Iklan

Pertanyaan

Jika 200 mL 0 , 2 M HCl dicampur dengan 100 mL H 2 ​ SO 4 ​ , ternyata pH campuran tersebut adalah 1 − lo g 2 , maka konsentrasi H 2 ​ SO 4 ​ yang dicampurkan adalah ...

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

  1. ...undefined

  2. ...undefined

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

00

:

21

:

17

:

02

Klaim

Iklan

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

konsentrasi yang dicampurkan adalah .

 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

Pembahasan

Pembahasan
lock

Larutan Larutan Menghitung Menghitung Menghitung Jadi,konsentrasi yang dicampurkan adalah .

  • 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

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

1

Iklan

Pertanyaan serupa

Sebanyak 200 mL larutan H 2 ​ SO 4 ​ 0,005 M, dimasukkan ke dalam gelas beker yang berisi 200 mL larutan HCOOH 0,05 M. pH campuran larutan yang terjadi sebesar ... ( K a ​ HCOOH = 1 , 8 × 1 0 − 4 )

2

5.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia