PERHITUNGAN
13
PERHITUNGAN V NaOH rata-rata(Untuk Larutan Blanko) = 3,53 ml Mol ekivalen HCl = mol ekivalen NaoH = 2 N x 3,53 ml = 7,06 mmol = 7,06 x 10 -3 mol Suhu ruang 30 0 C Pada label ρ etanol = 0,789gr/ml Pada label ρ asam asetat = 1,049gr/ml ERLENMEYER 1 V 1 = 6,5 x 10 -3 ml Mencari mol mula etanol ρ etanol = m V 1 etanol mol Etanol = (ρ etanol x V etanol)/Mr Etanol = (0,789gr/ml . 1 ml)/46gr/mL = 17,2x10 -3 mol Mencari mol mula mula asam asetat ρ asam asetat = m V As . Asetat mol As Asetat = (ρ As.Asetat x V As.Asetat)/Mr As.Asetat = (1,049gr/ml . 4 ml)/60gr/mL = 69,9x10 -3 mol mol ekivalen H + = mol ekivalen OH - = N NaOH x V NaOH = 2 N x 6,5x10 -3 L = 13x10 -3 mol
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Transcript of PERHITUNGAN
PERHITUNGAN
V NaOH rata-rata(Untuk Larutan Blanko) = 3,53ml
Mol ekivalen HCl = mol ekivalen NaoH
= 2 N x 3,53 ml = 7,06 mmol = 7,06 x 10-3 mol
Suhu ruang 300C
Pada label ρ etanol = 0,789gr/ml
Pada label ρ asam asetat = 1,049gr/ml
ERLENMEYER 1
V1 = 6,5 x 10-3 ml
Mencari mol mula etanol
ρ etanol = m
V 1etanol
mol Etanol = (ρ etanol x V etanol)/Mr Etanol
= (0,789gr/ml . 1 ml)/46gr/mL
= 17,2x10-3mol
Mencari mol mula mula asam asetat
ρ asam asetat = m
V As . Asetat
mol As Asetat = (ρ As.Asetat x V As.Asetat)/Mr As.Asetat
= (1,049gr/ml . 4 ml)/60gr/mL
= 69,9x10-3mol
mol ekivalen H+ = mol ekivalen OH-
= N NaOH x V NaOH
= 2 N x 6,5x10-3 L
= 13x10-3 mol
mol H+ sisa = mol H+ total – mol blanko
= 13x10-3mol – 7,06x10-3mol
= 5,94x10-3mol
CH3COOH(aq) + C2H5OH(aq) ↔ CH3COOC2H5(aq) + H2O(l)
M 69,9x10-3mol 17,2x10-3mol
R -63,96x10-3mol -63,96x10-3mol 63,96x10-3mol 63,96x10-3mol
___________________________________________________
S 5,94x10-3mol -46,76x10-3mol 63,96x10-3mol 63,96x10-3 mol
[CH3COOH] = -46,76x10-3mol/10-2 L
[ C2H5OH] = 5,94x10-3mol/10-2 L
[CH3COOC2H5] = 63,96x10-3mol/10-2 L
Kc =
[CH3COOC2H5][CH3COOH] [C2H5OH]
= 63,96 x10−3mol /10−2L(−46,76 x 10−3mol /10−2L)(5,94 x 10−3mol/10−2L)
=−2,303mol−1 L
ERLENMEYER 2
V1 = 6,1 x 10-3 ml
Mencari mol mula etanol
ρ etanol = m
V 1etanol
mol Etanol = (ρ etanol x V etanol)/Mr Etanol
= (0,789gr/ml . 2 ml)/46gr/mL
= 34,3x10-3mol
Mencari mol mula mula asam asetat
ρ asam asetat = m
VAs. Asetat
mol As Asetat = (ρ As.Asetat x V As.Asetat)/Mr As.Asetat
= (1,049gr/ml . 3 ml)/60gr/mL
= 52,4x10-3mol
mol ekivalen H+ = mol ekivalen OH-
= N NaOH x V NaOH
= 2 N x 6,1x10-3 L
= 12,2x10-3 mol
mol H+ sisa = mol H+ total – mol blanko
= 12,2x10-3mol – 7,06x10-3mol
= 5,14x10-3mol
CH3COOH(aq) + C2H5OH(aq) ↔ CH3COOC2H5((aq) + H2O(l)
M 52,4x10-3mol 34,3x10-3mol
R -47,26x10-3mol -47,26x10-3mol 47,26x10-3mol 47,26x10-3mol
___________________________________________________
S 5,14x10-3mol -12,96x10-3mol 47,26x10-3mol 47,26x10-3mol
[CH3COOH] = 5,14x10-3mol/10-2 L
[ C2H5OH] = -12,96x10-3mol/10-2 L
[CH3COOC2H5] = 47,26x10-3mol/10-2 L
Kc =
[CH3COOC2H5][CH3COOH] [C2H5OH]
= 47,26x 10−3mol /10−2L(5,14 x10−3mol/10−2L)(−12,96 x10−3mol /10−2L)
=−7,904mol−1 L
ERLENMEYER 3
V1 = 5,9 x 10-3 ml
Mencari mol mula etanol
ρ etanol = m
V 1etanol
mol Etanol = (ρ etanol x V etanol)/Mr Etanol
= (0,789gr/ml . 3 ml)/46gr/mL
= 51,5x10-3mol
Mencari mol mula mula asam asetat
ρ asam asetat = m
VAs. Asetat
mol As Asetat = (ρ As.Asetat x V As.Asetat)/Mr As.Asetat
= (1,049gr/ml . 2 ml)/60gr/mL
= 34,9x10-3mol
mol ekivalen H+ = mol ekivalen OH-
= N NaOH x V NaOH
= 2 N x 5,9x10-3 L
= 11,8x10-3 mol
mol H+ sisa = mol H+ total – mol blanko
= 11,8x10-3mol – 7,06x10-3mol
= 4,74x10-3mol
CH3COOH(aq) + C2H5OH(aq) ↔ CH3COOC2H5(aq) + H2O(l)
M 34,9x10-3mol 51,5x10-3mol
R -30,16x10-3mol -30,16x10-3mol 30,16x10-3mol 30,16x10-3mol
___________________________________________________
S 4,74x10-3mol 21,34x10-3mol 30,16x10-3mol 30,16x10-3 mol
[CH3COOH] = 4,74x10-3mol/10-2 L
[ C2H5OH] = 21,34x10-3mol/10-2 L
[CH3COOC2H5] = 30,16x10-3mol/10-2 L
Kc =
[CH3COOC2H5][CH3COOH] [C2H5OH]
= 30,1 6x 10−3mol /10−2L(4,74 x 10−3mol /10−2L)(21,3 4 x 10−3mol /10−2L)
=2,982mol−1 L
ERLENMEYER 4
V1 = 4,5 x 10-3 ml
Mencari mol mula etanol
ρ etanol = m
V 1etanol
mol Etanol = (ρ etanol x V etanol)/Mr Etanol
= (0,789gr/ml . 4 ml)/46gr/mL
= 68,6x10-3mol
Mencari mol mula mula asam asetat
ρ asam asetat = m
VAs. Asetat
mol As Asetat = (ρ As.Asetat x V As.Asetat)/Mr As.Asetat
= (1,049gr/ml . 1 ml)/60gr/mL
= 17,5x10-3mol
mol ekivalen H+ = mol ekivalen OH-
= N NaOH x V NaOH
= 2 N x 4,5x10-3 L
= 9x10-3 mol
mol H+ sisa = mol H+ total – mol blanko
= 9x10-3mol – 7,06x10-3mol
= 1,04x10-3mol
CH3COOH(aq) + C2H5OH(aq) ↔ CH3COOC2H5(aq) + H2O(l)
M 17,5x10-3mol 68,6x10-3mol
R -15,56x10-3mol -15,56x10-3mol 15,56x10-3mol 15,56x10-3mol
___________________________________________________
S 1,94x10-3mol 53,04x10-3mol 15,56x10-3mol 15,56x10-3 mol
[CH3COOH] = 1,94x10-3mol/10-2 L
[ C2H5OH] = 53,04x10-3mol/10-2 L
[CH3COOC2H5] = 15,56x10-3mol/10-2 L
Kc =
[CH3COOC2H5][CH3COOH] [C2H5OH]
= 15,5 6 x10−3mol/10−2L(1,94 x10−3mol/10−2 L)(53,04 x10−3mol /10−2 L)
=1,495mol−1L
Kc Rata rata = (−2,303+ (−7,094 )+2,982+1,495)mol−1L
4=−1,23mol−1L
1. Sebagai katalis, karena pada dasarnya reaksi esterifikasi berlangsung sangat
lambat. Sehingga diperlukan suatu katalis
2. C2H5OOH(aq) + CH3COOH(aq) ↔ CH3COOC2H5(aq) + H2O(aq)
3. Kc Erlenmeyer 1 = −0,0480
Kc Erlenmeyer 2 = −0,0038
Kc Erlenmeyer 3 = 0,0144
Kc Erlenmeyer 4 = 0,0031
LAMPIRAN
PERHITUNGAN
Larutan Blanko = V NAOH = 4.4ml
Mol ekivalen H+ = mol ekivalen NaoH
= 2 N x 4,4 ml = 8,8 mmol
Suhu ruang 300C
Pada label ρ etanol = 0,79
Pada label ρ asam asetat = 1,042
ERLENMEYER 1
V1 = 44,2 ml
Mencari mol mula etanol
ρ etanol = m
V 1etanol
M = ρ etanol x V etanol
= 0,798 . 1 ml
n = gMr
= 0,798
46 = 0,0171 mol
Mencari mol mula mula asam asetat
ρ asam asetat = mV
m = ρ asam asetat . v asam asetat
