Roboguru

Dalam suatu larutan terdapat ion Ag+, Ca2+, Ba2+, Sr2+, dan Pb2+ dengan konsentrasi yang sama. Jika larutan tersebut ditetesi dengan larutan K2​SO4​, zat yang mula-mula mengendap adalah ....

Pertanyaan

Dalam suatu larutan terdapat ion begin mathsize 14px style Ag to the power of plus sign end style, begin mathsize 14px style Ca to the power of 2 plus sign end style, begin mathsize 14px style Ba to the power of 2 plus sign end style, begin mathsize 14px style Sr to the power of 2 plus sign end style, dan begin mathsize 14px style Pb to the power of 2 plus sign end style dengan konsentrasi yang sama. Jika larutan tersebut ditetesi dengan larutan begin mathsize 14px style K subscript 2 S O subscript 4 end style, zat yang mula-mula mengendap adalah ....

  1. begin mathsize 14px style Ag subscript 2 S O subscript 4 space left parenthesis K subscript sp equals 1 comma 4 cross times 10 to the power of negative sign 5 end exponent right parenthesis end style 

  2. begin mathsize 14px style Sr S O subscript 4 space left parenthesis K subscript sp equals 2 comma 5 cross times 10 to the power of negative sign 7 end exponent right parenthesis end style 

  3. begin mathsize 14px style Pb S O subscript 4 space left parenthesis K subscript sp equals 1 comma 7 cross times 10 to the power of negative sign 8 end exponent right parenthesis end style 

  4. begin mathsize 14px style Ba S O subscript 4 space left parenthesis K subscript sp equals 1 comma 1 cross times 10 to the power of negative sign 10 end exponent right parenthesis end style 

  5. begin mathsize 14px style Ca S O subscript 4 space left parenthesis K subscript sp equals 9 cross times 10 to the power of negative sign 6 end exponent right parenthesis end style 

Pembahasan Video:

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.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Rochmawatie

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Kelarutan magnesium fluorida dalam air adalah 0,0086g/L. Berapa kelarutan magnesium fluorida dalam NaF0,02M?

Pembahasan Soal:

Kelarutan suatu larutan dapat mengalami perubahan ketika ada ion senama, namun tidak dengan nilai hasil kali kelarutan. Sehingga sebelum mencari kelarutan setelah ditambah dengan larutan yang memiliki ion yang sama, maka langkah awal adalah mencari nilai hasil kali kelarutan.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell s equals 0 comma 0086 space g space L to the power of negative sign 1 end exponent equals fraction numerator 0 comma 0086 space g space L to the power of negative sign 1 end exponent over denominator 62 space g space mol to the power of negative sign 1 end exponent end fraction equals 1 comma 38 cross times 10 to the power of negative sign 4 space end exponent mol space L to the power of negative sign 1 end exponent end cell row blank blank blank row blank blank cell K subscript sp space Mg F subscript 2 equals open square brackets Mg to the power of 2 plus sign close square brackets open square brackets F to the power of minus sign close square brackets squared equals s cross times left parenthesis 2 s right parenthesis squared equals 4 open parentheses 1 comma 38 cross times 10 to the power of negative sign 4 space end exponent close parentheses cubed equals 1 comma 067 cross times 10 to the power of negative sign 11 end exponent end cell end table end style 


Kelarutan begin mathsize 14px style Mg F subscript 2 end style setelah ditambah NaF 0,02 M


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Mg F subscript 2 end cell equals cell open square brackets Mg 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 1 comma 067 cross times 10 to the power of negative sign 11 end exponent end cell equals cell s cross times open parentheses 0 comma 02 close parentheses squared end cell row cell fraction numerator 1 comma 067 cross times 10 to the power of negative sign 11 end exponent over denominator 4 cross times 10 to the power of negative sign 4 end exponent end fraction end cell equals s row cell 2 comma 67 cross times 10 to the power of negative sign 8 end exponent end cell equals s end table end style  


Jadi, setelah penambahan ion senama kelarutan mengalami penurunan dari begin mathsize 14px style 1 comma 38 cross times 10 to the power of negative sign 4 end exponent end style  menjadi begin mathsize 14px style 2 comma 67 cross times 10 to the power of negative sign 8 end exponent end style.undefined 

0

Roboguru

Jika pada suhu tertentu dalam larutan jenuh AgCl terdapat  ion Cl− dengan konsentrasi 10−5 maka kelarutan AgCl di dalam larutan NaCl 0,1 M adalah ....

Pembahasan Soal:

Adanya ion senama di dalam larutan kan memperkecil kelarutan suatu zat. Untuk menyelesaikan soal diatas dapat dilakukan dengan cara sebagai berikut.

