Kimia Tekben
Transcript of Kimia Tekben
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Bab I. Materi dan Sifatnya
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Bab II.
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Praktik Pengenalan Bahan Kimia
• Tujuan Mengenal jenis, sifat dan kegunaan bahan – bahan kimia yang digunakan di pertanian.
• Bahan dan AlatBahan-bahan kimia seperti pupuk, pestisida, hormon, vitamin, asam amino
• Keselematan KerjaHindari kontak langsung anggota tubuh dengan bahan – bahan kimia
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Bab III. Sifat Periodik
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• Prosedur1. Catat nomor bahan kimia yang Anda amati
2. Beri nama bahan kimia sesuai dengan etiket yang ada
3. Golongkan bahan tersebut kedalam jenis anorganik atau organik
4. Tentukan bentuk/wujud bahan kimia tersebut: cairan, gas, butiran, serbuk atau tepung
5. Tentukan warna bahan kimia tersebut
6. Tentukan sifat sifat khusus seperti: mudah larut, mudah terbakar, dan lain-lain
7. Tentukan kegunanan bahan tersebut dalam pertanian
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Prinsip Aufbau
• Pengisian elektron pada kulit dimulai dari dalam keluarPengisian n = 1 sebelum 2, mengisi s sebelum p
• Elektron-elektron tidak menggunakan orbital yang sama terkecuali keadaan tersebut diperlukan
• Pengecualian yang penting:Pengisian 4S sebelum 3d Pengisian 5d1 sebelum 4fPengisian 5S sebelum 4d Pengisian 6d1 sebelum 5f
• Hal ini menjelaskan alasan mengapa kelompok logam transisisi dikelompokkan kedalam unsur-unsur kelompok B
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Prinsip Aufbau
Kecenderungan Umum Pengisian Elektron
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Bab IV. Stoikiometri I
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Persamaan Reaksi• Persamaan yang menyatakan perubahan materi dalam reaksi:
Zn + H2SO4 ZnSO4 + H2
N2 + H2 NH3
• Sebelah kiri tanda panah adalah reaktan, sebelah kanan tanda panah adalah produk
• Koefisien reaksi menyatakan perbandingan mol, pada reaksi dalam bentuk gas juga menyatakan perbandingan volume
• Menyetarakan koefisien reaksi : jumlah atom setiap unsur di sebelah kiri dan kanan tanda panah harus sama jumlahnya. Persamaan di atas menjadi:
Zn + H2SO4 ZnSO4 + H2
N2 + 3 H2 2 NH3
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Tentukanlah koefisien reaksi dari persamaan reaksi berikut ini : … Al + … HCl …AlCl3 + … H2
Jawaban : Kita misalkana koefisien reaksi kimianya adalah Al + b HCl c AlCl3 + d H2
misalkan a = 1, maka c = 1; b = 3c = 3x1 = 3;
b = 2d ; 2d = b = 1 d = ½; persamaan reaksi menjadi:
Al + 3 HCl AlCl3 + ½ H2 dibulatkan menjadi:
2 Al + 6 HCl 2 AlCl3 + H2
Atom Kiri = Kanan
Al a = C
H b = 2d
Cl b = 3c
Contoh soal-1
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Contoh Soal-2• Berapakah berat N2 yang terdapat pada pupuk urea C0(NH2)2
yang beratnya 50 kg? (Mr C=12, O=16, N=14, H=1)?
• Jawaban:
Berat N2 dalam urea = (Mr N x 2)/(Mr CO(NH2)2) x 50 kg
= (14 x 2)/(12+16+(14+1x2)x2) x 50 kg
= 28/(28+32) x 50 kg = 28/60 x 50 kg
= 23,33 kg
• Tentukanlah koefisien reaksi dari persamaan berikut
C8H14 + O2 → CO2 + H2O
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A. Konsentrasi larutan• Larutan terditri dari: zat terlarut dan pelarut (terutama air)
• Konsentrasi larutan menunjukkan banyaknya zat terlarut dalam suatu larutan yang didasarkan atas: masa zat terlarut/volume pelarut (m/v) dan volume zat terlarut/ volume pelarut (v/v)
• c = m/v atau v/V ; c = konsentrasi, m = masa terlrt, v = vol terlrt, V = vol lar
• Satuan konsentrasi: M, m, %, g/l, mg/l, ppm,
M = mol zat terlarut/volume larutan (l)
m = 1000/p x mol zat terlarut
mg/l = ppm (part per milion)
Bab V. Stoikiometri 2
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o Sebanyak 100 g H2SO4 dilarutkan kedalam air sampai volumenya 1000 ml, berapakah molaritas larutannya?
o Penyelesaian:mol H2SO4 = 100 g/Mr H2SO4
= 100/(1x2+32+16x4) = 100/98 = 1,02 mol. volume larutan = 1000 ml = 1000/1000 ltr = 1
l.
