Roboguru

Jika Ksp , maka kelarutan  dalam  0,1M adalah....

Pertanyaan

Jika Ksp begin mathsize 14px style Ag subscript 2 C O subscript 3 equals 1 cross times 10 to the power of negative sign 14 end exponent end style, maka kelarutan begin mathsize 14px style Ag subscript 2 C O subscript 3 end style dalam begin mathsize 14px style Ag Cl end style 0,1M adalah....undefined 

  1. begin mathsize 14px style 5 cross times 10 to the power of negative sign 13 end exponent space mol forward slash L end style 

  2. begin mathsize 14px style 1 cross times 10 to the power of negative sign 12 end exponent space mol forward slash L end style 

  3. begin mathsize 14px style 2 cross times 10 to the power of negative sign 12 end exponent space mol forward slash L end style 

  4. begin mathsize 14px style 5 cross times 10 to the power of negative sign 9 end exponent space mol forward slash L end style 

  5. begin mathsize 14px style 1 cross times 10 to the power of negative sign 8 end exponent space mol forward slash L end style 

Pembahasan Soal:

Langkah pertama adalah menentukan jumlah ion senama dari senyawa tersebut


begin mathsize 14px style Ag Cl equilibrium Ag to the power of plus sign and Cl to the power of minus sign 0 comma 1 M space space space space 0 comma 1 M space space 0 comma 1 M end style


Kemudian ditentukan rumus hasil kali kelarutan dari senyawa begin mathsize 14px style Ag subscript 2 C O subscript 3 end style


begin mathsize 14px style Ag subscript 2 C O subscript 3 equilibrium 2 Ag to the power of plus sign and C O subscript 3 to the power of 2 minus sign end exponent space space space space space s space space space space space space space space space space space space space 2 s space space space space space space space space space space space space s end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp end cell equals cell open square brackets Ag to the power of plus sign close square brackets squared space left square bracket C O subscript 3 to the power of 2 minus sign end exponent right square bracket end cell row cell 1 cross times 10 to the power of negative sign 14 end exponent end cell equals cell left parenthesis 2 s right parenthesis squared space space open parentheses s close parentheses end cell row cell 1 cross times 10 to the power of negative sign 14 end exponent end cell equals cell left parenthesis 0 comma 1 right parenthesis squared space open parentheses s close parentheses end cell row s equals cell fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 0 comma 01 end fraction end cell row blank equals cell 1 cross times 10 to the power of negative sign 12 end exponent end cell end table end style


Jadi, jawaban yang benar adalah B.space  

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 April 2021

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