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Kelarutan Pb(IO3​)2​ di dalam NaIO3​0,1M adalah 2,4×10−11. Hitunglah Ksp​Pb(IO3​)2​.

Pertanyaan

Kelarutan begin mathsize 14px style Pb open parentheses I O subscript 3 close parentheses subscript 2 end style di dalam begin mathsize 14px style Na I O subscript 3 space 0 comma 1 space M end style adalah begin mathsize 14px style 2 comma 4 cross times 10 to the power of negative sign 11 end exponent end style. Hitunglah begin mathsize 14px style K subscript sp space Pb open parentheses I O subscript 3 close parentheses subscript 2 end style.

  1. ...undefined 

  2. ...undefined 

Pembahasan Video:

Pembahasan Soal:

begin mathsize 14px style Na I O subscript 3 left parenthesis italic a italic q right parenthesis equilibrium Na to the power of plus sign left parenthesis italic a italic q right parenthesis plus I O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis space space space space 0 comma 1 space M space space space space space space space space space space 0 comma 1 space M space space space space space space space 0 comma 1 space M end style   

 

begin mathsize 14px style Pb open parentheses I O subscript 3 close parentheses subscript 2 left parenthesis italic s right parenthesis equilibrium Pb to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 I O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis space space space space space space space space italic s space space space space space space space space space space space space space space space space space space italic s space space space space space space space space space space space space space space 2 italic s end style 

 

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Jadi, begin mathsize 14px style K subscript sp italic space Pb open parentheses I O subscript 3 close parentheses subscript 2 equals 2 comma 4 cross times 10 to the power of negative sign 13 end exponent end styleundefined

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Rochmawatie

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Jika  Ksp​CaF2​=4×10−11, kelarutan CaF2​ dalam CaCl2​ 0,01 M adalah ...

Pembahasan Soal:

Larutan  Ca Cl subscript 2 0,01 M mengandung 0,01 M ion Ca to the power of 2 plus sign dan 0,02 M ion Cl to the power of minus sign.

Ca Cl subscript 2 open parentheses italic s close parentheses equilibrium Ca to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis space space 0 comma 01 space M space space space space space space 0 comma 01 space M space space space space space space 0 comma 02 space M space 

Jika ke dalam larutan ditambahkan  Ca F subscript 2 padat, kristal tersebut akan larut hingga larutan jenuh. Misa kelarutan Ca F subscript 2 double bond s space mol space L to the power of negative sign 1 end exponent, maka konsentrasi ion Ca to the power of 2 plus sign yang dihasilkan adalah s dan ion F to the power of minus sign adalah 2s.

Ca F subscript 2 left parenthesis italic a italic q right parenthesis equilibrium Ca to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 F to the power of minus sign left parenthesis italic a italic q right parenthesis italic space italic space italic space italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space 2 italic s 

Jadi konsentrasi ion Ca to the power of 2 plus sign equals 0 comma 01 plus italic s italic space mol space L to the power of negative sign 1 end exponent. Oleh karena nilai s relatif kecil, yaitu lebih kecil dari kelarutannya dalam air, maka konsentrasi ion Ca to the power of 2 plus sign dapat dianggap 0,01 M. Dalam larutan jenuh Ca F subscript 2 berlaku:

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ca F subscript 2 end cell equals cell open square brackets Ca to the power of 2 plus sign close square brackets open square brackets F to the power of minus sign close square brackets squared end cell row cell 4 cross times 10 to the power of negative sign 11 end exponent end cell equals cell open square brackets 0 comma 01 space M close square brackets open square brackets italic 2 italic s close square brackets squared end cell row cell 4 cross times 10 to the power of negative sign 11 end exponent end cell equals cell open square brackets 0 comma 01 space M close square brackets open square brackets 4 italic s to the power of italic 2 close square brackets end cell row cell s squared end cell equals cell fraction numerator 4 cross times 10 to the power of negative sign 11 end exponent over denominator 4 cross times 10 to the power of negative sign 2 end exponent end fraction end cell row italic s equals cell square root of 10 to the power of negative sign 9 end exponent end root end cell row italic s equals cell square root of 10 cross times 10 to the power of negative sign 10 end exponent end root end cell row italic s bold equals cell bold 3 bold comma bold 16 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent bold space bold mol bold space italic L to the power of bold minus sign bold 1 end exponent end cell end table 

Oleh karena itu, jawaban yang benar adalah B.space 

0

Roboguru

Pada suhu 25∘C Ksp​Ni(OH)2​=6×10−18. Hitunglah kelarutan Ni(OH)2​ pada larutan KOH 0,01 M.