1. Tentukan hasil kali kelarutan open parentheses K subscript sp close parentheses AgCl terlebih dahulu

table attributes columnalign right center left columnspacing 0px end attributes row cell Ag Cl space end cell rightwards harpoon over leftwards harpoon cell space Ag to the power of plus sign space plus space Cl to the power of minus sign end cell row blank blank cell space space space space space space space space space space space space space space space space space space space space space space space space space space 1 cross times 10 to the power of negative sign 5 end exponent end cell row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell open square brackets Cl to the power of minus sign close square brackets equals 1 cross times 10 to the power of negative sign 5 end exponent space left parenthesis karena space koefisien space sama right parenthesis end cell row blank blank blank row cell K subscript sp space Ag Cl end cell equals cell 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 blank equals cell left parenthesis 1 cross times 10 to the power of negative sign 5 end exponent right parenthesis left parenthesis 1 cross times 10 to the power of negative sign 5 end exponent right parenthesis end cell row blank equals cell 1 cross times 10 to the power of negative sign 10 end exponent end cell end table 


2. Tentukan kelarutan AgCl dalam larutan NaCl 0,1 M


table attributes columnalign right center left columnspacing 0px end attributes row cell Na Cl space end cell rightwards arrow cell space Na to the power of plus sign space plus space Cl to the power of minus sign end cell row blank blank cell 0 comma 1 space M space space space space 0 comma 1 space M space space space space 0 comma 1 space M end cell row blank blank blank row cell Ag Cl space end cell rightwards harpoon over leftwards harpoon cell space Ag to the power of plus sign space plus space Cl to the power of minus sign end cell row blank blank cell 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 plus 0 comma 1 space M end cell end table  

konsentrasi ion Cl to the power of minus sign dianggap hanya berasal dari  NaCl karena nilai s sangat kecil

table attributes columnalign right center left columnspacing 0px end attributes row cell Ag Cl space end cell rightwards harpoon over leftwards harpoon cell space Ag to the power of plus sign space plus space Cl to the power of minus sign end cell row blank blank cell space space space s space space space space space space space space space space s space space space space space space space space space 0 comma 1 space M end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ag Cl end cell equals cell open square brackets Ag to the power of plus sign close square brackets open square brackets Cl to the power of minus sign 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 left parenthesis 0 comma 1 right parenthesis end cell row s equals cell 1 cross times 10 to the power of negative sign 9 end exponent end cell end table   


Dengan demikian, kelarutan AgCl di dalam larutan NaCl 0,1 M adalah1 cross times 10 to the power of negative sign 9 end exponent space M.space 

0

Roboguru

Kelarutan Mn(OH)2​ di dalam air adalah 5×10−6M. Massa  yang dapat larut dalam 200 mL larutan Ca(OH)2​ 0,02 M adalah …. (Ar​  Mn = 55, O = 16, H = 1)

Pembahasan Soal:

Kelarutan begin mathsize 14px style Mn open parentheses O H close parentheses subscript 2 end style di dalam air adalah begin mathsize 14px style 5 cross times 10 to the power of negative sign 6 end exponent space M end style. Dengan demikian, K subscript sp space Mn open parentheses O H close parentheses subscript 2 dapat ditentukan sebagai berikut.
 

Mn open parentheses O H close parentheses subscript 2 equilibrium Mn to the power of 2 plus sign and 2 O H to the power of minus sign space space space space space space space s space space space space space space space space space space space space s space space space space space space space space space space 2 s 

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Mn open parentheses O H close parentheses subscript 2 end cell equals cell open square brackets Mn 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 K subscript sp space Mn open parentheses O H close parentheses subscript 2 end cell equals cell open parentheses s close parentheses left parenthesis 2 s right parenthesis squared end cell row cell K subscript sp space Mn open parentheses O H close parentheses subscript 2 end cell equals cell 4 s cubed end cell row cell K subscript sp space Mn open parentheses O H close parentheses subscript 2 end cell equals cell 4 left parenthesis 5 cross times 10 to the power of negative sign 6 end exponent right parenthesis cubed end cell row cell K subscript sp space Mn open parentheses O H close parentheses subscript 2 end cell equals cell 4 left parenthesis 125 cross times 10 to the power of negative sign 18 end exponent right parenthesis end cell row cell K subscript sp space Mn open parentheses O H close parentheses subscript 2 end cell equals cell 5 cross times 10 to the power of negative sign 16 end exponent end cell end table 
 

Langkah selanjutnya adalah menentukan konsentrasi O H to the power of minus sign dari Ca open parentheses O H close parentheses subscript 2 sebagai berikut.
 

Ca open parentheses O H close parentheses subscript 2 equilibrium Ca to the power of 2 plus sign and 2 O H to the power of minus sign space space space space 0 comma 02 space space space space space space space 0 comma 02 space space space space space 0 comma 04 
 

Langkah selanjutnya adalah menentukan kelarutan (s) Mn open parentheses O H close parentheses subscript 2 dalam Ca open parentheses O H close parentheses subscript 2 0,02 M.
 

Error converting from MathML to accessible text.  
 

