Roboguru

Larutan asam asetat  0,2 M sebanyak 200 mL ditambahkan padatan  (Mr = 82) 3,28 g hingga homogen. Ketika ketambahan 5 mL HCl 0,1 M, larutan memiliki perubahan pH dari ....

Pertanyaan

Larutan asam asetat left parenthesis C H subscript 3 C O O H comma space Ka equals 1 x 10 to the power of negative sign 5 end exponent right parenthesis space 0,2 M sebanyak 200 mL ditambahkan padatan C H subscript 3 C O O Na space (Mr = 82) 3,28 g hingga homogen. Ketika ketambahan 5 mL HCl 0,1 M, larutan memiliki perubahan pH dari ....

  1. 5 menjadi 6 - log 6,25space

  2. 5 menjadi 5 - log 1,25space space

  3. 5 menjadi 5 - log 1,025space space

  4. 5 - log1,025 menjadi 5space space

  5. 5 - log1,25 menjadi 5 - log 1,025space

Pembahasan Soal:

Campuran antara asam lemah dengan garam/ basa konjugasi akan terbentuk larutan penyangga asam.

Rumus Larutan Penyangga Asam:

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 mol space asam space lemah over denominator mol space basa space konjugasi end fraction end cell row pH equals cell negative sign log open square brackets H to the power of plus sign close square brackets end cell end table

 

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space C H subscript 3 C O O H end cell equals cell 0 comma 2 space M space x space 200 space mL equals 40 space mmol equals 0 comma 04 space mol space end cell row cell mol space C H subscript 3 C O O Na end cell equals cell fraction numerator 3 comma 28 space g space over denominator 82 space g forward slash mol end fraction equals 0 comma 04 space mol end cell row blank blank blank row cell open square brackets H to the power of plus sign close square brackets end cell equals cell left parenthesis 1 x 10 to the power of negative sign 5 end exponent right parenthesis space x fraction numerator space 0 comma 04 space mol over denominator 0 comma 04 space mol end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent end cell row pH equals cell negative sign log space left parenthesis 10 to the power of negative sign 5 end exponent right parenthesis end cell row blank equals 5 end table

 

Ketika penambahan HCl maka C H subscript 3 C O O Na space yang akan bereaksi.

mol space H Cl space equals space 0 comma 1 space M space x space 5 space mL equals space 0 comma 5 space mmol space equals space 0 comma 0005 space mol

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

 

 

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 fraction numerator space mol space C H subscript 3 C O O H over denominator space mol space C H subscript 3 C O O Na end fraction end cell row blank equals cell open parentheses 1 x 10 to the power of negative sign 5 end exponent close parentheses space x fraction numerator space 0 comma 0405 space mol over denominator 0 comma 0395 space mol end fraction end cell row blank equals cell 1 comma 025 space x 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 left parenthesis 1 comma 025 space x 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 025 end cell end table

Perubahan pH yang terjadi adalah dari 5 menjadi 5 - log 1,025.

 

Jadi, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

G. Suprobo

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 02 Mei 2021

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pH larutan dari campuran 100 mL larutan NaOH 0,2 M dengan 100 mL larutan CH3COOH 0,5 M (Ka CH3COOH = 104) adalah...

Pembahasan Soal:

menentukan mol masing-masing :

begin mathsize 14px style mol space Na O H double bond M cross times V space space space space space space space space space space space space space space space space space space equals 0 comma 2 M cross times 100 mL space space space space space space space space space space space space space space space space space equals 20 mmol  end style 

begin mathsize 14px style mol space C H subscript 3 C O O H double bond M cross times V space space space space space space space space space space space space space space space space space space space space space space equals 0 comma 5 M cross times 100 mL space space space space space space space space space space space space space space space space space space space space space space equals 50 mmol end style 

menentukan mol sisa dari MRS :

menentukan pH penyangga asam:

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space space equals Ka cross times fraction numerator nCH subscript 3 COOHsisa over denominator nCH subscript 3 C O O Na end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 5 end exponent cross times fraction numerator up diagonal strike 30 mmol over denominator up diagonal strike 20 mmol end fraction open square brackets H to the power of plus sign close square brackets equals 1 comma 5 cross times 10 to the power of negative sign 5 end exponent  pH space space equals minus sign log space left square bracket 1 comma 5 cross times 10 to the power of negative sign 5 end exponent right square bracket pH space space equals 5 minus sign log space 1 comma 5 end style 

Jadi, pH larutan dari campuran di atas adalah 5-log 1,5

 

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Dalam laboratorium kimia, larutan penyangga bisa dibuat dengan cara mencampurkan asam lemah dengan garamnya atau basa lemah dengan garamnya. Seorang laboran kimia ingin membuat larutan penyangga denga...

