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Pertanyaan

Diketahui kelarutan Ag Cl dalam air adalah   1 cross times 10 to the power of negative sign 5 end exponent space mol space L to the power of negative sign 1 end exponent. Kelarutan Ag Cl dalam Ca Cl subscript 2 0,05 M adalah ...space 

  1. 5 cross times 10 to the power of negative sign 10 end exponent space mol space L to the power of negative sign 1 end exponentspace  

  2. 1 cross times 10 to the power of negative sign 9 end exponent space mol space L to the power of negative sign 1 end exponentspace  

  3. 2 cross times 10 to the power of negative sign 9 end exponent space mol space L to the power of negative sign 1 end exponent   

  4. 1 cross times 10 to the power of negative sign 4 end exponent space mol space L to the power of negative sign 1 end exponent  

  5. 2 cross times 10 to the power of negative sign 4 end exponent space mol space L to the power of negative sign 1 end exponentspace  

L. Avicenna

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.space 

Pembahasan

Larutan Ca Cl subscript 2 0,05 M mengandung 0,05 ion Ca to the power of 2 plus sign dan 0,1 M ion Cl to the power of minus sign .

Ca Cl subscript 2 left parenthesis italic a italic q right parenthesis yields Ca to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis 0 comma 05 space M space space space space space space space space 0 comma 05 space M space space space space space space space space space 0 comma 1 space M 

Jika kedalam larutan ditambahkan Ag Cl padat, kristal tersebut akan larut hingga larutan jenuh. Misal kelarutan Ag Cl adalah s.

Ag Cl open parentheses italic s close parentheses equilibrium Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus Cl to the power of minus sign left parenthesis italic a italic q right parenthesis space italic space italic space italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic s 

Konsentrasi total ion Cl to the power of minus sign adalah 0 comma 1 plus italic s namun karena nilai s sangat kecil maka konsentrasi ion Cl to the power of minus sign dianggap 0,1 M, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ag Cl space dalam space air end cell equals cell open square brackets Ag to the power of plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets end cell row cell K subscript sp space Ag Cl space dalam space air end cell equals cell open square brackets 10 to the power of negative sign 5 end exponent close square brackets squared end cell row cell K subscript sp space Ag Cl space dalam space air end cell equals cell 10 to the power of negative sign 10 end exponent end cell row blank blank blank row cell K subscript sp space Ag Cl space dalam space dalam space Ca Cl subscript 2 end cell equals cell open square brackets Ag to the power of plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets end cell row cell 10 to the power of negative sign 10 end exponent end cell equals cell open square brackets Ag to the power of plus sign close square brackets open square brackets 10 to the power of negative sign 1 end exponent close square brackets end cell row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign 9 end exponent space mol space L to the power of negative sign 1 end exponent end cell end table 

Jadi kelarutan Ag Cl bold space bold dalam bold space Ca Cl subscript bold 2 adalah bold 10 to the power of bold minus sign bold 9 end exponent bold space bold mol bold space italic L to the power of bold minus sign bold 1 end exponent.

Oleh karena itu, jawaban yang benar adalah B.space 

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