= 1,042 . 4 ml
= 4,17 g
n = gmr
= 4,1760
=0,0695
mol H+ sisa = mol H+ total – mol blanko
= mol NaOH – mol Blanko
=V1.N – mol Blanko
= 88,4 – 8,8mmol
= 0,0796 mol
CH3COOH + C2H5OOH ↔ CH3COOC2H5 + H2O
M 0,0695 0,017
R -0,0101 -0,0101 -0,0101 -0,0101
___________________________________________________
S 0,0796 0,0272 -0,0101 -0,0101
[CH3COOH] = -0,0101mol/0,01 L = -1,01 M
[ C2H5OH] = 0,0277 mol/ 0,01 L = 2,72 M
[CH3COOC2H5] = 0,0796 mol/0,01 L = 7,96 M
Kc = CH3COOC2H5][CH3COOH] [C2H5OH]
= −1,0121,6512
=−0,0466
ERLENMEYER 2
V = 30,8 ml
Mencari mol mula etanol
ρ etanol = m
V 2etanol
m = ρ etanol x V etanol
= 0,798 . 2 ml
=1,58
n = gMr
= 1,596
46 = 0,0343 mol
Mencari mol mula mula asam asetat
ρ asam asetat = mV
m = ρ asam asetat . v asam asetat
= 1,042 . 3 ml
= 3,126 g
n = gmr
= 3,126
60=0,05121mol
mol H+ sisa
= mol H+ total – mol blanko
= mol NaOH – mol Blanko
= 30,8 . 2 – 8,8mmol
= 0,0528 mol
CH3COOH + C2H5OOH CH3COOC2H5 + H2O
M 0,05121 0,0343
R -0,0007 -0,0007 -0,0007 -0,0007
___________________________________________________
S 0,0528 0,035 -0,0007 -0,0007 [CH3COOH] = 0,0528mol/0,01 L = 5,28 M
[ C2H5OH] = 0,035mol/ 0,01 L = 3,5 M
[CH3COOC2H5] = -0,0007mol/0,01 l = −0,07 M
Kc = CH3COOC2H5][CH3COOH] [C2H5OH]
=−0,0718,48
=−0,0038
ERLENMEYER 3
V3 = 20,6 ml
Mencari mol mula etanol
ρ etanol = m
V 2etanol
M = ρ etanol x V etanol
= 0,798 . 3 ml
= 2,394
n = gMr
= 2,394
46 = 0,0515 mol
Mencari mol mula mula asam asetat
ρ asam asetat = mV
m = ρ asam asetat . v asam asetat
= 1,042 . 2 ml
= 2,084 g
n = gmr
= 2,084
61=0,0347mol
mol H+ sisa
= mol H+ total – mol blanko
= mol NaOH – mol Blanko
= V3.N – mol Blanko
= 20,6 . 2 – 8,8mmol
= 0,0324 mol
CH3COOH + C2H5OOH CH3COOC2H5 + H2O
M 0,0347 0,0515
R 0,0023 0,0023 0,0023 0,0023___________________________________________________
S 0,0324 0,0492 0,0023 0,0023
[CH3COOH] = 0,0324mol/0,01 L = 3,24 M
[ C2H5OH] = 0,0492mol/ 0,01 L = 4,92 M
[CH3COOC2H5] = 0,023mol/0,01 L = 0,23 M
Kc = CH3COOC2H5][CH3COOH] [C2H5OH]
= 0,2315,9408
=0,0144
ERLENMEYER 4
V4 = 12,9 ml
Mencari mol mula etanol
ρ etanol = m
V 2eta nol
M = ρ etanol x V etanol
= 0,798 . 12,9 ml
= 3,16 g
n = gMr
= 3,192
46 = 0,0687mol
Mencari mol mula mula asam asetat
ρ asam asetat = mV
m = ρ asam asetat . v asam asetat
= 1,042 . 1 ml
= 1.042 g
n = gmr
= 1,042
60=0,01736mol
mencari mol H+ sisa
= mol H+ total – mol blanko
= mol NaOH – mol Blanko
= V3.N – mol Blanko
= 12,9.2 – 8,8mmol
= 0,017 mol
CH3COOH + C2H5OOH CH3COOC2H5 + H2O
M 0,01736 0,0687
R 0,00036 0,00036 0,00036 0,00036
___________________________________________________
S 0,017 0,06834 0,00036 0,00036
[CH3COOH] = 0,017 mol/0,01 L = 1,7 M
[ C2H5OH] = 0,06834mol/ 0,01 L = 6,834 M
[CH3COOC2H5] = 0,00036mol/0,01 l = 0,036 M
Kc = CH3COOC2H5][CH3COOH] [C2H5OH]
= 0,03611,6365
=0,0031
Kc Rata rata = −0,0480+(−0,0038)+0,0144+0,0031
4=¿0,008575
