Roboguru

Kelarutan  pada suhu tertentu adalah 2 x 10-5mol/L. Hitunglah kelarutan zat tersebut di dalam 500 mL larutan yang mempunyai pH = 4 - log 2.

Pertanyaan

Kelarutan L open parentheses O H close parentheses subscript 2 pada suhu tertentu adalah 2 x 10-5mol/L. Hitunglah kelarutan zat tersebut di dalam 500 mL larutan yang mempunyai pH = 4 - log 2.

Pembahasan Soal:

Untuk mencari kelarutan zat tersebut, maka perlu mencari Ksp dari L open parentheses O H close parentheses subscript 2 terlebih dahulu.

  • Mencari Ksp dari L open parentheses O H close parentheses subscript 2
    L open parentheses O H close parentheses subscript 2 space rightwards harpoon over leftwards harpoon space L to the power of 2 plus sign space plus space 2 O H to the power of minus sign 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 2 s  K subscript sp space equals space s left parenthesis 2 s right parenthesis squared space K subscript sp space equals space 4 s cubed K subscript sp space equals space 4 left parenthesis 2 x 10 to the power of negative sign 5 end exponent right parenthesis cubed space K subscript sp space equals space 3 comma 2 space x 10 to the power of negative sign 14 end exponent 

Setelah diketahui Ksp dari L open parentheses O H close parentheses subscript 2, maka perlu mencari kelarutan L open parentheses O H close parentheses subscript 2 dalam 500 mL larutan dengan pH 4-log 2, 

space space space space space space space space space pH space equals minus sign log open square brackets H to the power of plus sign close square brackets 4 minus sign log space 2 space equals space minus sign space log space open square brackets H to the power of plus sign close square brackets space space space space space space open square brackets H to the power of plus sign close square brackets space equals space 2 space x space 10 to the power of negative sign 4 end exponent 

setelah ion Hdiketahuin, perlu untuk mencari nilai ion OH- dari persamaan kesetimbangan air.

space space space space space space space open square brackets H to the power of plus sign close square brackets open square brackets O H to the power of minus sign close square brackets space equals space 10 to the power of negative sign 14 end exponent 2 space x space 10 to the power of negative sign 4 end exponent open square brackets O H to the power of minus sign close square brackets space equals space 10 to the power of negative sign 14 end exponent space space space space space space space space space space space space space space open square brackets O H to the power of minus sign close square brackets space equals space fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 space x space 10 to the power of negative sign 4 end exponent end fraction space space space space space space space space space space space space space space open square brackets O H to the power of minus sign close square brackets space equals space 5 space x 10 to the power of negative sign 11 end exponent 

Maka kelarutan L open parentheses O H close parentheses subscript 2 adalah sebagai berikut:

space space space space space space space space space space space space space K subscript sp space equals space 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 space 3 comma 2 space x space 10 to the power of negative sign 14 end exponent space equals space s space left parenthesis 5 x 10 to the power of negative sign 11 end exponent right parenthesis squared space space space space space space space space space space space space space space space space space s space equals space fraction numerator 3 comma 2 space x space 10 to the power of negative sign 14 end exponent over denominator 25 space x space 10 to the power of negative sign 22 end exponent end fraction space space space space space space space space space space space space space space space space space s space equals space 1 comma 28 space x space 10 to the power of 8 space end exponent 

Jadi, kelarutan L open parentheses O H close parentheses subscript 2 dalam larutan dengan pH 4-log2 adalah 1 comma 28 space x space 10 to the power of 8.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Kartika

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Pada suhu . Hitunglah kelarutan pada larutan 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

Roboguru

Jika , maka kelarutan  dalam  0,001 M adalah ....

Pembahasan Soal:

Persamaan reaksi kesetimbangan kelarutan begin mathsize 14px style Ag subscript 2 C O subscript 3 end style dan begin mathsize 14px style Ag Cl end style adalah:

begin mathsize 14px style Ag subscript 2 C O subscript 3 open parentheses italic s close parentheses equilibrium 2 Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus C O subscript 3 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis 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 2 s space space space space space space space space space space space space s Ag Cl open parentheses italic s close parentheses equilibrium Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus Cl to the power of minus sign left parenthesis italic a italic q right parenthesis space space 0 comma 001 space space space space space space space 0 comma 001 space space space space space space 0 comma 001 end style 


begin mathsize 14px style K subscript sp Ag subscript 2 C O subscript 3 equals open square brackets Ag to the power of plus sign close square brackets squared left square bracket C O subscript 3 to the power of 2 minus sign end exponent right square bracket 1 cross times 10 to the power of negative sign 14 end exponent equals left parenthesis 2 s plus left parenthesis 1 cross times 10 to the power of negative sign 3 end exponent right parenthesis right parenthesis squared open parentheses s close parentheses end style


