Roboguru

Pada titrasi 50 mL   0,1 M   dengan NaOH 0,1 M. Tentukan nilai pH larutan setelah penambahan NaOH sebesar: a. 0 mL b. 25 mL

Pertanyaan

Pada titrasi 50 mL C H subscript 3 C O O H  0,1 M left parenthesis Ka space C H subscript 3 C O O H equals 1 comma 8 space x space 10 to the power of negative sign 5 end exponent right parenthesis  dengan NaOH 0,1 M. Tentukan nilai pH larutan setelah penambahan NaOH sebesar:

a. 0 mL

b. 25 mL

Pembahasan Soal:

Titrasi yang terjadi di atas adalah asam lemah dengan basa kuat.

a. Volume NaOH 0 mL

Nilai pH larutan setelah penambahan NaOH sebesar 0 mL artinya belum ada penambahan NaOH.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of Ka space X space M end root end cell row blank equals cell square root of left parenthesis 1 comma 8 space x space 10 to the power of negative sign 5 end exponent end root right parenthesis x space 10 to the power of negative sign 1 end exponent end cell row blank equals cell square root of 1 comma 8 space x space 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 1 comma 34 space x space 10 to the power of negative sign 3 end exponent end cell row pH equals cell negative sign log open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space left parenthesis 1 comma 34 space x space 10 to the power of negative sign 3 end exponent right parenthesis end cell row blank equals cell 3 minus sign log space 1 comma 34 end cell row blank equals cell 2 comma 87 end cell row blank blank blank end table

Jadi, nilai pH larutan setelah penambahan NaOH sebesar 0 mL adalah 2,87.

 

b.  Volume NaOH 25 mL

  • mol asam lemah = 0,1 M x 50 mL = 5 mmol
  • mol basa kuat = 0,1 M x 25 mL = 2,5 mmol

C H subscript 3 C O O H and Na O H yields C H subscript 3 C O O Na and H subscript 2 O 

 

Berdasarkan reaksi di atas terbentuk larutan penyangga asam karena jumlah mol asam lemah berlebih.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell Ka space x space fraction numerator mmol space asam space lemah over denominator mol space basa space konjugasi forward slash garam end fraction end cell row blank equals cell left parenthesis 1 comma 8 space x space 10 to the power of negative sign 5 end exponent right parenthesis space x space space fraction numerator 2 comma 5 space mmol over denominator 2 comma 5 space mmol end fraction end cell row blank equals cell 1 comma 8 space x space 10 to the power of negative sign 5 end exponent end cell row pH equals cell negative sign log open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space left parenthesis 1 comma 8 space x space 10 to the power of negative sign 5 end exponent right parenthesis end cell row blank equals cell 5 minus sign log space 1 comma 8 end cell row blank equals cell 4 comma 74 end cell end table 

Jadi, nilai pH larutan setelah penambahan NaOH sebesar 25 mL adalah 4,74.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

G. Suprobo

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 02 Mei 2021

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Pembahasan Soal:

Fenol merupakan asam lemah, maka rumus mencari pH asam lemah adalah


table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of K subscript a cross times M end root end cell row blank equals cell square root of 10 to the power of negative sign 10 end exponent cross times 10 to the power of negative sign 1 end exponent end root end cell row blank equals cell square root of 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell 10 to the power of negative sign 5 comma 5 end exponent end cell end table 


table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 10 to the power of negative sign 5 comma 5 end exponent end cell row blank equals cell 5 comma 5 end cell end table 


Jadi, jawaban yang tepat adalah D.

Roboguru

Untuk titrasi 25 ml larutan  0,15 M dengan larutan 0,1 M . Tentukan: perkıraan pH pada titik ekuivalen

Pembahasan Soal:

Pada saat titrasi berlangsung, keadaan secara stoikiometri titran dan titrat tepat habis bereaksi disebut titik ekuivalen. Artinya, saat titik ekuivalen jumlah mol ekuivalen asam sama dengan mol ekuivalen basa.

Pada soal ini, titik ekuivalen terjadi ketika jumlah mol undefined yang ada dalam larutan telah habis semuanya bereaksi dengan mol NaOH yang ditambahkan. Untuk menentukan jumlah NaOH yang ditambahkan, dapat dihitung dengan rumus titrasi sebagai berikut:
table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript C H subscript 3 C O O H end subscript cross times V subscript C H subscript 3 C O O H end subscript end cell equals cell M subscript NaOH cross times V subscript NaOH end cell row cell V subscript NaOH end cell equals cell fraction numerator M subscript C H subscript 3 C O O H end subscript cross times V subscript C H subscript 3 C O O H end subscript over denominator M subscript NaOH end fraction end cell row blank equals cell fraction numerator 0 comma 15 space M cross times 25 space mL over denominator 0 comma 1 space M end fraction end cell row blank equals cell 37 comma 5 space mL end cell end table 


