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Dalam larutan CaSO 4 ​ 0,1 M dan BaSO 4 ​ 0,1 M ditetesi sedikit demi sedikit dengan larutan Na 2 ​ SO 4 ​ . Volume dianggap tidak berubah dengan penambahan . ( Ksp CaSO 4 ​ = 9 × 1 0 − 6 , Ksp BaSO 4 ​ = 1 × 1 0 − 10 ) a. Ion manakah yang mengendap lebih dahulu Ca 2 + atau Ba 2 + ? Jelaskan! b. Berapakah [ Ca 2 + ] pada saat mulai mengendap? c. Berapakah [ Ba 2 + ] pada saat mulai mengendap?

Dalam larutan  0,1 M dan  0,1 M ditetesi sedikit demi sedikit dengan larutan . Volume dianggap tidak berubah dengan penambahan begin mathsize 14px style Na subscript 2 S O subscript 4 end style

a.  Ion manakah yang mengendap lebih dahulu  atau ? Jelaskan!

b.  Berapakah  pada saat begin mathsize 14px style Ba S O subscript 4 end style mulai mengendap?

c.  Berapakah  pada saat begin mathsize 14px style Ca S O subscript 4 end style mulai mengendap?

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D. Zharva

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a) Ion yang akan mengendap terlebih dahulu Yang akan mengendap terlebih dahulu adalah karena nilai kelarutannya (s) lebih kecil dari . b) Konsentrasi pada saat mulai mengendap c) Konsentrasi pada saat mulai mengendap

begin mathsize 14px style Ca S O subscript 4 yields Ca to the power of 2 plus sign and S O subscript 4 to the power of 2 minus sign end exponent 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 s end style 


begin mathsize 14px style Ba S O subscript 4 yields Ba to the power of 2 plus sign and S O subscript 4 to the power of 2 minus sign end exponent 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 s end style 


a)  Ion yang akan mengendap terlebih dahulu


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ba S O subscript 4 end cell equals cell 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 end exponent close square brackets end cell row cell 1 cross times 10 to the power of negative sign 10 end exponent end cell equals cell s cross times s end cell row cell 1 cross times 10 to the power of negative sign 10 end exponent end cell equals cell s squared end cell row s equals cell square root of 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 10 to the power of negative sign 5 space end exponent M end cell end table end style 


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ca S O subscript 4 end cell equals cell open square brackets Ca to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets end cell row cell 9 cross times 10 to the power of negative sign 6 end exponent end cell equals cell s cross times s end cell row cell 9 cross times 10 to the power of negative sign 6 end exponent end cell equals cell s squared end cell row s equals cell square root of 9 cross times 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 3 cross times 10 to the power of negative sign 3 end exponent space M end cell end table end style 


Yang akan mengendap terlebih dahulu adalah begin mathsize 14px style Ba S O subscript 4 end style karena nilai kelarutannya (s) lebih kecil dari begin mathsize 14px style Ca S O subscript 4 end style.


b)  Konsentrasi begin mathsize 14px style open square brackets Ca to the power of 2 plus sign close square brackets end style pada saat begin mathsize 14px style Ba S O subscript 4 end style mulai mengendap


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript sp space Ca S O subscript 4 end cell equals cell K subscript sp space Ba S O subscript 4 end cell row cell open square brackets Ca to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets end cell equals cell 1 cross times 10 to the power of negative sign 10 end exponent end cell row cell open square brackets Ca to the power of 2 plus sign close square brackets open parentheses 0 comma 1 close parentheses end cell equals cell 1 cross times 10 to the power of negative sign 10 end exponent end cell row cell open square brackets Ca to the power of 2 plus sign close square brackets end cell equals cell 1 cross times 10 to the power of negative sign 9 end exponent space M end cell end table end style 


c)  Konsentrasi  begin mathsize 14px style open square brackets Ba to the power of 2 plus sign close square brackets end style pada saat begin mathsize 14px style Ca S O subscript 4 end style mulai mengendap


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript sp space Ba S O subscript 4 end cell equals cell K subscript sp space Ca S O subscript 4 end cell row cell 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 end exponent close square brackets end cell equals cell 9 cross times 10 to the power of negative sign 6 end exponent end cell row cell open square brackets Ba to the power of 2 plus sign close square brackets open parentheses 0 comma 1 close parentheses end cell equals cell 9 cross times 10 to the power of negative sign 6 end exponent end cell row cell open square brackets Ba to the power of 2 plus sign close square brackets end cell equals cell 9 cross times 10 to the power of negative sign 5 end exponent space M end cell end table end style 

Konsep Kilat

Kelarutan

Hasil Kali Kelarutan (Kₛₚ)

Faktor-Faktor Kelarutan dan Pengendapan

Latihan Bab

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