Karena begin mathsize 14px style open square brackets Ag to the power of plus sign close square brackets end style yang berasal dari undefined sangat sedikit dibandingkan dengan begin mathsize 14px style open square brackets Ag to the power of plus sign close square brackets end style yang berasal dari undefined, maka undefined dari undefined dapat diabaikan. Dengan demikian dapat kita anggap:

begin mathsize 14px style 2 s plus left parenthesis 1 cross times 10 to the power of negative sign 3 end exponent right parenthesis almost equal to 1 cross times 10 to the power of negative sign 3 end exponent end style

sehingga:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 cross times 10 to the power of negative sign 14 end exponent end cell equals cell open parentheses 1 cross times 10 to the power of negative sign 3 end exponent close parentheses squared open parentheses s close parentheses end cell row cell 1 cross times 10 to the power of negative sign 14 end exponent end cell equals cell 1 cross times 10 to the power of negative sign 6 end exponent s end cell row s equals cell fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 6 end exponent end fraction end cell row s equals cell 1 cross times 10 to the power of negative sign 8 end exponent end cell end table end style 


Berdasarkan perhitungan tersebut, maka jawaban yang tepat adalah E

Roboguru

Dalam suatu larutan terdapat ion , , , , dan  dengan konsentrasi yang sama. Jika larutan tersebut ditetesi dengan larutan , zat yang mula-mula mengendap adalah ....

Pembahasan Soal:

Konstanta hasil kali kelarutan atau Ksp adalah hasil kali konsentrasi ion-ion tersebut dipangkatkan koefisien reaksi. Tidak semua garam bersifat larut dalam air, apabila garam yang sukar larut itu dikocok dalam air maka garam itu sebagian akan larut dalam air dan bagian yang larut ini akan terurai menjadi ion - ionnya. Suatu larutan dikatakan jenuh, tepat jenuh atau mengendap dapat dilakukan dengan membandingkan Qsp dan Ksp.

size 14px Qsp size 14px space size 14px less than size 14px space size 14px Ksp size 14px comma size 14px space size 14px tidak size 14px space size 14px terjadi size 14px space size 14px pengendapan size 14px Qsp size 14px space size 14px equals size 14px space size 14px Ksp size 14px comma size 14px space size 14px tepat size 14px space size 14px jenuh size 14px Qsp size 14px space size 14px greater than size 14px space size 14px Ksp size 14px comma size 14px space size 14px terjadi size 14px space size 14px endapan
 