Adapun jumlah mol zat masing-masing sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 3 C O O H end subscript space mula bond mula end cell equals cell M cross times V end cell row blank equals cell 0 comma 15 space M cross times 25 space mL end cell row blank equals cell 3 comma 75 space mmol end cell end table  

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript NaOH space mula bond mula end cell equals cell M cross times V end cell row blank equals cell 0 comma 1 space M cross times 37 comma 5 space mL end cell row blank equals cell 3 comma 75 space mmol end cell end table 

Sehingga diperoleh reaksi demikian:


Zat yang tersisa pada saat titik ekivalen adalah garam begin mathsize 14px style C H subscript 3 C O O Na end style yang dapat terhidrolisis dan bersifat basa. Maka pH larutan dapat dihitung dengan rumus hidrolisis garam basa sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets C H subscript 3 C O O Na close square brackets end cell equals cell mol over V subscript total end cell row blank equals cell fraction numerator 3 comma 75 mmol over denominator left parenthesis 25 plus 37 comma 5 right parenthesis mL end fraction end cell row blank equals cell fraction numerator 3 comma 75 over denominator 62 comma 5 end fraction end cell row blank equals cell 0 comma 06 space M end cell row blank blank blank row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times open square brackets C H subscript 3 C O O Na close square brackets end root end cell row blank equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent cross times 0 comma 06 end root end cell row blank equals cell square root of 6 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell 7 comma 7 cross times 10 to the power of negative sign 6 end exponent end cell row blank blank blank row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log 7 comma 7 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 5 comma 1 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign 5 comma 1 end cell row blank equals cell 8 comma 9 end cell end table 


Jadi, pH pada saat titik ekuivalen adalah 8,9.

Roboguru

Untuk titrasi 25 ml larutan  0,15 M dengan larutan NaOH 0,1 M Tentukan: perkiraan pH awal larutan

Pembahasan Soal:

pH awal larutan adalah pH larutan begin mathsize 14px style C H subscript 3 C O O H end style dan larutan NaOH sebelum dicampurkan.

Larutan begin mathsize 14px style C H subscript 3 C O O H end style merupakan suatu asam lemah

Jika Ka dianggap 1 middle dot 10 to the power of negative sign 5 end exponent, maka
table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of Ka cross times M end root end cell row blank equals cell square root of 1 middle dot 10 to the power of negative sign 5 end exponent cross times 0 comma 15 end root end cell row blank equals cell square root of 1 middle dot 10 to the power of negative sign 5 end exponent cross times 1 comma 5 middle dot 10 to the power of negative sign 1 end exponent end root end cell row blank equals cell square root of 1 comma 5 cross times 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 1 comma 2 cross times 10 to the power of negative sign 3 end exponent space M end cell row pH equals cell negative sign log open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log left parenthesis 1 comma 2 cross times 10 to the power of negative sign 3 end exponent right parenthesis end cell row blank equals cell 3 minus sign log space 1 comma 2 end cell end table   
JIka diketahui log 1,2 = 0,079, maka 
table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 3 minus sign 0 comma 079 end cell row blank equals cell 2 comma 921 end cell end table 

Larutan NaOH merupakan basa kuat

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell M cross times valensi end cell row blank equals cell 0 comma 1 cross times 1 end cell row blank equals cell 0 comma 1 space M end cell row blank equals cell 1 middle dot 10 to the power of negative sign 1 end exponent space M end cell row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log left parenthesis 1 middle dot 10 to the power of negative sign 1 end exponent right parenthesis end cell row blank equals cell 1 minus sign log 1 end cell row blank equals 1 row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign 1 end cell row blank equals 13 end table    


Jadi, pH awal larutan begin mathsize 14px style C H subscript 3 C O O H end style adalah 2,921 dan pH awal larutan NaOH adalah 13.

Roboguru

Pada titrasi 50 mL  0,1M  dengan  0,1M. Tentukan nilai pH larutan setelah penambahan  sebesar: a. 0 mL b. 25 mL c. 50 mL d. 75 mL

Pembahasan Soal:

a. sebelum penambahan basa berarti hanya ada asam sehingga untuk menentukan pH nya digunakan pH penentuan asam lemah


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of K subscript a cross times left square bracket Asam space lemah right square bracket end root end cell row blank equals cell square root of 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times 0 comma 1 M end root end cell row blank equals cell square root of 1 comma 8 cross times 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell square root of 1 comma 8 end root cross times 10 to the power of negative sign 3 end exponent end cell row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space square root of 1 comma 8 end root cross times 10 to the power of negative sign 3 end exponent end cell row blank equals cell 3 minus sign log space square root of 1 comma 8 end root end cell end table end style


b. Untuk menentukan pH setelah penambahan begin mathsize 14px style Na O H end style 0,25mL maka perlu dilakukan perhitungan stoikiometri


begin mathsize 14px style mol subscript C H subscript 3 C O O H end subscript equals 0 comma 1 M cross times 50 mL equals 5 mmol mol subscript NaOH equals 0 comma 1 M cross times 25 mL equals 2 comma 5 mmol end style



Pada reaksi tersebut yang bersisa adalah asam lemah dan garamnya sehingga untuk menentukan pH nya digunakan pH larutan penyangga


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell K subscript a cross times fraction numerator mol space asam over denominator mol space garam end fraction end cell row blank equals cell 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 2 comma 5 over denominator 2 comma 5 end fraction end cell row blank equals cell 1 comma 8 cross times 10 to the power of negative sign 5 end exponent end cell row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 1 comma 8 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 1 comma 8 end cell end table end style


c. Untuk menentuka pH setelah penambahan 50mL larutan begin mathsize 14px style Na O H end style 0,1M maka dilakukan perhitungan stoikiometri terlebih dahulu


begin mathsize 14px style mol subscript NaOH equals 0 comma 1 M cross times 50 mL equals 5 mmol end style


 


Pada perhitungan tersebut yang tersisa adalah garamnya saja sehingga untuk menentukan pH larutannya digunakan pH hidrolisis


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Garam close square brackets end cell equals cell fraction numerator 5 mmol over denominator 100 mL end fraction end cell row blank equals cell 0 comma 05 M end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times open square brackets garam close square brackets end root end cell row blank equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 1 comma 8 cross times 10 to the power of negative sign 5 end exponent end fraction cross times 0 comma 05 end root end cell row blank equals cell square root of 2 comma 8 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell square root of 2 comma 8 end root cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log space square root of 2 comma 8 end root cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row blank equals cell 5 comma 5 minus sign log space square root of 2 comma 8 end root end cell row pH equals cell 14 minus sign left parenthesis 5 comma 5 minus sign log space square root of 2 comma 8 end root right parenthesis end cell row blank equals cell 8 comma 5 plus log square root of 2 comma 8 end root end cell end table end style 


d. Untuk menentuka pH setelah penambahan 75mL larutan begin mathsize 14px style Na O H end style 0,1M maka dilakukan perhitungan stoikiometri terlebih dahulu


begin mathsize 14px style mol subscript NaOH equals 75 mL cross times 0 comma 1 M equals 7 comma 5 mmol end style


 


Dari hasil perhitungan yang bersisa adalah basa kuat dan garam maka untuk menentukan pH larutan tersebut digunakan konsentrasi dari basa kuat


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript NaOH end cell equals cell fraction numerator 1 comma 5 mmol over denominator 75 mL and 25 mL end fraction end cell row blank equals cell 0 comma 015 M end cell row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log space 1 comma 5 cross times 10 to the power of negative sign 2 end exponent end cell row blank equals cell 2 minus sign log space 1 comma 5 end cell row pH equals cell 14 minus sign left parenthesis 2 minus sign log 1 comma 5 right parenthesis end cell row blank equals cell 12 plus log space 1 comma 5 end cell end table end style


Jadi, pH pada kondilsi a, b, c, dan d berturut-turut begin mathsize 14px style bottom enclose bold 3 bold minus sign bold log square root of bold 1 bold comma bold 8 end root end enclose end style; 5-log1,8; begin mathsize 14px style bottom enclose bold 8 bold comma bold 5 bold plus bold log square root of bold 2 bold comma bold 8 end root end enclose end style dan 12+log1,5.space 

Roboguru

Larutan sebanyak 20 ml dititrasi dengan larutan 0,1 M dengan data sebagai berikut: Berdasarkan data tersebut, konsentrasi larutan adalah ...

Pembahasan Soal:

Titrasi menggunakan konsep dasar penetralan asam dan basa.

M o l space a s a m cross times v a l e n s i space a s a m equals m o l space b a s a cross times v a l e n s i space b a s a  M space C H subscript 3 C O O H cross times V space C H subscript 3 C O O H cross times v a l e n s i space a s a m equals space M space B a left parenthesis O H right parenthesis subscript 2 cross times V space B a left parenthesis O H right parenthesis subscript 2 cross times v a l e n s i space b a s a space B a left parenthesis O H right parenthesis subscript 2  M space C H subscript 3 C O O H equals fraction numerator 0 comma 1 space M cross times 15 space m L cross times 2 over denominator 20 space m L end fraction equals 0 comma 150 space M

Roboguru

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