Pembahasan Soal:

Menentukan nilai molaritas begin mathsize 14px style H to the power of plus sign end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic p H end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row cell 5 minus sign log space 2 end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row cell negative sign log space 2 cross times 10 to the power of negative sign 5 end exponent end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 

Menentukan nilai mol larutan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 3 C O O H end subscript end cell equals cell M cross times V end cell row blank equals cell 0 comma 025 cross times 200 end cell row blank equals cell 5 space mmol end cell row blank equals cell 5 cross times 10 to the power of negative sign 3 end exponent space mol end cell end table end style 

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 n subscript C H subscript 3 C O O H end subscript over n subscript C H subscript 3 C O O to the power of minus sign end subscript end cell row cell 2 cross times 10 to the power of negative sign 5 end exponent end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 5 cross times 10 to the power of negative sign 3 end exponent over denominator n subscript C H subscript 3 C O O to the power of minus sign end subscript end fraction end cell row cell n subscript C H subscript 3 C O O to the power of minus sign end subscript end cell equals cell fraction numerator 10 cross times 10 to the power of negative sign 8 end exponent over denominator 2 cross times 10 to the power of negative sign 5 end exponent end fraction end cell row blank equals cell 5 cross times 10 to the power of negative sign 3 end exponent end cell end table end style 

  

Menentukan nilai begin mathsize 14px style italic A subscript r end style X:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 3 COOX end subscript end cell equals cell m over italic M subscript r end cell row cell 5 cross times 10 to the power of negative sign 3 end exponent end cell equals cell fraction numerator 0 comma 49 over denominator italic M subscript r end fraction end cell row cell italic M subscript r end cell equals cell 98 space g space mol to the power of negative sign 1 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space C H subscript 3 COOX end cell equals cell left parenthesis 2 cross times italic A subscript r space C right parenthesis plus left parenthesis 3 cross times italic A subscript r space H right parenthesis plus left parenthesis 2 cross times italic A subscript r space O right parenthesis plus left parenthesis 1 cross times italic A subscript r space X right parenthesis end cell row 98 equals cell left parenthesis 2 cross times 12 right parenthesis plus left parenthesis 3 cross times 1 right parenthesis plus left parenthesis 2 cross times 16 right parenthesis plus left parenthesis 1 cross times italic A subscript r space X right parenthesis end cell row 98 equals cell 24 plus 3 plus 32 plus italic A subscript r space X end cell row 98 equals cell 59 plus italic A subscript r space X end cell row cell italic A subscript r space X end cell equals 39 end table end style 

Sehingga unsur X adalah unsur K dengan nilai begin mathsize 14px style italic A subscript r end style 39.

Jadi, unsur X yang terkandung dalam garam tersebut adalah K.

0

Roboguru

Sebanyak 100 mL larutan mengandung HCOOH dan HCOONa masing-masing 0,1 M   a. Hitung pH larutan awal b. Hitung pH setelah ditambah 1 mL HCl 0,1 M (tulis reaksi yang terjadi) c. Hitung pH setelah dit...

Pembahasan Soal:

Larutan penyangga asam:

a. pH larutan awal

  • mmol HCOOH = 100 mL x 0,1 M= 10 mmol
  • mmol HCOONa= 10 ml x 0,1 M = 10 mmol
  • 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 Ka space x fraction numerator space mol space asam space lemah over denominator mol space basa space lemah end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent space x space fraction numerator 10 space mmol over denominator 10 space mmol end fraction end cell row blank equals cell 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 left parenthesis 10 to the power of negative sign 5 end exponent right parenthesis end cell row blank equals 5 row blank blank blank end table end style  

Jadi, nilai pH larutan awal adalah 5.

 

b. pH setelah ditambah 1 mL HCl 0,1 M 

  • mmol begin mathsize 14px style H to the power of plus sign end style atau HCl = 1 mL x 0,1 M = 0,1 mmol
  • mmol HCOOH atau basa konjugasi = 100 mL x 0,1 M= 10 mmol
  • mmol HCOONa= 10 ml x 0,1 M = 10 mmol

HCl yang ditambahkan akan bereaksi basa konjugasi.

Reaksi: begin mathsize 14px style H to the power of plus sign and H C O O to the power of minus sign equilibrium H C O O H end style  

 

 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 10 to the power of negative sign 5 end exponent space x space fraction numerator 10 comma 1 space mmol over denominator 0 comma 1 space mmol end fraction end cell row blank equals cell 1 comma 01 x 10 to the power of negative sign 3 end exponent end cell row pH equals cell negative sign log open parentheses 1 comma 01 x 10 to the power of negative sign 3 end exponent close parentheses end cell row blank equals cell 2 comma 99 end cell end table end style

Jadi, pH setelah penambahan 1 mL HCl  0,1 M adalah 2,99.

