Roboguru

Hasil kali kelarutan . Dalam 1 liter larutan terdapat 0,01 mol . Bila ke dalam larutan ditambah  sedikit demi sedikit, maka endapan mulai terbentuk setelah pH mencapai ....

Pertanyaan

Hasil kali kelarutan Ca open parentheses O H close parentheses subscript 2 equals 1 cross times 10 to the power of negative sign 6 end exponent. Dalam 1 liter larutan terdapat 0,01 mol Ca to the power of 2 plus sign. Bila ke dalam larutan ditambah Na O H sedikit demi sedikit, maka endapan mulai terbentuk setelah pH mencapai .... 

  1. 2space 

  2. 4space

  3. 10space 

  4. 12space 

  5. 13space 

Pembahasan Soal:

Suatu senyawa akan mulai mengendap ( tepat jenuh) jika hasil kali konsentrasi ion open parentheses Q close parentheses sama dengan hasil kali kelarutan open parentheses K subscript italic s italic p end subscript close parenthesessenyawa tersebut. Endapan Ca open parentheses O H close parentheses subscript 2 akan mulai terbentuk pada pH sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets Ca to the power of 2 plus sign close square brackets equals mol over Liter equals fraction numerator 0 comma 01 space mol over denominator 1 space L end fraction equals 0 comma 01 space M equals 10 to the power of negative sign 2 space end exponent M 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 Ksp double bond Q 1 cross times 10 to the power of negative sign 6 end exponent equals open square brackets Ca 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 open square brackets O H to the power of minus sign close square brackets equals square root of 10 to the power of negative sign 6 end exponent over 10 to the power of negative sign 2 end exponent end root equals 10 to the power of negative sign 2 end exponent  pOH equals minus sign log open square brackets O H to the power of minus sign close square brackets equals minus sign log space 10 to the power of negative sign 2 end exponent equals 2 pH equals 14 minus sign pOH equals 14 minus sign 2 equals 12 end cell row blank blank blank end table 

Jadi, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Krisna

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 04 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Pada suhu , . Hitunglah kelarutan  pada : a.  larutan  0,001 M, b.  larutan  0,001 M, c.  larutan yang mempunyai pH = 11, d.  larutan yang berisi  0,1 M dan  0,1 M

Pembahasan Soal:

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 end style 


Dalam larutan jenuh begin mathsize 14px style Ni open parentheses O H close parentheses subscript 2 end style terdapat kesetimbangan 


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 space S space space space space space space space space 2 S end style 


Maka harga Ksp nya dinyatakan dalam :


Error converting from MathML to accessible text. 


a)  Penambahan 0,001 M NaOH

begin mathsize 14px style Na O H yields Na to the power of plus sign and O H to the power of minus sign 0 comma 001 space space space 0 comma 001 space space 0 comma 001 end style   


begin mathsize 14px style open square brackets ion space sejenis close square brackets equals open square brackets O H to the power of minus sign close square brackets equals 0 comma 001 equals 10 to the power of negative sign 3 end exponent end style 


begin mathsize 14px style 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 middle dot open square brackets 2 O H to the power of minus sign close square brackets end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell S middle dot open square brackets 10 to the power of negative sign 3 end exponent 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 middle dot 10 to the power of negative sign 6 end exponent end cell row S equals cell 6 cross times 10 to the power of negative sign 12 end exponent space M end cell end table end style 


Jadi, kelarutan dalam 0,001 M NaOH adalah begin mathsize 14px style 6 cross times 10 to the power of negative sign 12 end exponent space M end style 

b)  Penambahan 0,001 M begin mathsize 14px style Ni Cl subscript 2 end style 


begin mathsize 14px style Ni Cl subscript 2 yields Ni to the power of 2 plus sign and 2 Cl to the power of minus sign 0 comma 001 space space space 0 comma 001 space space 0 comma 002 end style 


begin mathsize 14px style open square brackets ion space sejenis close square brackets equals open square brackets Ni to the power of 2 plus sign close square brackets equals 0 comma 001 equals 10 to the power of negative sign 3 end exponent end style 


 begin mathsize 14px style 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 middle dot open square brackets 2 O H to the power of minus sign close square brackets end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell 10 to the power of negative sign 3 end exponent middle dot open parentheses 2 S close parentheses squared end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell 10 to the power of negative sign 3 end exponent middle dot 4 S squared end cell row cell 6 cross times 10 to the power of negative sign 15 end exponent end cell equals cell 4 S squared 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 3 comma 87 cross times 10 to the power of negative sign 8 end exponent space M end cell end table end style 