Pembahasan Soal:

Persamaan reaksi yang terjadi adalah


Ni open parentheses O H close parentheses subscript 2 left parenthesis italic a italic q right parenthesis equilibrium Ni to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 O H to the power of minus sign left parenthesis italic a italic q right parenthesis K O H left parenthesis italic a italic q right parenthesis equilibrium K to the power of plus sign left parenthesis italic a italic q right parenthesis plus O H to the power of minus sign left parenthesis italic a italic q right parenthesis


Misalkan kelarutan Ni open parentheses O H close parentheses subscript 2 dalam K O H adalah s mol/L, maka pada sistem terdapat:

  • open square brackets Ni to the power of 2 plus sign close square brackets equals s space mol forward slash L
  • open square brackets O H to the power of minus sign close square brackets equals open parentheses 0 comma 01 plus 2 s close parentheses space mol forward slash L

Oleh karena open square brackets O H to the power of minus sign close square brackets dari Ni open parentheses O H close parentheses subscript 2 jauh lebih kecil daripada besarnya open square brackets O H to the power of minus sign close square brackets dari K O H, maka open square brackets O H to the power of minus sign close square brackets dari Ni open parentheses O H close parentheses subscript 2 dapat diabaikan.


table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp end cell equals cell open square brackets Ni to the power of 2 plus sign close square brackets open square brackets O H to the power of minus sign close square brackets squared end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell s cross times open parentheses 0 comma 01 close parentheses squared end cell row s equals cell 6 cross times 10 to the power of negative sign 14 end exponent space mol forward slash L end cell end table


Jadi, kelarutannya adalah bold 6 bold cross times bold 10 to the power of bold minus sign bold 14 end exponent mol/liter.space

0

Roboguru

Pada suhu 25∘C Ksp​Ni(OH)2​=6×10−18. Hitunglah kelarutan Ni(OH)2​ pada larutan NiCl2​ 0,001 M.

Pembahasan Soal:

Persamaan reaksi yang terjadi adalah


Ni open parentheses O H close parentheses subscript 2 left parenthesis italic a italic q right parenthesis equilibrium Ni to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 O H to the power of minus sign left parenthesis italic a italic q right parenthesis Ni Cl subscript 2 left parenthesis italic a italic q right parenthesis equilibrium Ni to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis


Misalkan kelarutan Ni open parentheses O H close parentheses subscript 2 dalam Ni Cl subscript 2 adalah s mol/liter, maka di dalam sistem terdapat:

  • open square brackets Ni to the power of 2 plus sign close square brackets equals open parentheses 10 to the power of negative sign 3 end exponent plus s close parentheses space mol forward slash L
  • open square brackets O H to the power of minus sign close square brackets equals 2 s space mol forward slash L

Oleh karena ion Ni dari Ni open parentheses O H close parentheses subscript 2 jauh lebih kecil dari ion Ni dari Ni Cl subscript 2, maka besarnya ion Ni dari Ni open parentheses O H close parentheses subscript 2 dapat diabaikan.


table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp end cell equals cell open square brackets Ni to the power of 2 plus sign close square brackets open square brackets O H to the power of minus sign close square brackets squared end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell open parentheses 10 to the power of negative sign 3 end exponent close parentheses cross times open parentheses 2 s close parentheses squared end cell row s equals cell 3 comma 87 cross times 10 to the power of negative sign 8 end exponent space mol forward slash L end cell end table


Jadi, kelarutannya adalah bold 3 bold comma bold 87 bold cross times bold 10 to the power of bold minus sign bold 8 end exponent mol/L.space

0

Roboguru

Hasil kali kelarutan L(OH)2​ pada suhu tertentu =1,2×10−11. Bila ke dalam larutan LCl2​0,2M ditambahkan NaOH padat, maka  mulai mengendap pada pH ...

