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Titrasi Asam Basa : 25 Contoh Soal dan Pembahasanya

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Kelas XI >> Pada kesempatan kali ini, kita akan membahas contoh soal mengenai titrasi asam basa yang mana pada soal titrasi asam basa ini kita juga akan membahas mengenai Kurva Grafik asam basa, Menghitung PH titrasi  asam basa dan banyak materi lainya. Selamat Mengerjakan !

1. Untuk menentukan konsentrasi larutan HNO3, diambil sekitar 50 ml larutan tersebut kemudian dititrasi dengan larutan NaOH 0,2 M. Data proses titrasi yang didapat ialah sebagai berikut :

Percobaan 1 :

Volume larutan HNO3 = 10 ml
Volume larutan NaOH = 12 ml

Percobaan 2 :

Volume larutan HNO3 =  10 ml
Volume larutan NaOH = 18 ml

Berdasarkan data tersebut, maka tentukanlah konsentrasi larutan HNO3 !

Pertama, kita cari volume rata-rata dari larutan HNO3 dan NaOH !

Vrata-rata HNO3 = 10 + 10/2 = 10 ml
Vrata-rata NaOH = 12 + 18/2 = 15 ml

Untuk mencari molaritas dari larutan HNO3, kita gunakan rumus :

V1 . M1 . n1 =  V2 . M2 . n2

Ket : V = Volume Larutan
M = Molaritas
n = Valensi dari larutan asam atau basa ( jumlah atom H atau OH )

Maka :

V1 . M1 . n1 = V2 . M2 . n2
15 . 0,2 . 1 = 10 . M2 . 1
3 = 10.M2
M2 = 3/10 = 0,3 M konsentrasi dari HCl

2. Berikut kurva Titrasi larutan Asam Klorida dengan larutan NaOH
Jika 15 ml larutan HCl diteteskan dengan larutan NaOH 0,1 M , maka konsentrasi larutan HCl yang dititrasi ialah.... M

- Untuk mencari molaritas dari HCl maka rumus yang kita gunakan ialah....

V1 . M1 . n1 = V2 . M2 . n2
30 . 0,1 . 1 = 15 . M2 . 1
3 = 15 M2
M2 = 3/15 = 0,2 M

3. Perhatikkan data hasil titrasi larutan HBr dengan larutan Ba(OH)2 berikut !

Percobaan 1 :

Volume HCl = 25 ml
Volume Ba(OH)2 = 22 ml

Percobaan 2 :

Volume HCl = 25 ml
Volume Ba(OH)2 = 18 ml

Percobaan 3 :

Volume HCl = 25 ml
Volume Ba(OH)2 = 20 ml

Berdasarkan data tersebut, konsentrasi larutan HCl adalah....

Pertama, kita cari volume rata-rata dari HCl dan Ba(OH)2 !

- Volume HCl = 25+25+25 / 3 = 25 ml
- Volume Ba(OH)2 = 22+18+20 / 3 = 20 ml

Kemudian kita cari molaritas HCl dengan menggunakan rumus berikut :

V1 . M1 . n1 = V2 . M2 . n2
25 . 0,1 . 2 = 20 . M2 . 1
5 = 20 M2
M2 = 0,25 M

4. Sebanyak 10 ml Asam Klorida diperlukan untuk menetralkan 20 ml larutan Ca(OH)2 0,1 M. Molaritas Asam Klorida adalah..... M

- Untuk mencari molaritas dari larutan HCl, kita dapat menggunakan rumus :

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

5. Sebanyak 25 ml asam cuka dititrasi dengan larutan NaOH 0,01 M. data hasil titrasi yang diperoleh ialah sebagai berikut :

1. Volume NaOH = 51 ml
2. Volume NaOH = 49 ml
3. Volume NaOH = 50 ml

Berdasarkan data tersebut, Konsentrasi Asam Cuka ialah sebesar .... M

Untuk mencari konsentrasi dari Asam Cuka, kita gunakan rumus :

V1 . M1 . n1 = V2 . M2 . n2
50 . 0,01 . 1 = 25 . M2 . 1
0,5 = 25 M2
M = 0,02

6. Sebanyak 0,5 gr cuplikan KOH ( Mr KOH = 56 ) dilarutkan ke dalam air dan memerlukan 25 ml larutan H2SO4 0,1 M untuk menetralkanya. Kadar KOH dalam cuplikan tersebut adalah....

