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Titrasi Asam Basa : Contoh Soal 11-25

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span id="goog_1146892688">11. 31 ml larutan Ba(OH)2 0,35 M dititrasi dengan HI 0,690 M. Berapakah volume HF yang digunakan untuk mencapai titik ekuivalen titrasi ?
  
V1 . M1 . n1 = V2 . M2 . n2
31 . 0,135 . 2 = V2 . 0,690 . 1
8,37 = 0,69 . V2 
V2 = 8,37 / 0,69 = 12,13 ml

12. Di dalam air soda terdapat Asam Fosfat dengan konsentrasi 0,00015 M. Bila air soda tersebut dapat dinetralkan oleh 31,67 ml asam cuka 0,118 M maka berapakah volume dari air soda tersebut ?

V1 . M1 . n1 = V2 . M2 . n2 
V1 . 0,00015 . 2 = 31,67 . 0,118 .1 
0,003 . V1 = 3,73
V1 = 1245 ml = 1,245 L

13. Sebanyak 250 ml larutan Asam Nitrat dinetralkan oleh 150 ml larutan KOH 0,15 M. konsentrasi larutan Asam Nitrat tersebut ialah....

V1 . M1 . n1 = V2 . M2 . n2
250 . M1 . 1 = 150 . 0,15 . 1
250 . M1 = 22,5
M = 22,5 / 250 = 0,09 M

14. 10 gr Tablet yang mengandung Vitamin C dilarutkan ke dalam 200 ml air. Bila ternyata 100 ml larutan vitamin C tersebut dapat menetralkan 50 ml larutan basa yang memiliki konsentrasi 0,02 M. Maka tentukanlah kadar vitamin C yang terdapat di dalam tablet tersebut ( dalam % ). ( Perbandingan ekuivalen 1 : 1 )

Pertama, kita cari konsentrasi dari larutan vitamin C tersebut

V1 . M1 . n1 = V2 . M2 . n2
100 . M1 . 1 = 50 . 0,02 . 1
100 . M1 = 1
M = 0,01

Dalam 200 ml vitamin C terdapat  = 200 . 0,01 / 1000 ml = 0,002 mol vitamin C

Maka massa vitamin C = Mol vitamin C . Mr Vitamin C
                                      = 0,002 . 128 = 0,256 gr

Maka kadar dari vitamin C = 0,256 x 100 / 10 = 2,56 %

15. Ke dalam 200 ml air, dialirkan gas SO2, lalu sebanyak 20 ml larutan tersebut dititrasi dan tepat mencapai titik ekuivalen dengan 120 ml larutan NaOH 0,2 M. Maka tentukanlah volume gas SO2  yang dialirkan  ( diukur pada keadaan STP ) !

Pertama, kita buat reaksinya terlebih dahulu !

SO2 + H2O ⇄ H2SO3

Penetralan oleh NaOH

H2SO3 + 2 NaOH → K2SO3 + 2 H2O

nNaOH = 120 ml x 0,2  = 24 mmol
nH2SO3 = 1/2 x 24 mmol = 12 mmol  ( dalam 20 ml larutan )

Dalam 200 ml larutan H2SO3 = 200/20 x 12 = 120 mmol 

nSO2 = nH2SO3 ( koofesien sama ) = 120 mmol = 0,12 mol
V SO2 = 0,12 x 22,4 = 2,688 Liter  

16. 20 gr Tablet mengandung Asam Askorbat yang dilarutkan ke dalam 400 ml air. Bila ternyata 200 ml larutan asam askorbat tersebut dapat menetralkan 100 ml larutan basa yang memiliki konsentrasi 0,04 M . Maka tentukanlah kadar asam askorbat yang terdapat di dalam tablet tersebut ( dalam % ). ( perbandingan ekuivalen 1 : 1 ) 

Pertama, kita cari konsentrasi dari larutan Asam Askorbat tersebut !

V1 . M1 . n1 = V2 . M2 . n2  
200 . M1 . 1 = 100 . 0,04 . 1
200 . M1 = 4
M = 0,02

Dalam 400 ml, Asam Askorbat terdapat = 400. 0,02 / 1000 = 0,008 mol asam askorbat

Maka massa vitamin C = Mol Asam Askorbat x Mr Asam Askorbat
                                      = 0,008 x 256 = 2,048 gr

Maka kadar vitamin C = 2,048 x 100 / 20 = 10,24 %

17. Ke dalam 400 ml Air, dialirkan gas NO2. Lalu sebanyak 40 ml larutan tersebut dititrasi dan dapat mencapai titik ekuivalen dengan 240 ml larutan KOH 0,4 M. Maka tentukanlah volume gas NO2 yang dialirkan.

Pertama, kita buat reaksinya terlebih dahulu !

