Roboguru

Hitunglah kelarutan Ag2​SO4​(Ksp​=1,5×10−5) dalam: b.  0,2 M Na2​SO4​

Pertanyaan

Hitunglah kelarutan Ag subscript 2 S O subscript 4 space left parenthesis K subscript sp equals 1 comma 5 cross times 10 to the power of negative sign 5 end exponent right parenthesis dalam:

b.  0,2 M Na subscript 2 S O subscript 4 

Pembahasan Soal:

stack Ag subscript 2 S O subscript 4 with s below rightwards harpoon over leftwards harpoon stack 2 Ag to the power of plus sign with 2 s below plus stack S O subscript 4 to the power of 2 minus sign end exponent with s below stack Na subscript 2 S O subscript 4 with 0 comma 2 space M below yields 2 Na to the power of plus sign plus stack S O subscript 4 to the power of 2 minus sign end exponent with 0 comma 2 space M below 

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ag subscript 2 S O subscript 4 end cell equals cell 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 end exponent right square bracket end cell row cell 1 comma 5 cross times 10 to the power of negative sign 5 end exponent end cell equals cell left parenthesis 2 s right parenthesis squared left parenthesis s plus 0 comma 2 right parenthesis end cell row cell 1 comma 5 cross times 10 to the power of negative sign 5 end exponent end cell equals cell left parenthesis 4 s squared right parenthesis left parenthesis 0 comma 2 right parenthesis end cell row cell 1 comma 5 cross times 10 to the power of negative sign 5 end exponent end cell equals cell 0 comma 8 s squared end cell row cell s squared end cell equals cell fraction numerator 1 comma 5 cross times 10 to the power of negative sign 5 end exponent over denominator 0 comma 8 end fraction end cell row s equals cell square root of 1 comma 875 cross times 10 to the power of negative sign 5 end exponent end root end cell row s equals cell 4 comma 33 cross times 10 to the power of negative sign 3 end exponent end cell end table 


Jadi, kelarutan bottom enclose Ag subscript bold 2 S O subscript 4 end enclose dalam bottom enclose Na subscript bold 2 S O subscript bold 4 end enclose 0,2 M sebesar bottom enclose bold 4 bold comma bold 33 bold cross times bold 10 to the power of bold minus sign bold 3 end exponent end enclose.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Hidayati

Mahasiswa/Alumni Universitas Indonesia

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

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 kelarutan Ag2​SO4​(Ksp​=1,5×10−5) dalam: a.  0,2 M AgNO3​

Pembahasan Soal:

stack Ag subscript 2 S O subscript 4 with s below rightwards harpoon over leftwards harpoon stack 2 Ag to the power of plus sign with 2 s below plus stack S O subscript 4 to the power of 2 minus sign end exponent with s below stack Ag N O subscript 3 with 0 comma 2 space M below rightwards arrow stack Ag to the power of plus sign with 0 comma 2 space M below plus N O subscript 3 to the power of minus sign 

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ag subscript 2 S O subscript 4 end cell equals cell open square brackets Ag to the power of plus sign close square brackets squared left square bracket S O subscript 4 end subscript to the power of 2 minus sign end exponent right square bracket end cell row cell 1 comma 5 cross times 10 to the power of negative sign 5 end exponent end cell equals cell left parenthesis 2 s plus 0 comma 2 right parenthesis squared open parentheses s close parentheses 1 comma 5 cross times 10 to the power of negative sign 5 end exponent equals left parenthesis 0 comma 2 right parenthesis squared open parentheses s close parentheses end cell row cell 1 comma 5 cross times 10 to the power of negative sign 5 end exponent end cell equals cell left parenthesis 4 cross times 10 to the power of negative sign 2 end exponent right parenthesis open parentheses s close parentheses end cell row s equals cell fraction numerator 1 comma 5 cross times 10 to the power of negative sign 5 end exponent over denominator 4 cross times 10 to the power of negative sign 2 end exponent end fraction end cell row s equals cell 3 comma 75 cross times 10 to the power of negative sign 4 end exponent end cell end table 


Jadi, kelarutan bottom enclose Ag subscript bold 2 S O subscript 4 end enclose dalam bottom enclose Ag N O subscript bold 3 end enclose 0,2 M sebesar bottom enclose bold 3 bold comma bold 75 bold cross times bold 10 to the power of bold minus sign bold 4 end exponent end enclose.

