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Sebanyak 100 mL larutan HCl yang mempunyai pH=1 dicampur dengan 100 mL larutan HCl lain yang mempunyai pH=2. Hitung pH campuran kedua larutan.

Pertanyaan

Sebanyak 100 mL larutan HCl yang mempunyai pH=1 dicampur dengan 100 mL larutan HCl lain yang mempunyai pH=2. Hitung pH campuran kedua larutan. space

Pembahasan Soal:

Larutan HCl merupakan senyawa asam kuat bervalensi satu karena saat terionisasi menhasilkan satu buah ion H to the power of plus sign. pH campuran dari kedua larutan HCl yang berbeda pH dapat ditentukan dengan cara sebagai berikut.

  • Menentukan molaritas dari kedua larutan HCl


table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell negative sign log open square brackets H to the power of plus sign close square brackets end cell row blank blank blank row pH equals cell 1 comma space maka space open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 1 end exponent end cell row pH equals cell 2 comma space maka space open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 2 end exponent 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 M cross times valensi end cell row M equals cell open square brackets H to the power of plus sign close square brackets space left parenthesis karena space valensi space H Cl equals 1 right parenthesis end cell row blank blank blank row pH equals cell 1 comma space maka space M equals 10 to the power of negative sign 1 end exponent end cell row pH equals cell 2 comma space maka space M equals 10 to the power of negative sign 2 end exponent end cell end table

 

  • Menentukan molaritas campuran


table attributes columnalign right center left columnspacing 0px end attributes row cell misal colon space H Cl space pH end cell equals cell 1 equals senyawa space 1 end cell row cell H Cl space pH end cell equals cell 2 equals senyawa space 2 end cell row blank blank blank row cell M subscript campuran end cell equals cell fraction numerator left parenthesis M subscript 1 cross times V subscript 1 right parenthesis space plus left parenthesis M subscript 2 cross times V subscript 2 right parenthesis space over denominator left parenthesis V subscript 1 and V subscript 2 right parenthesis end fraction end cell row blank equals cell fraction numerator left parenthesis 10 to the power of negative sign 1 end exponent space M cross times 100 space mL right parenthesis space plus space left parenthesis 10 to the power of negative sign 2 end exponent space M cross times 100 space mL right parenthesis over denominator left parenthesis 100 plus 100 right parenthesis mL end fraction end cell row blank equals cell fraction numerator left parenthesis 10 plus 1 right parenthesis mmol over denominator 200 space mL end fraction end cell row blank equals cell fraction numerator 11 space mmol over denominator 200 space mL end fraction end cell row blank equals cell 0 comma 055 space end cell row blank equals cell 5 comma 5 cross times 10 to the power of negative sign 2 end exponent space M end cell end table

 

  • Menentukan konsentrasi ion H to the power of plus sign dan pH campuran


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 subscript campuran end cell equals cell M subscript campuran space left parenthesis karena space valensi equals 1 right parenthesis end cell row blank equals cell 5 comma 5 cross times 10 to the power of negative sign 2 end exponent space M end cell row blank blank blank 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 5 comma 5 cross times 10 to the power of negative sign 2 end exponent right parenthesis end cell row blank equals cell 2 minus sign log space 5 comma 5 end cell end table


Jadi, pH campuran kedua larutan HCl tersebut adalah 2 - log 5,5.space​​​​​​​ 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Susanti

Mahasiswa/Alumni Universitas Jayabaya

Terakhir diupdate 30 April 2021

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Sebanyak 100 mL larutan  0,004 M dicampurkan dengan 100 mL larutan  0,012 M. Tentukan pH campuran. Anggap kedua jenis larutan tersebut mengion sempurna.

Pembahasan Soal:

Larutan H subscript 2 S O subscript 4 merupakan senyawa asam kuat bervalensi 2, sementara larutan H Cl merupakan asam kuat bervalensi 1. Untuk menentukan pH campuran, dapat dilakukan dengan cara sebagai berikut.