M = mol H2SO4/vol larutan = 1,02 mol/1 l
= 1,02 mol/l = 1,02 M
Contoh Soal-1
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o Berapa gram HCl yang terlarut di dalam 50 ml larutan HCl 5 M ?
o Penyelesaian:Diketahui: v = 50 ml, c = 5 MDitanyakan: berapa m?c = m/v atau
m = c x v = 50 ml x 5 mol/lt = 50 ml x 5 Mr HCl gram/1000 ml = 5 x (1+35)/20 gram = 5 x 36 gr/20 = 9 gram
Contoh Soal-2
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• Kedalam 200 ml air ditambahkan 2 gram NaOH, berapakah molalitas. Berapakah molalitas, fraksi mol, dan persen berat larutan tersebut?
• Jawaban: mol NaOH = 2/Mr NaOH = 2/(23+16+1) = 0,05 molmol air = 200 /Mr H2O = 200/(1x2+16) = 11,11 molmolalitas (m) = 1000 x mol zat terlarut /pelarut
= 1000 gr x 0,05 mol /200 gr = 0,25 m
Fraksi mol = 0,05/(0,05+11,11) = 0,0045
Persen berat = 2 gr/(2+200) gr = 0,9 %
Contoh Soal-3
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B. Pengenceran Larutan
Pengenceran larutan berarti:
- memperkecil konsentrasi larutan - memperbesar volume larutan - masa (berat) zat terlarut sebelum dan sesudah pengenceran adalah sama
Rumus
V1C1 = V2C2
V1, V2 = volume sebelum dan sesudah pengenceranC1, C2 = konsentrasi sebelum dan sesudah pengenceran
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10 ml H2SO4 5 M diencerkan menjadi H2SO4 0,2 M, berapakah jumlah air yang harus ditambahkan?Penyelesaian:
V1xC1 = V2xC2V1, V2 = vol H2SO4 sebelum dan sesudah diencerkanC1, C2 = konsentrasi H2SO4 sebelum dan sesudah diencerkan
10 x 5 = V2 x 0,2V2 = 250 ml
Jadi harus ditambahkan air sampai volume larutan mencapai 250 ml.
Contoh Soal-4
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2. Pencampuran Larutan
M campuran = (V1xC1+V2xC2)/(V1 +V2)
Contoh-soal-5:
Jika 1 liter HCl 0,3 M dicampurkan dengan 110 ml HCl 1 M, berapakah konsentrasi larutan HCl yang dihasilkan?
• Jawaban:
V1 = 1 l, V2 = 110 ml = 0,11 l;
M1 = 0,3 M, M2 = 1 M
Konsentrasi campuran
= (V1xM1)+(V2xM2)/(V1 + V2)
= (1x0,3)+(0,110x1)/(1+0,11)
= 0,37 M
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Praktik Pembuatan Larutan
• Tujuan Membuat larutan kimia yang biasa digunakan dalam pembuatan media kultur jaringan tanaman
• Bahan dan AlatBahan-bahan kimia untuk bahan pembuatan larutan stok media kultur jaringan tanaman seperti: bahan sumber unsur hara (makro, mikro, hormon dan vitamin. Alat-alat gelas seperti beaker glass, labu takar, erlenmeyer. Timbangan, pipet.
• Keselematan KerjaHindari kontak langsung anggota tubuh dengan bahan – bahan kimia
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• Prosedur1. Ambil bahan kimia yang akan digunakan untuk membuat larutan
2. Tentukan kebutuhan bahan untuk pembuatan larutan
3. Timbang / pipet bahan kimia yang sudah diketahui kebutuhannya
4. Tentukan volume akhir larutan
5. Isi beaker glass dengan aquades kurang lebih ½ volume akhir larutan
6. Larutkan bahan kimia kedalam beaker glass dengan bantuan pengaduk atau pemanas yang dilengkapi pengaduk
7. Pindahkan larutan kedalam labu takar yang ukurannya sesuai
8. Tambahakan aquades sampai dengan tanda tera
9. Tuangkan larutan tersebut kedalam wadah penyimpanan
10. Beri etiket/label: nama larutan, konsentrasi larutan, tanggal pembuatan, nama pembuat,
11. Simpan larutan ditempat yang aman
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Bab VI. Praktik Perhitungan Kimia
• Tujuan: 1. Menentukan %(m/v), % (v/v), M, m,
fraksi mol, persen mol2. Menyiapkan larutan kimia
• Bahan dan Alat :
ATK, tabel sistem periodik unsur kimia
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• ProsedurLakukanlah perhitungan - perhitungan di bawah ini!