begin mathsize 14px style italic a bold right parenthesis bold space bold Ksp bold space Ag subscript bold 2 S O subscript bold 4 bold space bold equals bold space bold 1 bold space italic x bold space bold 10 to the power of bold minus sign bold 5 end exponent Ag subscript 2 S O subscript 4 open parentheses italic s close parentheses equilibrium 2 Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus S O subscript 4 to the power of 2 minus sign space end exponent left parenthesis italic a italic q right parenthesis s space space space space space space space space space space space space space space space space space space space space space 2 s space space space space space space space space space space space space space space space space space space s K subscript italic s italic p end subscript italic space Ag subscript 2 S O subscript 4 equals space italic Q subscript italic s italic p end subscript Ag subscript italic 2 S O subscript italic 4 K subscript italic s italic p end subscript Ag subscript 2 S O subscript 4 equals space open square brackets Ag to the power of plus sign close square brackets squared left square bracket S O subscript 4 to the power of 2 minus sign space end exponent right square bracket K subscript italic s italic p end subscript space Ag subscript 2 S O subscript 4 equals left parenthesis 2 s right parenthesis squared open parentheses s close parentheses 1 space x space 10 to the power of negative sign 5 end exponent space space space space equals 4 s cubed space space space space space space space space space space space space space space space space space space s space equals 0 comma 0135 space mol forward slash L  italic b bold right parenthesis bold space bold Ksp bold space Sr S O subscript bold 4 bold space bold equals bold space bold 2 bold comma bold 5 bold space italic x bold space bold 10 to the power of bold minus sign bold 7 end exponent Sr S O subscript 4 bold space left parenthesis italic s right parenthesis equilibrium Sr to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus S O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis 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 s K subscript italic s italic p end subscript space Sr S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space Sr S O subscript 4 K subscript italic s italic p end subscript space Sr S O subscript 4 equals space open square brackets Sr to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space Sr S O subscript 4 equals s squared 2 comma 5 space x space 10 to the power of negative sign 7 end exponent space equals s squared space space space space space space space space space space space space space s space equals space 5 x 10 to the power of negative sign 4 end exponent space mol forward slash L  italic c bold right parenthesis bold space bold Ksp bold space Pb S O subscript bold 4 bold space bold equals bold space bold 1 bold comma bold 7 bold space italic x bold space bold 10 to the power of bold minus sign bold 8 end exponent bold space Pb S O subscript 4 open parentheses italic s close parentheses equilibrium Pb to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus S O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis 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 s K subscript italic s italic p end subscript space space Pb S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space space Pb S O subscript 4 K subscript italic s italic p end subscript space Pb S O subscript 4 equals space open square brackets Pb to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space space Pb S O subscript 4 equals s squared 1 comma 7 space x space 10 to the power of negative sign 8 end exponent space equals s squared bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space italic s bold space bold equals bold space bold 1 bold comma bold 3 italic x bold 10 to the power of bold minus sign bold 4 end exponent bold space bold mol bold forward slash italic L  italic d bold right parenthesis bold space bold Ksp bold space Ba S O subscript bold 4 bold space bold equals bold space bold 1 bold comma bold 1 bold space italic x bold space bold 10 to the power of bold minus sign bold 10 end exponent space Ba S O subscript 4 open parentheses italic s close parentheses equilibrium Ba to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus S O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis 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 s K subscript italic s italic p end subscript space space Ba S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space space Ba S O subscript 4 K subscript italic s italic p end subscript space Ba S O subscript 4 equals space open square brackets Ba to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space space Ba S O subscript 4 equals s squared 1 comma 1 space x space 10 to the power of negative sign 10 end exponent space equals s squared space space space space space space space space space space space space space italic s bold space bold equals bold space bold 1 bold comma bold 0 italic x bold 10 to the power of bold minus sign bold 5 end exponent bold space bold mol bold forward slash italic L  italic e bold right parenthesis bold space bold Ksp bold space Ca S O subscript bold 4 bold space bold equals bold space bold 9 bold space italic x bold space bold 10 to the power of bold minus sign bold 6 end exponent space Ca S O subscript 4 open parentheses italic s close parentheses equilibrium Ca to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus S O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis 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 s K subscript italic s italic p end subscript space space Ca S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space space Ca S O subscript 4 K subscript italic s italic p end subscript space Ca S O subscript 4 equals space open square brackets Ca to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space space Ca S O subscript 4 equals s squared 9 space x space 10 to the power of negative sign 6 end exponent space equals s squared bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space italic s bold space bold equals bold space bold 3 bold comma bold 0 italic x bold 10 to the power of bold minus sign bold 3 end exponent bold space bold mol bold forward slash italic L  bold Semakin bold space bold kecil bold comma bold space bold kelarutannya bold space bold maka bold space bold akan bold space bold semakin bold space bold mudah bold space bold pengendap    space end style

Jadi, jawaban yang benar adalah D.

Roboguru

Pada suhu . Hitunglah kelarutan pada larutan 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

Roboguru

Pada suhu tertentu, 1 L larutan  memiliki . Massa  ( = 461) yang dapat larut jika ditambahkan larutan  0,02 M adalah ….

Pembahasan Soal:

Larutan begin mathsize 14px style Pb I subscript 2 end style mempunyai ion senama dengan larutan undefined yaitu ion begin mathsize 14px style Pb to the power of 2 plus end exponent end styleKonsentrasi ion begin mathsize 14px style Pb to the power of 2 plus end exponent end style adalah begin mathsize 14px style 2 cross times 10 to the power of negative sign 2 end exponent end style. Reaksi yang terjadi di soal adalah sebagai berikut.

begin mathsize 14px style Pb I subscript 2 equilibrium Pb to the power of 2 plus sign and 2 I to the power of minus sign space space s space space space space space space space space space s space space space space space space space space 2 s end style 

Nilai kelarutan (s) dapat dihitung sebagai berikut.

Error converting from MathML to accessible text.  

Menentukan massa yang dapat larut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row s equals cell n over V end cell row s equals cell m over M subscript r cross times 1 over V end cell row m equals cell s cross times M subscript r cross times V end cell row m equals cell 2 cross times 10 to the power of negative sign 4 end exponent space mol forward slash L cross times 461 space g forward slash mol cross times 1 space L end cell row m equals cell 922 cross times 10 to the power of negative sign 4 end exponent space g end cell row m equals cell 9 comma 22 cross times 10 to the power of negative sign 2 end exponent space g end cell end table end style 

Jadi, jawaban yang benar adalah B.undefined

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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