  • Menentukan masing-masing mol ion H to the power of plus sign dari H subscript 2 S O subscript 4 dan H Cl melalui perbandingan koefisien


table attributes columnalign right center left columnspacing 0px end attributes row n equals cell M cross times V end cell row blank blank blank row cell n space H subscript 2 S O subscript 4 end cell equals cell 0 comma 004 cross times 100 equals 0 comma 4 space mmol end cell row cell n space H Cl end cell equals cell 0 comma 012 cross times 100 equals 1 comma 2 space mmol end cell row blank blank blank row cell H subscript 2 S O subscript 4 space end cell rightwards arrow cell space 2 H to the power of plus sign space plus space S O subscript 4 to the power of 2 minus sign end exponent end cell row cell n space H to the power of plus sign end cell equals cell 2 cross times n space H subscript 2 S O subscript 4 equals 2 cross times 0 comma 4 space mmol equals 0 comma 8 space mmol end cell row blank blank blank row cell H Cl space end cell rightwards arrow cell space H to the power of plus sign space plus space Cl to the power of minus sign end cell row cell n space H to the power of plus sign end cell equals cell n space H Cl equals 1 comma 2 space mmol end cell end table

 

  • Menentukan konsentrasi ion H to the power of plus sign


table attributes columnalign right center left columnspacing 0px end attributes row cell n space H to the power of plus sign space total end cell equals cell left parenthesis n space H to the power of plus sign space H subscript 2 S O subscript 4 right parenthesis space plus space left parenthesis n space H to the power of plus sign space H Cl right parenthesis end cell row blank equals cell left parenthesis 0 comma 8 plus 1 comma 2 right parenthesis mmol end cell row blank equals cell 2 space mmol end cell row blank blank blank row cell open square brackets H to the power of plus sign close square brackets total end cell equals cell fraction numerator n space H to the power of plus sign space total over denominator V space total end fraction end cell row blank equals cell fraction numerator 2 space mmol over denominator left parenthesis 100 plus 100 right parenthesis mL end fraction end cell row blank equals cell fraction numerator 2 space mmol over denominator 200 space mL end fraction end cell row blank equals cell 10 to the power of negative sign 2 end exponent space M end cell end table

 

  • Menentukan pH campuran


table attributes columnalign right center left columnspacing 0px end attributes 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 2 end exponent right parenthesis end cell row blank equals 2 end table


Jadi, pH campuran kedua larutan tersebut adalah 2.space 

0

Roboguru

Perhatikan kedua pernyataan berikut! Jika larutan  pH = 1 diencerkan 100 kali, pH larutan berubah menjadi 3. Konsentrasi  berubah jika larutan diencerkan. Hubungan pernyataan 1 dan 2 yang tepat ...

Pembahasan Soal:

Pembuktian Pernyataan (1)

  1. Konsentrasi awal sebelum pengenceran

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals 1 row cell open square brackets H to the power of plus sign close square brackets end cell equals cell a cross times M end cell row cell 10 to the power of negative sign 1 end exponent end cell equals cell 1 cross times M end cell row M equals cell 10 to the power of negative sign 1 end exponent end cell end table end style 
     
  2. Konsentrasi setelah pengenceran

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 1 middle dot V subscript 1 end cell equals cell M subscript 2 middle dot V subscript 2 end cell row cell 10 to the power of negative sign 1 end exponent middle dot V subscript 1 end cell equals cell M subscript 2 middle dot 100 V subscript 1 end cell row cell M subscript 2 end cell equals cell fraction numerator 10 to the power of negative sign 1 end exponent middle dot V subscript 1 over denominator 100 V subscript 1 end fraction end cell row cell M subscript 2 end cell equals cell 10 to the power of negative sign 3 end exponent end cell end table end style 
     
  3. pH larutan setelah pengenceran

    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 a cross times M end cell row blank equals cell 1 cross times 10 to the power of negative sign 3 end exponent end cell row blank equals cell 10 to the power of negative sign 3 end exponent end cell end table end style 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell negative sign log space 10 to the power of negative sign 3 end exponent end cell row blank equals cell 3 space open parentheses bold benar close parentheses end cell end table end style 


Pembuktian pernyataan (2)

Ketika larutan diencerkan maka konsentrasi H+ akan berubah (benar)

Sehingga berdasarkan penjelasan diatas pernyataan benar dan alasan benar dan keduanya ada hubungan sebab akibat.

Jadi, jawaban yang benar adalah A.undefined 

0

Roboguru

Hitunglah konsentrasi ion H+ asam kuat berikut : HBr 0,5 M

Pembahasan Soal:

Larutan HBr merupakan larutan asam kuat. Untuk menghitung konsentrasi ion H+ dari larutan HBr dengan mengunakan rumus :

open square brackets H to the power of plus sign close square brackets equals M cross times a

a = valensi asam (jumlah H+)

open square brackets H to the power of plus sign close square brackets equals 0 comma 5 cross times 1 space space space space space space space equals 0 comma 5 space M

Maka, konsentrasi ion H to the power of plus sign dari larutan HBr adalah 0,5 M.space

0

Roboguru

larutan  yang pH nya  dicampur dengan  larutan   yang pH nya , pH campuran adalah...

Pembahasan Soal:

  • Larutan undefined 

Error converting from MathML to accessible text. 

 

  • Larutan undefined 

Error converting from MathML to accessible text. 

 

  • Menghitung begin mathsize 14px style open square brackets H to the power of plus sign close square brackets subscript camp 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 subscript camp end cell equals cell fraction numerator open square brackets H to the power of plus sign close square brackets subscript 1 cross times V subscript 1 plus open square brackets H to the power of plus sign close square brackets subscript 2 cross times V subscript 2 over denominator V subscript 1 and V subscript 2 end fraction end cell row blank equals cell fraction numerator 2 cross times 10 to the power of negative sign 1 end exponent cross times 20 plus 2 cross times 10 to the power of negative sign 3 end exponent cross times 80 over denominator 20 plus 80 end fraction end cell row blank equals cell fraction numerator 4 plus 0 , 16 over denominator 100 end fraction end cell row blank equals cell 4.16 cross times 10 to the power of negative sign 2 end exponent end cell end table end style 

 

  • Menghitung pH campuran

Error converting from MathML to accessible text. 

 

Jadi, pH campuran larutan tersebut adalah Error converting from MathML to accessible text. 

0

Roboguru

Sebanyak 200 mL larutan  0,005 M, dimasukkan ke dalam gelas beker yang berisi 200 mL larutan HCOOH 0,05 M. pH campuran larutan yang terjadi sebesar ... ()

Pembahasan Soal:

Asam sulfat (H subscript 2 S O subscript 4) merupakan suatu asam kuat, sedangkan asam format (H C O O H) merupakan asam lemah. Campuran kedua larutan ini akan menghasilkan suatu larutan asam. Perhitungan pH campuran kedua larutan ini dapat dilakukan dengan langkah-langkah sebagai berikut.

  1. Menentukan begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style larutan undefined 
    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 a middle dot M subscript a end cell row blank equals cell 2 middle dot 0 comma 005 end cell row blank equals cell 0 comma 01 end cell row blank equals cell 1 cross times 10 to the power of negative sign 2 end exponent end cell end table 
     
  2. Menentukan begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style larutan HCOOH
    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 middle dot M subscript a end root end cell row blank equals cell square root of 1 comma 8 cross times 10 to the power of negative sign 4 end exponent middle dot 0 comma 05 end root end cell row blank equals cell square root of 9 cross times 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 3 cross times 10 to the power of negative sign 3 end exponent end cell end table
     
  3. Menentukan begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style campuran
    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 space campuran end cell equals cell fraction numerator open square brackets H to the power of plus sign close square brackets subscript 1 middle dot V subscript 1 end subscript plus open square brackets H to the power of plus sign close square brackets subscript 2 middle dot V subscript 2 end subscript over denominator V subscript 1 and V subscript 2 end fraction end cell row blank equals cell fraction numerator left parenthesis 1 cross times 10 to the power of negative sign 2 end exponent middle dot 200 right parenthesis plus left parenthesis 3 cross times 10 to the power of negative sign 3 end exponent middle dot 200 right parenthesis over denominator 200 plus 200 end fraction end cell row blank equals cell fraction numerator 2 plus 0 comma 6 over denominator 400 end fraction end cell row blank equals cell 0 comma 0065 end cell row blank equals cell 6 comma 5 cross times 10 to the power of negative sign 3 end exponent end cell end table

     
  4. Menentukan pH campuran
    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 6 comma 5 cross times 10 to the power of negative sign 3 end exponent end cell row blank equals cell 3 minus sign log space 6 comma 5 end cell end table 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

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