1. Buatlah larutan HCl 1% sebanyak 100 ml dari larutan HCl 1M yang tersedia sebanyak 50 ml
2. Buatlah suatu larutan alkohol 70% dari larutan alkohol 96% dengan mudah dan tepat!
3. Larutan etanol-air terdiri dari 20 ml etanol yang berbobot jenis 0,64 g/cm3 dilarutkan kedalam air dengan volume akhir 200 ml dan bobot jenisnya 0,8 g/cm3, hitunglah:
a) % etanol (v/v); b) % etanol (m/v); c) % etanol (v/v); d) M etanol; e) m etanol; f) fraksi mol etanol; g) % mol etanol
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Buatlah larutan HCl 1% sebanyak 100 ml dari larutan HCl 1M yang tersedia sebanyak 50 ml!Jawaban:Diketahui: C1 = 1 M, C2 = 1% C1 = 0,01 M V2 = 100 mlDitanyakan: cara membuatnya?Jawab:
V1xC1 = V2xC2V1 x 1M = 100 ml x 0,01 M
V1 = 1 mlCara membuatnya ……..
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Cara membuatnya:
1. Ambil larutan HCl 1 M
2. Tuangkan/pipet sebanyak 1 ml larutan HCl 1 M
3. Masukkan kedalam labu takar/gelas ukur 100 ml
4. Tambahkan akuades sampai tanda tera 100 ml
5. Goyang perlahan
6. Tempatkan kedalam wadah dan tutup rapat
7. Beri label larutan HCl 1% tersebut
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Buatlah suatu larutan alkohol 70% dari larutan alkohol 96% dengan mudah dan tepat!Jawaban:Diketahui: C1 = 96%
C2 = 70%Ditanyakan: cara membuat alkohol 70%Jawab:
V1xC1 = V2xC2V1 x 96% = V2 x 70% M
V2 = V1 + A (air)V1 x 96% = (V1+A) x 70% M
Misalkan V1 = 70 ml
70 x 96 = (70+A) x 70
A = 26 ml
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Cara membuatnya:
1. Ambil larutan alkohol 96%
2. Tuangkan larutan tersebut sebanyak 70 ml kedalam gelas ukur 100 ml
3. Masukkan kedalam labu takar/gelas ukur 100 ml
4. Tambahkan akuades sampai tanda tera 100 ml
5. Goyang perlahan
6. Tempatkan kedalam wadah dan tutup rapat
7. Beri label larutan alkohol 70% tersebut
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Larutan etanol-air terdiri dari 20 ml etanol yang berbobot jenis 0,64 g/cm3 dilarutkan kedalam air dengan volume akhir 200 ml dan bobot jenisnya 0,8 g/cm3, hitunglah:
a)% etanol (v/v); b) % etanol (m/v); c) % etanol (v/v); d) M etanol; e) m etanol; f) fraksi mol etanol; g) % mol etanolJawaban:a) % etanol (v/v) = 20/200 x 100% = 10%
b) % etanol (m/v)
m etanol = 20 ml x 0,64 g/ml = 12,8 g
% (m/v) = 12,8 g/200 ml x 100% = 6,4%
c) % (m/m) =
m larutan = 200 ml x 0,8 g/ml = 160 g
% (m/m) = 12,8 g/160 g x 100% = 8%
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d) M etanol
mol etanol = 12,8 g/Mr etanol = 12,8 g/ Mr C2H5OH
= 12,8/(12x2+1x6+16x1) = 12,8 g/46 g/mol
= 0,28 mol
volume lar = 200 ml = 0,2 l
M etanol = 0,28 mol/0,2 l = 1,4 M
e) m etanol (m/v)
m = (1000 x mol terlarut)/berat pelarut
= 1000 x 0,28/(berat air)
berat air = 160 g – 12, 8 g = 147,2 g
= 1000 x 0,28/147,2
= 280/147,2 = 1,9 m
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f) Fraksi mol etanol
frak mol = mol etanol/(mol etanol+mol pelarut)
= 0,28 mol/(0,28 mol+mol air)
mol air = m air/Mr air = 147,2/Mr H2O
=147,2/18 = 8,2 mol
frak mol = 0,28/(0,28+8,2) = 0,033
g) % mol etanol
= mol etanol/(mol etanol+mol pelarut) x 100%
= 0,28/(0,28+8,2) x 100%
= 0,033 x 100%
= 3,3%
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