0

Roboguru

Pada 25∘C, Ksp Ni(OH)2​=6×10−18. Hitunglah, kelarutan Ni(OH)2​ dalam: b. Larutan NaOH 0,001 M

Pembahasan Soal:

Kelarutan(s) adalah konsentrasi maksimum yang mungkin dari zat terlarut dalam larutan pada suhu dan tekanan tertentu. Hasil kali kelarutan (Ksp) adalah hasil kali konsentrasi ion-ion yang terdapat dalam larutan jenuh zat yang sukar larut dipangkatkan koefisien.

Ni open parentheses O H close parentheses subscript 2 left parenthesis italic s 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 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 space space space space space space space space space space space space space space 2 s

Hubungan antara Ksp dan s pada reaksi diatas adalah sebagai berikut:

Error converting from MathML to accessible text.

Adanya ion senama (sejenis) yang ditambahkan pada zat yang sukar larut akan memperkecil kelarutan zat tersebut, karena menggeser kesetimbangan ke arah kiri (pembentukan endapan).

Mencari kelarutan Ni open parentheses O H close parentheses subscript 2 spacedalam Na O H space0,001 M

  • Menentukan begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets end style dalam larutan Na O H space

Na O H left parenthesis italic a italic q right parenthesis yields Na 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 0 comma 001 space M 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 001 space M space

  • Mencari kelarutan (s) Ni open parentheses O H close parentheses subscript 2 space

table attributes columnalign right center left columnspacing 0px end attributes row cell Ksp space Ni open parentheses O H close parentheses subscript 2 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 s close parentheses left parenthesis 10 to the power of negative sign 3 end exponent right parenthesis squared space end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell open parentheses s close parentheses left parenthesis 10 to the power of negative sign 6 end exponent right parenthesis end cell row s equals cell fraction numerator 6 cross times 10 to the power of negative sign 18 end exponent over denominator 10 to the power of negative sign 6 end exponent end fraction end cell row s equals cell 6 cross times 10 to the power of negative sign 12 end exponent space space space space end cell end table

Jadi, kelarutan Ni open parentheses O H close parentheses subscript 2 spacedalam larutan Na O H space0,001 M adalah 6 cross times 10 to the power of negative sign 12 end exponent. 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 25∘C, Ksp Ni(OH)2​=6×10−18. Hitunglah, kelarutan Ni(OH)2​ dalam: c. Larutan NiCl2​ 0,001 M

Pembahasan Soal:

Hasil kali kelarutan (Ksp) merupakan sebuah tetapan yang didapatkan dari hasil kali konsentrasi ion pada kondisi larutan tersebut jenuh setelah dipangkatkan dengan nilai koefisien dari persamaan ionisasinya. Secara umum hubungan antara kelarutan (s) dengan tetapan hasil kali kelarutan (Ksp) untuk larutan elektrolit begin mathsize 14px style Ni open parentheses O H close parentheses subscript 2 end style dapat dinyatakan sebagai berikut:

begin mathsize 14px style Ni open parentheses O H close parentheses subscript 2 yields Ni to the power of 2 plus sign and 2 O H to the power of minus sign space 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 2 s end style

sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row Ksp 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 space end cell end table


Adanya pengaruh ion senama pada Ni open parentheses O H close parentheses subscript 2 dan Ni Cl subscript 2, yaitu ion begin mathsize 14px style Ni squared to the power of plus end style, maka dapat ditentukan konsentrasi begin mathsize 14px style Ni squared to the power of plus end style pada kelarutan begin mathsize 14px style Ni Cl subscript 2 end style 0,001 M sebagai berikut:

Sehingga diperoleh kelarutan Ni open parentheses O H close parentheses subscript 2 sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row Ksp 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 left parenthesis 10 to the power of negative sign 3 end exponent right parenthesis left parenthesis 2 s right parenthesis squared end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell 4 cross times 10 to the power of negative sign 3 end exponent left parenthesis s squared right parenthesis end cell row cell s squared end cell equals cell 1 comma 5 cross times 10 to the power of negative sign 15 end exponent end cell row s equals cell space square root of 1 comma 5 cross times 10 to the power of negative sign 15 end exponent end root end cell row blank equals cell square root of 15 cross times 10 to the power of negative sign 16 end exponent end root end cell row blank equals cell 3 comma 87 cross times 10 to the power of negative sign 8 end exponent space M end cell end table

Jadi, kelarutan Ni bold open parentheses O H bold close parentheses subscript bold 2 dalam larutan Ni Cl subscript bold 2 0,001 M adalah Error converting from MathML to accessible text..

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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