Pembahasan Soal:

Hasil kali kelarutan ialah hasil kali konsentrasi ion- ion dari larutan jenuh garam yang sukar larut dalam air, setelah masing- masing dipangkatkan koefisien nya sesuai dengan persamaan reaksi ionisasinya.

table attributes columnalign right center left columnspacing 0px end attributes row cell LCl subscript 2 end cell rightwards arrow cell L to the power of 2 plus sign and 2 Cl to the power of minus sign end cell row cell open square brackets L to the power of 2 plus sign close square brackets end cell equals cell open square brackets LCl subscript 2 close square brackets equals 0 comma 2 space M end cell row blank blank blank row cell L open parentheses O H close parentheses subscript 2 end cell rightwards arrow cell L to the power of 2 plus sign and 2 O H to the power of minus sign end cell row Ksp equals cell open square brackets L to the power of 2 plus sign close square brackets open square brackets O H to the power of minus sign close square brackets squared end cell row cell 1 comma 2 cross times 10 to the power of negative sign 11 end exponent end cell equals cell left parenthesis 0 comma 2 right parenthesis left square bracket O H to the power of minus sign right square bracket squared end cell row cell left square bracket O H to the power of minus sign right square bracket squared end cell equals cell fraction numerator 1 comma 2 cross times 10 to the power of negative sign 11 end exponent over denominator 2 cross times 10 to the power of negative sign 1 end exponent end fraction end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 6 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row blank blank blank row pOH equals cell negative sign log space 2 comma 5 cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row blank equals cell 5 comma 5 minus sign log space 2 comma 5 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign left parenthesis 5 comma 5 minus sign log space 2 comma 5 right parenthesis end cell row blank equals cell 8.5 plus log space 2 comma 5 end cell row blank equals cell 8 comma 5 plus 0 comma 4 end cell row blank equals cell 8 comma 9 end cell end table 

Jadi, L open parentheses O H close parentheses subscript 2 mulai mengendap pada pH 8,9.

0

Roboguru

Tentukan konsentrasi ion Ag+ minimal yang diperlukan untuk mengendapkan AgCl dari suatu larutan CaCl2​ 0,1 M. Jika diketahui Ksp​AgCl=1,8×10−10.

Pembahasan Soal:

Reaksi disosiasi dari AgCl dan CaCl2

Ca Cl subscript bold 2 bold space bold space bold rightwards arrow bold space bold space Ca to the power of bold 2 bold plus sign bold space bold space bold space bold plus bold space bold space bold space bold space bold 2 Cl to the power of bold minus sign 0 comma 1 M space space space space space space space space space space space 0 comma 1 M space space space space space space space space space space 0 comma 2 M  

Ag Cl bold space bold rightwards arrow bold space Ag to the power of bold plus sign bold space bold plus bold space bold space Cl to the power of bold minus sign bold space xM space space space space space space space space space xM space space space space space space space space 0 comma 2 M  

Hitung konsentrasi Ag+ berdasarkan nilai Ksp

table attributes columnalign right center left columnspacing 0px end attributes row cell space space space space space space K subscript sp space end subscript Ag Cl end cell equals cell space open square brackets Ag to the power of plus sign close square brackets space open square brackets Cl to the power of minus sign close square brackets end cell row cell 1 comma 8 space cross times space 10 to the power of negative sign 5 end exponent space end cell equals cell space open square brackets Ag to the power of plus sign close square brackets space left square bracket 0 comma 2 right square bracket end cell row cell space space space space space space open square brackets Ag to the power of plus sign close square brackets space space space space end cell equals cell space fraction numerator 1 comma 8 space cross times space 10 to the power of negative sign 5 end exponent over denominator 0 comma 2 end fraction end cell row cell space space space space space space open square brackets Ag to the power of plus sign close square brackets space space space space end cell equals cell space 9 space cross times space 10 to the power of negative sign 5 end exponent space M end cell end table  

Jadi, konsentrasi minimum ion begin mathsize 14px style Ag to the power of plus sign end style adalah begin mathsize 14px style 9 space cross times space 10 to the power of negative sign 5 end exponent end style M.

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