon the rodenticide warfarin, C 19H16O4 (308.34 g/mol),
results in the formation of 1 mol of iodoform, CHI 3 (393.73 g/mol), for each mole of the parent compound reacted. Analysis for warfarin can then be based upon the reaction between CHI 3 and +
Ag : +
+
CHI3 + 3Ag + H2O 3AgI(s) + 3H + CO(g) The CHI3 produced from a 13.96-g sample was treated with 25.00 mL of 0.02979 M AgNO 3, and the +
excess Ag was then titrated with 2.85 mL of 0.05411 M KSCN. Calculate the percentage of the th
warfarin in the sample. (Skoog, sample. (Skoog, Fundamentals of Analytical Chemistry 7 ed, hlm. 277) Solusi :
Mol AgNO3 mula-mula = 25 mL x 0.02979 M = 0.74475 mmol Mol KSCN = 2.85 mL x 0.05411 M = 0.1542 mmol mol AgNO3 yang bereaksi = mol AgNO 3 mula-mula - mol KSCN = 0.74475 mmol - 0.1542 mmol = 0.5905 mmol
+
Mol AgNO3 = mol Ag = 0.5905 mmol +
mol CHI3 = ⅓ x mol Ag = ⅓ x 0.5905 mmol = 0.1968 mmol Massa CHI3 = 0.1968 mmol x 308.34 mg/mmol = 60 .695 mg Massa warfarin = massa CHI 3 = 60.695 mg Persentase warfarin pada sampel =
x 100% = 0.4348%
14-14 The Tl in a 9.76 g sample of rodenticide was oxidized to the trivalent state and treated with an
unmeasured excess of Mg/EDTA solution. The reaction is 3+
2-
-
Tl + MgY TlY + Mg
2+
Titration of the liberated Mg2+ required 13.34 mL of 0.03560 M EDTA. Calculate the percentage of th
Tl2SO4 (504.8 g/mol) in the sample. (Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 301) Solution :
Mole of EDTA = 0.0356 M x 13.34 mL = 0.475 mmol Mol Tl2SO4 : mol EDTA = 1:2, then mol of Tl 2SO4 = 0.5 x mol EDTA = 0.5 x 0.475 mmol = 0.237 mmol Mr Tl2SO4 = (2x204+32+4x16)g/mol = 504 g/mol -3
Mass of Tl 2SO4 = 0.237x10 mol x 504 g/mol = 0.1197 g Percentage of Tl2SO4 in the sample =
x 100% = 1.226%
Filino (NPM: 2012620100) 6-21 A 0.3396-g sample that assayed 96.4% Na 2SO4 required 37.70 mL of a barium chloride solution.
Reaction:
+ -> (s) Calculate the analytical molarity of BaCl 2 in the solution. (Skoog, Fundamentals of Analytical th
7 ed, hlm. 120) Chemistry 7 Solusi :
Mr Na2SO4 = 142.04 g/mol Massa Na2SO4 = 0.964 x 0.3396 g = 0.3274 g -3
Mol Na2SO4 = 0.3274 g/(142.04 g/mol) = 2.3048 x 10 mol = 2.3048 mmol Konsentrasi = 2.3048 mmol/ 41.25mL = 0.0611 M BaCl2
13-12 A 0.1064-g sample of pesticide was decomposed by the action of sodium biphenyl in toluene. -
The liberated Cl was extracted with water and titrated with 23.28 ml of 0.03337 M AgNO 3 using an adsorption indicator. Express the results of this analysis in terms of percent aldrin, C 12H8Cl6(364.92 th
g/mol). (Skoog, g/mol). (Skoog, Fundamentals of Analytical Chemistry 7 7 ed, hlm. 276) Mol AgNO3 yang bereaksi = 23.28 mL x 0.03337 M = 0.7768 mmol -
Mol AgNO3 yang bereaksi = mol Cl = 0.7768 mmol -
Mol aldrin= mol Cl = 0.7768 mmol Massa aldrin = 0.7768 mmol x 364.92 mg/mmol = 283.489 mg 3
Persentase aldrin = (283.489 mg/(106.4 x 10 mg)) x 100% = 0.266%
