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Diketahui: K sp ​ BaCO 3 ​ = 9 × 1 0 − 6 K sp ​ BaC 2 ​ O 4 ​ = 1 , 6 × 1 0 − 7 K sp ​ BaF 2 ​ = 1 , 7 × 1 0 − 6 K sp ​ BaCrO 4 ​ = 1 , 1 × 1 0 − 6 K sp ​ BaSO 4 ​ = 9 × 1 0 − 9 Berdasarkan data tersebut, senyawa yang paling besar kelarutannya dalam air adalah ...

Diketahui:

 

Berdasarkan data tersebut, senyawa yang paling besar kelarutannya dalam air adalah ...space space space

  1. Ba F subscript 2 

  2. Ba C O subscript 3 space space space

  3. Ba S O subscript 4 space space space

  4. Ba Cr O subscript 4 space space space

  5. Ba C subscript 2 O subscript 4 space space space

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

jawaban yang tepat adalah A.space space space

Pembahasan

Untuk mengetahui kelarutan senyawa yang paling tinggi dalam air, kita hitung kelarutannya masing-masing: Kelarutan yang paling tinggi dimiliki oleh senyawa , yaitu sebesar . Jadi, jawaban yang tepat adalah A.

Untuk mengetahui kelarutan senyawa yang paling tinggi dalam air, kita hitung kelarutannya masing-masing:

  1. Ba C O subscript 3 

    Ba C O subscript 3 equilibrium Ba to the power of 2 plus sign and C O subscript 3 to the power of 2 minus sign space space space space space s space space space space space space space space space space s space space space space space space space space space space s  K subscript sp space Ba C O subscript 3 double bond open square brackets Ba to the power of 2 plus sign close square brackets open square brackets C O subscript 3 to the power of 2 minus sign close square brackets K subscript sp space Ba C O subscript 3 double bond open parentheses s close parentheses open parentheses s close parentheses 9 cross times 10 to the power of negative sign 6 end exponent equals s squared s equals 3 cross times 10 to the power of negative sign 3 end exponent   
     
  2. Ba F subscript 2 

    Ba F subscript 2 equilibrium Ba to the power of 2 plus sign and 2 F to the power of minus sign space space space s space space space space space space space space space space s space space space space space space space space 2 s  K subscript sp space Ba F subscript 2 double bond open square brackets Ba to the power of 2 plus sign close square brackets open square brackets F to the power of minus sign close square brackets squared space K subscript sp space Ba F subscript 2 double bond open parentheses s close parentheses left parenthesis 2 s right parenthesis squared space 1 comma 7 cross times 10 to the power of negative sign 6 end exponent equals 4 s cubed s equals 7 comma 5 cross times 10 to the power of negative sign 3 end exponent   
     
  3. Ba S O subscript 4 

    Ba S O subscript 4 equilibrium Ba to the power of 2 plus sign and S O subscript 4 to the power of 2 minus sign space space space space space s space space space space space space space space space space s space space space space space space space space space space s  K subscript sp space Ba S O subscript 4 double bond open square brackets Ba to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign close square brackets K subscript sp space Ba S O subscript 4 double bond open parentheses s close parentheses open parentheses s close parentheses 9 cross times 10 to the power of negative sign 9 end exponent equals s squared s equals 3 cross times 10 to the power of negative sign 4 comma 5 end exponent  
     
  4. Ba C subscript 2 O subscript 4 

    Ba C subscript 2 O subscript 4 equilibrium Ba to the power of 2 plus sign and C subscript 2 O subscript 4 to the power of 2 minus sign space space space space space s space space space space space space space space space space s space space space space space space space space space space s  K subscript sp space Ba C subscript 2 O subscript 4 double bond open square brackets Ba to the power of 2 plus sign close square brackets open square brackets C subscript 2 O subscript 4 to the power of 2 minus sign close square brackets K subscript sp space Ba C subscript 2 O subscript 4 double bond open parentheses s close parentheses open parentheses s close parentheses 1 comma 6 cross times 10 to the power of negative sign 7 end exponent equals s squared s equals 4 cross times 10 to the power of negative sign 4 end exponent  
     
  5. Ba Cr O subscript 4  

    Ba Cr O subscript 4 equilibrium Ba to the power of 2 plus sign and Cr O subscript 4 to the power of 2 minus sign space space space space space s space space space space space space space space space space s space space space space space space space space space space s  K subscript sp space Ba Cr O subscript 4 double bond open square brackets Ba to the power of 2 plus sign close square brackets open square brackets Cr O subscript 4 to the power of 2 minus sign close square brackets K subscript sp space Ba Cr O subscript 4 double bond open parentheses s close parentheses open parentheses s close parentheses 1 comma 1 cross times 10 to the power of negative sign 6 end exponent equals s squared s equals 1 comma 05 cross times 10 to the power of negative sign 3 end exponent  


Kelarutan yang paling tinggi dimiliki oleh senyawa Ba F subscript 2, yaitu sebesar 7 comma 5 cross times 10 to the power of negative sign 3 end exponent space bevelled mol over L.


Jadi, jawaban yang tepat adalah A.space space space

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