Untuk mencari kadar KOH, Pertama mari kita cari dulu molaritas dari KOH dengan menggunakan rumus :

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

Setelah itu, kita cari molaritas dari KOH dengan cara menggunakan volume dari hasil titrasi KOH

N = gr/Mr x 1000/Vtitrasi = 0,5/56 x 1000/25 = 0,357

Setelah itu, kita bagi molaritas hasil titrasi dengan hasil dari pencarian molaritas basa :

0,2/0,357 = 0,56 x 100 = 56%

7. Sebanyak 2 gram cuplikan NaOH ( Mr = 40 ) dialrutkan ke dalam air dan memerlukan 100 ml larutan H2SeO4 0,4 M untuk menetralkanya. Kadar NaOH dalam cuplikan tersebut adalah.... %

- Sama seperti soal sebelumnya, langkah pertama yang harus kita lakukan ialah mencari molaritas dari NaOH :

V1 . M1 . n1 = V2 .M2 . n2
100 . 0,4 . 2 = 100 . M2 .1
8 = 100 . M2
M = 0,08 M

Setelah itu, kita cari molaritas dari NaOH dengan menggunakan volume dari titrasi NaOH :

M = gr/Mr x 1000/VTitrasi = 2/40 x 1000/100 = 2000/4000 = 0,5 M

Setelah itu, kita bagi molaritas hasil titrasi dengan hasil dari pencarian molaritas biasa :

0,08 / 0,5 = 0,16 x 100 = 16 %

8. Sebanyak 25 ml NaOH 0,5 M dapat dinetralkan oleh HCl 20 ml. Hitunglah molaritas dari HCl !

Rumus : V1 . M1 . n1 = V2 . M2 . n2
25 . 0,5 . 1 = 20 . M . 1
12,5 = 20 M
M = 0,625 M

9. 35 ml larutan HBr 0,02 M dapat digunakan untuk mentitrasi 25 ml larutan KOH sampai pada titik ekuivalen, berapakah konsentrasi dari larutan KOH ?

Seperti biasa, rumus yang kita gunakan ialah :

V1 . M1 . n1 = V2 . M2 . n2
35 . 0,02 . 1 = 25 . M2 . 1
0,7 = 25 . M2
M2 = 0,7 /25 = 0,028 M

10. Berapakah volume 0,2 M HBr untuk mentitrasi larutan berikut sampai pada titik ekuivalen ?

A. 30 ml larutan NaOH 0,15 M
B. 25 ml larutan NH3 0,5 M

Rumus yang kita gunakan adalah :

V1 . M1 . n1 = V2 . M2 . n2

A. V1 . 0,2 . 1 = 30 . 0,15 . 1
     V1 . 0,2 = 4,5
     V1 = 22,5 ml

B.  V1 . 0,2 . 1 = 25 . 0,5 . 1
      V1 . 0,2 = 12,5
      V1 = 62,5 ml

Soal Selanjutnya >> 11-25
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5 Responses to "Titrasi Asam Basa : 25 Contoh Soal dan Pembahasanya"

  1. halo kak salam kenal boleh minta nomor wa untuk belajar saya

    ibnu +62-83823027421

    ReplyDelete
  2. soal dan jawabannya

    cukup bagus tapi yang saya tanyakan emang semua soal dalam kategori titrasi

    meskipun untuk mencari molaritas dan sebagainya menggunakan M1.V1.n1=M2.V2.n2

    mohon segera di tanggapi
    mohon maaf apabila saran/kritikan saya menyinggung perasaan
    terima kasih

    ReplyDelete
  3. Cukup bagus

    tapi saya tanya untuk soal dalam kategori titrasi
    meskipun untuk mencari molaritas dan molalitas dsb
    apakah menggunakan rumus yang sama yaitu M1.V1.n1=M2.V2.n2

    mohon segera di tanggapi

    apabila kritikan/saran saya menyingggung perasaan anda

    saya mohon maaf yang sebesar besarnya
    terima kasih

    ReplyDelete
  4. Kenapa gaada soal menetukan pH titrasi

    ReplyDelete

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