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Untuk mendapatkan ion  dari 340 mL larutan jenuh  diperlukan 31 mL .  adalah...

Pertanyaan

Untuk mendapatkan ion begin mathsize 14px style Ag to the power of plus sign end style dari 340 mL larutan jenuh begin mathsize 14px style Ag Br O subscript 3 end style diperlukan 31 mL begin mathsize 14px style H subscript 2 S space left parenthesis 25 degree C comma space 1 space atm right parenthesis end stylebegin mathsize 14px style K subscript italic s italic p end subscript space dari space Ag Br O subscript 3 end style adalah...undefined 

Pembahasan Soal:

Reaksi pengendapan yang terjadi adalah sebagai berikut.

begin mathsize 14px style 2 Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus H subscript 2 S left parenthesis italic a italic q right parenthesis yields Ag subscript 2 S open parentheses italic s close parentheses and 2 H to the power of plus sign left parenthesis italic a italic q right parenthesis end style  

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H subscript 2 S end cell equals cell fraction numerator 31 cross times 10 to the power of negative sign 3 end exponent space L over denominator 22 comma 4 space L end fraction end cell row cell mol space H subscript 2 S end cell equals cell 1 comma 38 cross times 10 to the power of negative sign 3 end exponent space mol end cell row blank blank blank end table end style 

Koefisien begin mathsize 14px style H subscript 2 S colon Ag to the power of plus sign end style adalah 1:2 maka, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Ag to the power of plus sign end cell equals cell 2 cross times mol space H subscript 2 S end cell row cell mol space Ag to the power of plus sign end cell equals cell 2 cross times 1 comma 38 cross times 10 to the power of negative sign 3 end exponent space mol end cell row cell mol space Ag to the power of plus sign end cell equals cell 2 comma 76 cross times 10 to the power of negative sign 3 end exponent space mol end cell row blank blank blank row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell fraction numerator mol space Ag to the power of plus sign over denominator V space Ag Br O subscript 3 end fraction end cell row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell fraction numerator 2 comma 76 cross times 10 to the power of negative sign 3 end exponent space mol over denominator 0 comma 34 space L end fraction end cell row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell 8 comma 1 cross times 10 to the power of negative sign 3 end exponent space M end cell row blank blank blank end table end style  

Pada reaksi berikut:

begin mathsize 14px style Ag Br O subscript 3 left parenthesis italic a italic q right parenthesis equilibrium Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus Br O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell open square brackets Br O subscript 3 to the power of minus sign close square brackets end cell row cell K subscript italic s italic p end subscript space Ag Br O subscript 3 end cell equals cell open square brackets Ag to the power of plus sign close square brackets cross times open square brackets Br O subscript 3 to the power of minus sign close square brackets end cell row cell K subscript italic s italic p end subscript space Ag Br O subscript 3 end cell equals cell open square brackets 8 comma 1 cross times 10 to the power of negative sign 3 space end exponent M close square brackets cross times open square brackets 8 comma 1 cross times 10 to the power of negative sign 3 end exponent space M close square brackets end cell row cell K subscript italic s italic p end subscript space Ag Br O subscript 3 end cell equals cell 6 comma 56 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 

Jadi begin mathsize 14px style K subscript italic s italic p end subscript Ag Br O subscript bold 3 bold space bold adalah bold space bold 6 bold comma bold 56 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end styleundefined 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 April 2021

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