 

c. pH setelah ditambah 1 mL NaOH 0,1 M

  • mmol begin mathsize 14px style O H to the power of minus sign end style atau NaOH= 1 mL x 0,1 M = 0,1 mmol
  • mmol HCOOH = 100 mL x 0,1 M= 10 mmol
  • mmol HCOONa= 10 ml x 0,1 M = 10 mmol

NaOH akan bereaksi dengan HCOOH.

Reaksi: begin mathsize 14px style H C O O H and O H to the power of minus sign equilibrium H C O O to the power of minus sign and H subscript 2 O end style

 

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 10 to the power of negative sign 5 end exponent space x space fraction numerator 10 comma 1 space mmol over denominator 0 comma 1 space mmol end fraction end cell row blank equals cell 1 comma 01 x 10 to the power of negative sign 3 end exponent end cell row pH equals cell negative sign log open parentheses 1 comma 01 x 10 to the power of negative sign 3 end exponent close parentheses end cell row blank equals cell 2 comma 99 end cell end table end style 

Jadi, pH setelah penambahan 1 mL NaOH 0,1 M adalah 2,99.

0

Roboguru

Larutan buffer di bawah ini yang memiliki pH = 5 adalah ...

Pembahasan Soal:

Larutan buffer dengan pH = 5 adalah campuran dari asam lemah dengan garamnya. Oleh karena pada pilihan jawaban larutan penyangga asam adalah campuran HCN dan NaCN serta nilai Ka HCN = 10-5, maka pH = pKa yang berarti nilai mol HCN sama dengan NaCN. 

Perhitungannya sebagai berikut:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets equals K subscript a cross times fraction numerator n space H C N over denominator n space C N to the power of minus sign end fraction equals 10 to the power of negative sign 5 end exponent cross times fraction numerator 12 comma 5 space mmol over denominator 12 comma 5 space mmol end fraction equals 10 to the power of negative sign 5 end exponent end cell row blank blank blank row blank blank cell pH equals minus sign log space open square brackets H to the power of plus sign close square brackets equals minus sign log space 10 to the power of negative sign 5 end exponent equals 5 end cell end table end style 
 

Jadi, jawaban yang benar adalah B.space 

0

Roboguru

Di dalam 250 mL larutan terdapat 8,2 gram  () dan 0,1 mol  (). pH larutan tersebut adalah ....

Pembahasan Soal:

Campuran dari kedua larutan diatas akan membentuk larutan penyangga asam, karena terdiri dari senyawa asam lemah open parentheses C H subscript 3 C O O H close parentheses dan garam / basa konjugasinya open parentheses C H subscript 3 C O O Na close parentheses.

Untuk menyelesaikan soal diatas dapat dilakukan dengan cara sebagai berikut.

1. Tentukan mol (n) C H subscript 3 C O O Na terlebih dahulu


table attributes columnalign right center left columnspacing 0px end attributes row cell n space C H subscript 3 C O O Na end cell equals cell g over M subscript r end cell row blank equals cell fraction numerator 8 comma 2 space g over denominator 82 space g forward slash mol end fraction end cell row blank equals cell 0 comma 1 space mol end cell end table


2. Tentukan konsentrasi ion H to the power of plus sign menggunakan rumus penyangga asam


table attributes columnalign right center left columnspacing 0px end attributes row cell C H subscript 3 C O O Na space end cell rightwards arrow cell space C H subscript 3 C O O to the power of minus sign space plus space Na to the power of plus sign end cell row blank blank cell space space space space 0 comma 1 space mol space space space space space space space space space space space space 0 comma 1 space mol end cell row blank blank blank 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 space lemah over denominator mol space basa space konjugasi end fraction end cell row blank equals cell K subscript a cross times fraction numerator n space C H subscript 3 C O O H over denominator n space C H subscript 3 C O O to the power of minus sign 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 0 comma 1 space mol over denominator 0 comma 1 space mol end fraction end cell row blank equals cell 1 comma 8 cross times 10 to the power of negative sign 5 end exponent space M end cell end table 


3. Tentukan harga pH


begin mathsize 14px style 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 left parenthesis 1 comma 8 cross times 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 end table end style 


Dengan demikian, pH larutan tersebut adalah 5 - log 1,8.

Jadi, jawaban yang tepat adalah A.begin mathsize 14px style space end style 

0

Roboguru

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