2b Soal Soal Titrasi Redoks

Soal-soal Titrasi Redoks

sample mass= 04185g

KMnO4 titration= 0025M 4127mL

Fe2O3=

mol Fe2O3=12mol Fe=125mol KMNO4

=52(VM)KMnO4

25794mmol

massa Fe2O3= 4127mg

9861

sample= 04185 g= 4185 mg

Titration Fe2+

KMnO4 Vol= 4127 mLM= 0025 M

Mol KMnO4= (MV)= 103175 mmol

Reaction 5 Fe2+ + KMnO4 === 5 Fe3+ + Mn2+

mol Fe2+ = 5 mol KMnO4= 515875 mmol

Reaction 2 Fe2+ === 2 Fe3+ ===== Fe2O3

mol Fe2O3 = 05mol Fe2+ = 2579375 mmol

mass Fe2O3= molMr Fe2O3= 4127 mg

Fe2O3= mass Fe2O3mass sample100= 9861

Walden reductor UO22+ --gt U4+

Addition of Fe3+ U4++ 2Fe3+ ---gt U6+ + 2Fe2+

Titration 6 Fe2+ + K2Cr2O7 --gt 6 Fe3+ + 2 Cr3+

sample= 0315 g= 315 mgTitration

Vol= 1052 mLM= 000987 M

mmol K2Cr2O7= (VM)= 0103832 mmol

mmol Fe2+ =6 mmol K2Cr2O7= 0622994 mmol

mmol U4+ =05 mmol Fe2+ = 0311497 mmolmass U= mmol U4+ Ar (U)= 7414534 mgU= mass Umass sample 100= 2354

Ar U= 2380289

Cr= 519961 Fe 558Area= 30 cm2 density Cr= 72 gcm3

oxidizing 2Cr3+ === Cr2O72-

Addition of Fe(NH4)2(SO4)26H2O= 5000 mg

Reaction Cr2O72- + 6Fe2+ === 2Cr3+ + 6Fe3+

Backtitration

Reaction Cr2O72- + 6Fe2+ === 2Cr3+ + 6Fe3+

vol= 1829 mLM= 000389 M

mmol back titration = Fe2+ excess= (VM) backtitration== 0071148 mmol

mmol Fe total = mmol Fe-amm-sulfat=massMr Fe-amm-sulfat= 1276015 mmol

mmol Fe2+ bereaksi dengan Cr2O72- sampel= 120487 mmol

mmol Cr2O72-= 16 mmol Fe2+= 020081 mmol

mmol Cr3+ = 2 x mmol Cr2O72-= 040162 mmol

mass Cr3+= 209 mg= 002088 g

vol Cr3+= massdensity= 00029 cm3thickness=volumearea= 97E-05 cm

= 09668 μmMr Fe-amm-sulfat= 39185 gmol

• Soal-soal Titrasi Redoks
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32