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Jika diketahui K sp ​ Hg 2 ​ Cl 2 ​ = 1 x 1 0 − 18 , kelarutan Hg 2 ​ Cl 2 ​ dalam larutan CaCl 2 ​ 0,01 M adalah ...

Jika diketahui , kelarutan dalam larutan  0,01 M adalah ...

  1. begin mathsize 14px style 2 comma 5 space x space 10 to the power of negative sign 15 end exponent end style  

  2. begin mathsize 14px style 5 comma 0 space x space 10 to the power of negative sign 15 end exponent end style  

  3. begin mathsize 14px style 1 comma 0 space x space 10 to the power of negative sign 14 end exponent end style  

  4. begin mathsize 14px style 2 comma 5 space x space 10 to the power of negative sign 14 end exponent end style  

  5. begin mathsize 14px style 5 comma 0 space x space 10 to the power of negative sign 14 end exponent end style 

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S. Lubis

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Jawaban

jawaban yang tepat adalah opsi A.

jawaban yang tepat adalah opsi A.

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. . Jadi kelarutan dalam larutan adalah . Jadi, jawaban yang tepat adalah opsi A.

begin mathsize 14px style Ca Cl subscript 2 left parenthesis italic a italic q right parenthesis rightwards harpoon over leftwards harpoon space Ca to the power of 2 plus sign left parenthesis italic a italic q right parenthesis space plus space space 2 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis space space space space 0 comma 01 space space space space space space space space space space space space space 0 comma 01 space space space space space space space space space space space space space 0 comma 02 space space Hg subscript 2 Cl subscript 2 left parenthesis italic a italic q right parenthesis rightwards harpoon over leftwards harpoon Hg subscript 2 to the power of 2 plus end exponent left parenthesis italic a italic q right parenthesis space plus space 2 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis space space space space s space space space space space space space space space space space 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 space 2 s end style . 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Ksp space Hg subscript 2 Cl subscript 2 end cell equals cell Qsp space open square brackets Hg to the power of 2 plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets squared end cell row cell 1 space x space 10 to the power of negative sign 18 end exponent end cell equals cell open parentheses s close parentheses left parenthesis 0 comma 02 right parenthesis squared space end cell row cell 1 space x space 10 to the power of negative sign 18 end exponent end cell equals cell open parentheses s close parentheses left parenthesis 0 comma 0004 right parenthesis end cell row s equals cell fraction numerator 1 cross times 10 to the power of negative sign 18 end exponent over denominator 4 cross times 10 to the power of negative sign 4 end exponent end fraction end cell row s equals cell 2 comma 5 cross times 10 to the power of negative sign 15 end exponent end cell row blank blank blank end table end style.  

Jadi kelarutan begin mathsize 14px style Hg subscript 2 Cl subscript 2 end style dalam larutan begin mathsize 14px style Ca Cl subscript 2 space 0 comma 1 space M end style adalah Error converting from MathML to accessible text.

Jadi, jawaban yang tepat adalah opsi A.

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