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Larutan jenuh MgCl 2 ​ sebanyak 100 mL diuapkan pada suhu 1 9 ∘ C dan diperoleh 15,2 mg padat. Berapakah nilai K sp ​ ? ( A r ​ Mg = 24, Cl = 35,5)

Larutan jenuh  sebanyak 100 mL diuapkan pada suhu  dan diperoleh 15,2 mg begin mathsize 14px style Mg Cl subscript 2 end style padat. Berapakah nilai  begin mathsize 14px style Mg Cl subscript 2 end style? ( Mg = 24, Cl = 35,5)

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A. Chandra

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

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nilai yang benar adalah .

nilai begin mathsize 14px style K subscript bold sp end style begin mathsize 14px style Mg Cl subscript bold 2 end style yang benar adalah begin mathsize 14px style bold 10 bold comma bold 24 bold cross times bold 10 to the power of bold minus sign bold 6 end exponent end style.

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Pembahasan

Menentukan : Menentukan nilai mol : Menentukan kelarutan : Menentukan : Jadi, nilai yang benar adalah .

Menentukan begin mathsize 14px style italic M subscript r space Mg Cl subscript 2 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space Mg Cl subscript 2 end cell equals cell left parenthesis 1 cross times italic A subscript r space Mg right parenthesis plus left parenthesis 2 cross times italic A subscript r space Cl right parenthesis end cell row blank equals cell left parenthesis 1 cross times 24 right parenthesis plus left parenthesis 2 cross times 35 comma 5 right parenthesis end cell row blank equals cell 24 plus 71 end cell row blank equals cell 95 space g space mol to the power of negative sign 1 end exponent end cell end table end style 

Menentukan nilai mol begin mathsize 14px style Mg Cl subscript 2 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript Mg Cl subscript 2 end subscript end cell equals cell m over italic M subscript r end cell row blank equals cell fraction numerator 0 comma 0152 space g over denominator 95 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 00016 space mol end cell end table end style 

Menentukan kelarutan begin mathsize 14px style Mg Cl subscript 2 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell s subscript Mg Cl subscript 2 end subscript end cell equals cell n over V end cell row blank equals cell fraction numerator 0 comma 00016 space mol over denominator 0 comma 1 space L end fraction end cell row blank equals cell 0 comma 0016 space M end cell row blank equals cell 1 comma 6 cross times 10 to the power of negative sign 3 end exponent space M end cell end table end style 

Menentukan begin mathsize 14px style K subscript sp end style begin mathsize 14px style Mg Cl subscript 2 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Mg Cl subscript 2 end cell equals cell open square brackets Mg 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 blank equals cell s cross times open parentheses 2 s close parentheses squared end cell row blank equals cell 4 s cubed end cell row blank equals cell 4 cross times open parentheses 1 comma 6 cross times 10 to the power of negative sign 3 end exponent close parentheses squared end cell row blank equals cell 4 cross times 2 comma 56 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 10 comma 24 cross times 10 to the power of negative sign 6 end exponent end cell end table end style 

Jadi, nilai begin mathsize 14px style K subscript bold sp end style begin mathsize 14px style Mg Cl subscript bold 2 end style yang benar adalah begin mathsize 14px style bold 10 bold comma bold 24 bold cross times bold 10 to the power of bold minus sign bold 6 end exponent end style.

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