Roboguru

Diketahui :     Terdapat 4 tabung reaksi masing...

Diketahui :

begin mathsize 14px style Ksp space Mg C O subscript 3 space equals space 3 comma 5 cross times 10 to the power of negative sign 10 end exponent Ksp space Sr C O subscript 3 space equals space 1 comma 1 space cross times 10 to the power of negative sign 10 end exponent Ksp space Ca C O subscript 3 space equals space 2 comma 8 space cross times 10 to the power of negative sign 9 end exponent Ksp space Mn C O subscript 3 space equals space 1 comma 8 space cross times 10 to the power of negative sign 11 end exponent end style 

begin mathsize 14px style Mr space Mg C O subscript 3 space equals space 84 comma space Ca C O subscript 3 space equals space 100 comma space Sr C O subscript 3 space equals space 148 comma space Mn C O subscript 3 space equals space 115 end style 

Terdapat 4 tabung reaksi masing-masing tabung reaksi berisi begin mathsize 14px style 10 space mL space 0 comma 0001 space M space zat space Mg Cl subscript 2 comma space Sr Cl subscript 2 comma space Ca Cl subscript 2 comma space dan space Mn Cl subscript 2 point end style 

Kedalam 4 tabung tersebut masing-masing ditambahkan begin mathsize 14px style K subscript 2 C O subscript 3 end style sebanyak begin mathsize 14px style 10 space mL space 0 comma 0002 space M end style, maka massa endapan yang terjadi adalah . . .

Jawaban:

Tahap 1 : Analisis apakah benar endapan terjadi dengan cara mencari nilai Qsp, jika nilai Qsp > Ksp maka akan terjadi endapan.space 

Semua garam klorida yang ada di masing-masing tabung reaksi memiliki volume dan konsentrasi yang sama, maka :space 

begin mathsize 14px style open square brackets Mg to the power of 2 plus sign close square brackets space equals space open square brackets Sr to the power of 2 plus sign close square brackets space equals space open square brackets Ca to the power of 2 plus sign close square brackets space equals space open square brackets Mn to the power of 2 plus sign close square brackets space equals 10 to the power of negative sign 4 end exponent M end style 

Maka masing-masing mmol garam adalah = 10 to the power of negative sign 4 end exponent M space cross times space 10 mL space equals space 10 to the power of negative sign 3 end exponent space mmol space .

Untuk mmol undefined = 10 space mL space cross times space 0 comma 0002 space M space equals space 2 cross times 10 to the power of negative sign 3 end exponent space mmol space . mol space C O subscript 3 to the power of 2 minus sign end exponent space dapat space dicari space melalui space reaksi space ionisasi space K subscript 2 C O subscript 3 K subscript 2 C O subscript 3 yields 2 K to the power of plus sign and C O subscript 3 to the power of 2 minus sign end exponent space 2 cross times 10 to the power of negative sign 3 end exponent space space space space space space space space space space space space 2 cross times 10 to the power of negative sign 3 end exponent space mmol 

Jika tebentuk garam karbonat maka konsentrasi garam juga akan sama karena mmol kation dan anion berjumlah sama, sehingga :space 

open square brackets Mg close square brackets space equals space open square brackets Sr close square brackets space equals open square brackets Ca close square brackets space equals open square brackets Mn close square brackets space equals space fraction numerator 10 to the power of negative sign 3 end exponent mol over denominator 20 space mL end fraction equals 5 cross times 10 to the power of negative sign 5 end exponent space M 
open square brackets O H to the power of minus sign close square brackets equals fraction numerator 2 cross times 10 to the power of negative sign 3 end exponent space mmol over denominator 20 space mL end fraction equals 10 to the power of negative sign 4 end exponent space M   

10 to the power of negative sign 3 end exponent mmol merupakan mmol kation (yang menjadi pereaksi pembatas) dan 20 mL merupakan volume total kation + anion.space 

Qsp juga akan sama yaitu begin mathsize 14px style Qsp space equals space open square brackets Kation close square brackets open square brackets Anion close square brackets end style = left square bracket 5 cross times 10 to the power of negative sign 5 end exponent right square bracket left square bracket 10 to the power of negative sign 4 end exponent right square bracket equals 5 cross times 10 to the power of negative sign 9 end exponent    
Karena nilai Qsp > Ksp maka semua garam mengendapspace 

Tahap 2 : Menghitung masing-masing jumlah endapan garam karbonat space 

mmol masing-masing garam karbonat adalah 1 mmol, maka :space 

begin mathsize 14px style mg space equals space mmol space cross times space Mr end style 

  1. mg space Mg C O subscript 3 space end subscript equals space 0 comma 001 space cross times space 84 space equals space 0 comma 084 space mg  
  2. mg space Ca C O subscript 3 space equals space 0 comma 001 space cross times space 100 space equals space 0 comma 1 space mg  
  3. mg space Sr C O subscript 3 space equals space 0 comma 001 space cross times space 148 space equals space 0 comma 148 space mg 
  4. mg space Mn C O subscript 3 space equals space 0 comma 001 cross times space 115 space equals space 0 comma 115 space mg  

Maka massa total adalah = space 
left parenthesis 0 comma 084 plus 0 comma 1 plus 0 comma 148 plus 0 comma 115 right parenthesis space mg space equals space 0 comma 447 space mg space equals space 0 comma 0004470 space gram equals 4 comma 47 cross times 10 to the power of negative sign 4 end exponent  
Jadi, massa endapan yang terjadi sebesar 4 comma 47 cross times negative sign 10 to the power of negative sign 4 end exponent  gram

0

Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved