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Jika 100 mL larutan  0,005 M dicampurkan dengan 100 ml larutan HCI 0,004 M, maka pH campuran ....

Pertanyaan

Jika 100 mL larutan H subscript 2 S O subscript 4 0,005 M dicampurkan dengan 100 ml larutan HCI 0,004 M, maka pH campuran ....

  1. 3 - log 2,5 space

  2. 2 - log 4,5 space

  3. 2 - log 7 space

  4. 3 - log 4,5 space

  5. 3 - log 7space 

Pembahasan Soal:

Pada pencampuran asam kuat, berlaku rumus berikut:

open square brackets H to the power of plus sign close square brackets subscript 1 cross times V subscript 1 end subscript plus open square brackets H to the power of plus sign close square brackets subscript 2 cross times V subscript 2 equals open square brackets H to the power of plus sign close square brackets subscript camp cross times V subscript camp

Kita anggap H subscript 2 S O subscript 4 sebagai asam pertama, 

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 subscript a end cell row blank equals cell 2 cross times 0 comma 005 end cell row blank equals cell 0 comma 01 end cell end table 

Sedangkan HCl asam kedua,

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 subscript a end cell row blank equals cell 1 cross times 0 comma 004 end cell row blank equals cell 0 comma 004 end cell end table 

Masukkan ke persamaan awal tadi,

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 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 end cell equals cell open square brackets H to the power of plus sign close square brackets subscript camp cross times V subscript camp end cell row cell open square brackets H to the power of plus sign close square brackets subscript camp end cell equals cell fraction numerator open parentheses open square brackets H to the power of plus sign close square brackets subscript 1 cross times V subscript 1 close parentheses plus open parentheses open square brackets H to the power of plus sign close square brackets subscript 2 cross times V subscript 2 close parentheses over denominator V subscript camp end fraction end cell row blank equals cell fraction numerator left parenthesis 0 comma 01 cross times 100 right parenthesis plus left parenthesis 0 comma 004 cross times 100 right parenthesis over denominator 100 plus 100 end fraction end cell row blank equals cell fraction numerator 1 plus 0 comma 4 over denominator 200 end fraction end cell row blank equals cell 7 cross times 10 to the power of negative sign 3 end exponent end cell end table 

Persamaan pH dapat dicari seperti biasa,

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 7 cross times 10 to the power of negative sign 3 end exponent end cell row blank equals cell 3 minus sign log space 7 end cell end table 

Sehingga pH campuran asam kuat tersebut adalah 3 - log 7.

Jadi, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Q. 'Ainillana

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

Jika  dicampur dengan , ternyata pH campuran tersebut adalah , maka konsentrasi  yang dicampurkan adalah...

Pembahasan Soal:

  • Larutan begin mathsize 14px style H Cl end style 

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 campuran end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH subscript campuran end cell equals cell 1 minus sign log space 2 end cell row cell open square brackets H to the power of plus sign close square brackets subscript campuran end cell equals cell 2 cross times 10 to the power of negative sign 1 end exponent end cell row blank blank blank end table end style 
 

  • Menghitung Error converting from MathML to accessible text.  

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 cell 2 cross times 10 to the power of negative sign 1 end exponent end cell equals cell fraction numerator 2 cross times 10 to the power of negative sign 1 end exponent cross times 200 plus open square brackets H to the power of plus sign close square brackets subscript 2 cross times 100 over denominator 200 plus 100 end fraction end cell row cell 2 cross times 10 to the power of negative sign 1 end exponent end cell equals cell fraction numerator 40 plus open square brackets H to the power of plus sign close square brackets subscript 2 cross times 100 over denominator 300 end fraction end cell row 60 equals cell 40 plus open square brackets H to the power of plus sign close square brackets subscript 2 cross times 100 end cell row 20 equals cell open square brackets H to the power of plus sign close square brackets subscript 2 cross times 100 end cell row cell open square brackets H to the power of plus sign close square brackets subscript 2 end cell equals cell fraction numerator 20 over denominator 100 end fraction equals 0 comma 2 end cell end table end style
 

  • Menghitung begin mathsize 14px style M space H subscript 2 S O subscript 4 end style 

Error converting from MathML to accessible text.


Jadi, konsentrasi begin mathsize 14px style H subscript bold 2 S O subscript bold 4 end style yang dicampurkan adalah begin mathsize 14px style 0 comma 1 space M end style

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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.

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Roboguru

lonisasi larutan asam lemah bergantung harga tetapan ionisasi () atau derajat ionisasi (alpha). Konsentrasi ion  yang terdapat dalam 200 mL larutan  0,5 M dengan  sebesar ...

Pembahasan Soal:

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

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K subscript a cross times M end root open square brackets H to the power of plus sign close square brackets equals square root of 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times 0 comma 5 end root open square brackets H to the power of plus sign close square brackets equals square root of 9 cross times 10 to the power of negative sign 6 end exponent end root open square brackets H to the power of plus sign close square brackets equals 3 cross times 10 to the power of negative sign 3 end exponent end style  


Jadi, konsentrasi ion begin mathsize 14px style H to the power of plus sign end style begin mathsize 14px style C H subscript 3 C O O H end style adalah begin mathsize 14px style 3 cross times 10 to the power of negative sign 3 end exponent end style.

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Roboguru

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

Pilihlah Jawaban: A. Jika pernyataan benar, alasan benar, dan keduanya menunjukan hubungan sebab akibat. B. Jika pernyataan benar, alasan benar, tetapi keduanya tidak menunjukan hubungan sebab akibat...

Pembahasan Soal:

Karena volume sama, maka V = 100 mL

pH awal = pH HCl

[H+]HCl = 0,1 M

pH = 1

 

 [H+] HCl awal = 0,1 M

[H+] CH3COOH = Ka x Ma

                          = 10-5x 0,1

                          = 10-6

                          = 10–3 M

pH campuran

M1xV1 + M2xV2 = M3xV3

0,1x100mL +0,001x100mL = M3x200mL

10+0,1 = 200M3

M3 = (10,1)/(200)

       = 0,0505 M

[H+]campuran = 0,0505 M

                          = 5,05 x 10–2 M

pH campuran = - log 5,05 x 10–2

                                       = 2 – log 5,05

                         = 2 – 0,7

                         = 1,3

pH akhir > pH awal

sehingga akibat salah

pH CH3COOH > pH HCl

Sehingga sebab salah

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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