Langkah berikutnya adalah menentukan mol.
 

table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell s cross times V end cell row mol equals cell 3 comma 125 cross times 10 to the power of negative sign 13 end exponent space mol forward slash L cross times 0 comma 2 space L end cell row mol equals cell 0 comma 625 cross times 10 to the power of negative sign 13 end exponent space mol end cell row mol equals cell 6 comma 25 cross times 10 to the power of negative sign 14 end exponent space mol end cell end table  
 

Setelah itu, massa Ca open parentheses O H close parentheses subscript 2 dapat ditentukan sebagai berikut.
 

table attributes columnalign right center left columnspacing 0px end attributes row massa equals cell mol cross times italic M subscript r end cell row massa equals cell 6 comma 25 cross times 10 to the power of negative sign 14 end exponent space mol cross times 89 space g forward slash mol end cell row massa equals cell 556 comma 25 cross times 10 to the power of negative sign 14 end exponent space g end cell row massa equals cell 5 comma 56 cross times 10 to the power of negative sign 12 end exponent space g end cell end table  
 

Jadi, jawaban yang tepat adalah E.space 

0

Roboguru

Hitunglah perbandingan kelarutan PbSO4​ (Ksp​=2,25×10−8) dalam air murni dan dalam larutan Na2​SO4​0,01M

Pembahasan Soal:

Kelarutan adalah kemampuan zat terlarut untuk larut dalam pelarut.
Kelarutan dalam air

Error converting from MathML to accessible text.

Kelarutan dalam begin mathsize 14px style Na subscript 2 S O subscript 4 space 0 comma 01 M end style

begin mathsize 14px style Na subscript 2 S O subscript 4 space space space rightwards arrow space space 2 Na to the power of plus space end exponent plus space S O subscript 4 to the power of 2 minus sign end exponent 0 comma 01 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 0 comma 01 Pb S O subscript 4 space space space space space rightwards arrow space Pb to the power of 2 plus sign space space plus space 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 space space space space space space space italic s space space space space space space space space space space space space space space K subscript sp space space space equals space open square brackets s close square brackets space left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket 2 comma 25 space cross times space 10 to the power of negative sign 8 end exponent space space equals space italic s italic space italic cross times italic space 0 comma 01 space space space space space space space space space space space space space space space space space space italic s italic space italic space italic space italic space equals italic space fraction numerator 2 comma 25 space cross times space 10 to the power of negative sign 8 end exponent over denominator 0 comma 01 end fraction space space space space space space space space space space space space space space space space space space italic s italic space italic space italic space italic space equals italic space 2 comma 25 space cross times space 10 to the power of negative sign 6 end exponent end style 

Perbandingan kelarutan

Error converting from MathML to accessible text.   

0

Roboguru

Jika diketahui Ksp​Hg2​Cl2​=1×10−18, kelarutan Hg2​Cl2​ dalam larutan CaCl2​ 0,01 M adalah ....

Pembahasan Soal:

Adanya ion senama di dalam larutan akan memperkecil atau mengurangi kelarutan suatu zat tersebut.

Langkah pertama, menentukan reaksi ionisasi Ca Cl subscript 2.


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


langkah kedua, menentukan reaksi ionisasi Hg subscript 2 Cl subscript 2.


Hg subscript 2 Cl subscript 2 open parentheses italic s close parentheses yields 2 space Hg to the power of 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 s space M space space space space space space space space space space space space space space 2 s space M space space space space space space space space space space space 0 comma 02 space M


Langkah ketiga, menentukan nilai kelarutan dengan mengunakan rumus K subscript sp.


table attributes columnalign right center left columnspacing 0px end attributes row cell Ksp space Hg subscript 2 Cl subscript 2 end cell equals cell open square brackets Hg to the power of plus sign close square brackets to the power of 2 space end exponent open square brackets Cl to the power of minus sign close square brackets squared space space space space space end cell row cell 1 cross times 10 to the power of negative sign 18 end exponent end cell equals cell left parenthesis 2 s right parenthesis squared left parenthesis 0 comma 02 right parenthesis squared end cell row cell 1 cross times 10 to the power of negative sign 18 end exponent end cell equals cell left parenthesis 4 s squared right parenthesis left parenthesis 0 comma 0004 right parenthesis end cell row cell 4 s squared space end cell equals cell fraction numerator 1 cross times 10 to the power of negative sign 18 end exponent over denominator 4 cross times 10 to the power of negative sign 4 end exponent end fraction end cell row cell 4 s squared end cell equals cell 0 comma 25 cross times 10 to the power of negative sign 14 end exponent end cell row cell s squared end cell equals cell fraction numerator 0 comma 25 cross times 10 to the power of negative sign 14 end exponent over denominator 4 end fraction end cell row blank equals cell 0 comma 0625 cross times 10 to the power of negative sign 14 end exponent end cell row s equals cell square root of 0 comma 0625 cross times 10 to the power of negative sign 14 end exponent end root end cell row blank equals cell 0 comma 25 cross times 10 to the power of negative sign 7 end exponent end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 8 end exponent end cell end table 


Jadi, kelarutannya adalah 2 comma 5 cross times 10 to the power of negative sign 8 end exponent.

0

Roboguru

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