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Jika K sp ​ Ag 2 ​ CO 3 ​ = 1 × 1 0 − 14 , maka kelarutan Ag 2 ​ CO 3 ​ dalam AgCl 0,001 M adalah ....

Jika , maka kelarutan  dalam  0,001 M adalah ....

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

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

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

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

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

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S. Lubis

Master Teacher

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Pembahasan

Persamaan reaksi kesetimbangan kelarutan dan adalah: Karena yang berasal dari sangat sedikit dibandingkan dengan yang berasal dari , maka dari dapat diabaikan. Dengan demikian dapat kita anggap: sehingga: Berdasarkan perhitungan tersebut, maka jawaban yang tepat adalah E

Persamaan reaksi kesetimbangan kelarutan begin mathsize 14px style Ag subscript 2 C O subscript 3 end style dan begin mathsize 14px style Ag Cl end style adalah:

begin mathsize 14px style Ag subscript 2 C O subscript 3 open parentheses italic s close parentheses equilibrium 2 Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus C O subscript 3 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis space space space space space space space s space space space space space 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 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 space 0 comma 001 space space space space space space space 0 comma 001 space space space space space space 0 comma 001 end style 


begin mathsize 14px style K subscript sp Ag subscript 2 C O subscript 3 equals open square brackets Ag to the power of plus sign close square brackets squared left square bracket C O subscript 3 to the power of 2 minus sign end exponent right square bracket 1 cross times 10 to the power of negative sign 14 end exponent equals left parenthesis 2 s plus left parenthesis 1 cross times 10 to the power of negative sign 3 end exponent right parenthesis right parenthesis squared open parentheses s close parentheses end style


Karena begin mathsize 14px style open square brackets Ag to the power of plus sign close square brackets end style yang berasal dari undefined sangat sedikit dibandingkan dengan begin mathsize 14px style open square brackets Ag to the power of plus sign close square brackets end style yang berasal dari undefined, maka undefined dari undefined dapat diabaikan. Dengan demikian dapat kita anggap:

begin mathsize 14px style 2 s plus left parenthesis 1 cross times 10 to the power of negative sign 3 end exponent right parenthesis almost equal to 1 cross times 10 to the power of negative sign 3 end exponent end style

sehingga:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 cross times 10 to the power of negative sign 14 end exponent end cell equals cell open parentheses 1 cross times 10 to the power of negative sign 3 end exponent close parentheses squared 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 1 cross times 10 to the power of negative sign 6 end exponent s end cell row s equals cell fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 6 end exponent end fraction end cell row s equals cell 1 cross times 10 to the power of negative sign 8 end exponent end cell end table end style 


Berdasarkan perhitungan tersebut, maka jawaban yang tepat adalah E

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