Jadi, kelarutan dalam 0,001 M undefined adalah begin mathsize 14px style 3 comma 87 cross times 10 to the power of negative sign 8 end exponent space M end style.

c)  Penambahan larutan pH 11


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals 11 row pOH equals 3 row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 3 end exponent space M end cell end table end style 


begin mathsize 14px style open square brackets ion space sejenis close square brackets equals open square brackets O H to the power of minus sign close square brackets equals 0 comma 001 equals 10 to the power of negative sign 3 end exponent end style 


begin mathsize 14px style 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 middle dot open square brackets 2 O H to the power of minus sign close square brackets end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell S middle dot open square brackets 10 to the power of negative sign 3 end exponent 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 middle dot 10 to the power of negative sign 6 end exponent end cell row S equals cell 6 cross times 10 to the power of negative sign 12 end exponent space M end cell end table end style 


Jadi, kelarutan dalam larutan pH 11 adalah begin mathsize 14px style 6 cross times 10 to the power of negative sign 12 end exponent space M end style.

d)  Penambahan larutan larutan yang berisi begin mathsize 14px style N H subscript 3 end style 0,1 M dan begin mathsize 14px style N H subscript 4 Cl end style 0,1 M undefined


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell Kb cross times fraction numerator open square brackets basa space lemah close square brackets over denominator open square brackets garam close square brackets end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times 1 over 1 end cell row blank equals cell 10 to the power of negative sign 5 end exponent space M end cell end table end style 


begin mathsize 14px style open square brackets ion space sejenis close square brackets equals open square brackets O H to the power of minus sign close square brackets equals open square brackets 10 to the power of negative sign 5 end exponent close square brackets end style 


begin mathsize 14px style 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 middle dot open square brackets 2 O H to the power of minus sign close square brackets end cell row cell 6 cross times 10 to the power of negative sign 18 end exponent end cell equals cell S middle dot open square brackets 10 to the power of negative sign 5 end exponent 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 middle dot 10 to the power of negative sign 10 end exponent end cell row S equals cell 6 cross times 10 to the power of negative sign 8 end exponent space M end cell end table end style 


Jadi, kelarutan dalam larutan penyangga adalah begin mathsize 14px style 6 cross times 10 to the power of negative sign 8 end exponent space M end style.

0

Roboguru

Sulfida timah () sangat tidak larut dalam air, tetapi kelarutannya meningkat sangat tajam dalam suasana asam. Sifat ini tercermin dari nilai  dan  yang berturut-turut sebesar  dan . a. Tuliskan reaks...

Pembahasan Soal:

Di dalam air, garam Sn S akan mengalami kesetimbangan. Reaksi kesetimbangan:

Sn S equilibrium Sn to the power of 2 plus sign and S to the power of 2 minus sign 


Jadi, reaksi kesetimbangannya adalah bottom enclose Sn S bold equilibrium Sn to the power of bold 2 bold plus sign bold and S to the power of bold 2 bold minus sign bold point end enclose  

0

Roboguru

Larutan  memiliki konsentrasi 0,1 M ditambahkan dengan larutan NaOH sedikit demi sedikit untuk menaikkan pH-nya. Jika diketahui , endapan akan terbentuk pada pH .... (log 3 = 0,5)

Pembahasan Soal:

Larutan begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell Ni left parenthesis C O subscript 3 right parenthesis subscript 2 end cell end table end style memiliki konsentrasi 0,1 M.


begin mathsize 14px style Ni left parenthesis C O subscript 3 right parenthesis subscript 2 equilibrium Ni to the power of 2 plus sign plus 2 C O subscript 3 to the power of 2 minus sign end exponent space 0 comma 1 space M space space space space space space space 0 comma 1 space M end style
 

Hitung konsentrasi begin mathsize 14px style O H to the power of minus sign end style dari persamaan undefined begin mathsize 14px style Ni open parentheses O H close parentheses subscript 2 end style.
 