NO2 + H2O → HNO3

HNO3 + KOH → KNO3 + H2O

nKOH = 240 x 0,4 = 96 mmol
nHNO3 = 1/1 x 96 = 96 mmol ( dalam 40 ml )

Dalam 400 larutan HNO3 = 400/40 x 96 = 960 mmol

nNO2 = N HNO3 ( koofesienya sama ) = 960 mmol = 0,96 mol

V NO2 = 0,96 x 22,4 = 21,5 Liter

18. Dilakukan percobaan titrasi 25 ml larutan KOH 0,1 M dengan larutan HCl 0,1 M menurut reaksi berikut :

KOH + HCl → KCl + H2O

Buatlah kurva grafik yang menunjukan hasil titrasi tersebut....

Pertama, kita cari volume dari HCl :

V1 . M1 . n1 = V2 . M2 . n2
25 . 0,1 . 1 = V2 . 0,1 . 1
2,5 = 0,1 . V2
V2 = 25 ml

Maka kurva titrasinya ialah :


19. Dilakukan percobaan titrasi 15 ml larutan CH3COOH 0,2 M dengan larutan NaOH 0,1 M menurut reaksi berikut.

CH3COOH + NaOH → CH3COONa + H2O

Buatlah kurva grafik yang menunjukan hasil titrasi tersebut !

Pertama, kita cari volume dari NaOH!

V1 x M1 x n1 = V2 x M2 x n2
15 x 0,1 x 1 = V2  x 0,1 x 1
1,5 = 0,1
1,5 =  0,1
V2 = 30 ml

Maka kurva titrasinya ialah ...
20.  Sebanyak 20 ml Asam Nitrat diperlukan untuk menetralkan 40 ml Ca(OH)2 0,1 M. Molaritas Asam nitrat tersebut adalah.... M

V1 . M1 . n1 = V2 . M2 . n2
40 . 0,1 . 2 = 20 . M2 . 1
8 = 20 . M2
M = 0,4 M

21. Jika pada titrasi 25 mL NaOH membutuhkan 15 mL HCl 0,1 M, maka berapakah massa NaOH yang terlarut dalam 100 mL larutan NaOH tersebut? (Ar Na=23 ,Ar O= 16,dan Ar H= 1)

Pembahsan : Va x Ma = Vb x Mb
15 x 0,1 = 25 x Mb
 1,5 = 25Mb
Mb = 0,06

M = Massa/mr x 1000/v
massa = M x Mr x V/1000
massa = 0,06 x 40 x 100 / 1000
massa = 0,24 gr

22.. Jika pada titrasi 50 mL NaOH membutuhkan 40 mL HCl 0,1 M . Maka berapakah massa NaOH yang terdapat dalam larutan tersebut? (diketahui Ar Na=23 ,Ar O= 16,dan Ar H= 1).

Jawaban : 0, 16 gr

Pembahasan :
 
Va x Ma = Vb x Mb
40 x 0,1 = 50 x Mb
4 = 50Mb
Mb = 0,08

Massa = M x Mr x V / 1000
Massa = 0,08 x 40 x 50 / 1000  =  0,16 gr

23. Jika pada titrasi 70 mL Mg(OH)2 membutuhkan 65 mL HCl 0,1 M .maka berapakah massa Mg(OH)2 yang terdapat dalam larutan tersebut? (Ar Mg= 24 ,Ar O= 16,dan Ar H= 1)

Jawaban : 0 ,1885 gr
 
Pembahasan :
 
Va x Ma = Vb x Mb x indeks basa
65 x 0,1 = 70 x Mb x 2
6,5 = 140 Mb
Mb =  0,04

Massa = M x Mr x V / 1000
Massa = 0,04 x 58 x 70 / 1000  =  0,18 gr

24. Jika pada titrasi 35 mL NaOH membutuhkan 10,15 mL HNO3 0,1 M dengan indikator fenolftalein, maka berapakah massa NaOH yang terlarut dalam 100 mL larutan NaOH tersebut? (Ar Na=23 ,Ar O= 16,dan Ar H= 1)

Jawaban : 0,116 gr
 
Pembahasan :
 
Va x Ma = Vb x Mb
10,15 x 0,1 = 35 x Mb
1,015 = 35 Mb
Mb = 0,029

Massa = M x Mr x V / 1000
Massa = 0,029 x 40 x 100 / 1000  =  0,116 gr

25. Jika pada titrasi 25,15 mL Ca(OH)2 membutuhkan 15,3 mL H 2SO4 0,1 M dengan indicator fenolftalein, Berapakah Massa Ca(OH)2 yang diperlukan untuk membuat 100 mL larutan Ca(OH)2 tersebut? (Ar Ca=40 ,Ar O= 16,dan Ar H= 1)

Jawaban : 0,4 502 gr
 
Pembahasan :
 
Va x Ma x i.a = Vb x Mb x i.b
15,3 x 0,1 x 2 = 25,15 x Mb x 2
3,06 = 50,3 Mb
Mb = 0,06


Massa = M x Mr x V / 1000
Massa = 0,029 x 74 x 100 / 1000  =  0,4502 gr
 
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