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

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp 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 space open square brackets O H to the power of minus sign close square brackets squared end cell row cell 8 comma 8 cross times 10 to the power of negative sign 7 end exponent end cell equals cell left square bracket 10 to the power of negative sign 1 end exponent right square bracket space open square brackets O H to the power of minus sign close square brackets squared end cell row cell open square brackets O H to the power of minus sign close square brackets squared end cell equals cell fraction numerator 8 comma 8 cross times 10 to the power of negative sign 7 end exponent over denominator 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 8 comma 8 cross times 10 to the power of negative sign 6 end exponent end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 comma 97 cross times 10 to the power of negative sign 3 end exponent end cell row cell open square brackets O H to the power of minus sign close square brackets end cell almost equal to cell 3 cross times 10 to the power of negative sign 3 end exponent end cell end table end style


Kemudian, hitung pH larutan jenuh.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space 3 cross times 10 to the power of negative sign 3 end exponent end cell row pOH equals cell 3 minus sign log space 3 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign left parenthesis 3 minus sign log space 3 right parenthesis end cell row pH equals cell 11 plus log space 3 end cell row pH equals cell 11 plus 0 comma 5 end cell row pH equals cell 11 comma 5 end cell end table end style 


Larutan tepat jenuh pada pH 11,5. Oleh karena itu, endapan akan terbentuk pada pH = 11,6.

Jadi, jawaban yang benar adalah D.

0

Roboguru

Di dalam setiap 1 liter air terlarut 0,01 mol ion . Untuk mendapatkan ion tersebut, maka pH air dinaikkan. Ion mulai mengendap sebagai pada pH ....

Pembahasan Soal:

begin mathsize 14px style K subscript sp equals open square brackets Ca to the power of 2 plus sign close square brackets middle dot open square brackets O H to the power of minus sign close square brackets squared 4 cross times 10 to the power of negative sign 6 end exponent equals 0 comma 01 space middle dot space open square brackets O H to the power of minus sign close square brackets squared open square brackets O H to the power of minus sign close square brackets equals 0 comma 02 pH equals 14 minus sign log space open square brackets O H to the power of minus sign close square brackets pH equals 14 minus sign left parenthesis 2 minus sign log space 2 right parenthesis pH equals 12 plus log space 2 end style

Jadi, jawaban yang benar adalah A.undefinedspace

0

Roboguru

pH larutan jenuh pada adalah 9 + log 2, maka pada suhu tersebut adalah ....

Pembahasan Soal:

Pada soal diketahui data-data sebagai berikut:

  • begin mathsize 14px style pH space Pb open parentheses O H close parentheses subscript 2 equals 9 plus log 2 end style 
  • Larutan begin mathsize 14px style Pb open parentheses O H close parentheses subscript 2 end style jenuh, sehingga hasil kali kelarutan (begin mathsize 14px style K subscript sp end style) akan sama nilainnya dengan hasil kali ion (begin mathsize 14px style Q subscript sp end style)

Hal yang akan dicari tahu adalah begin mathsize 14px style K subscript sp space Pb open parentheses O H close parentheses subscript 2 end style.

Sebelum menentukan begin mathsize 14px style K subscript sp space Pb open parentheses O H close parentheses subscript 2 end style, maka perlu mencari konsentrasi begin mathsize 14px style O H to the power of minus sign end style dari data pH.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH and pOH end cell equals 14 row pOH equals cell 14 minus sign pH end cell row pOH equals cell 14 minus sign left parenthesis 9 plus log space 2 right parenthesis end cell row pOH equals cell 5 minus sign log space 2 end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 

Menentukan begin mathsize 14px style K subscript sp space Pb open parentheses O H close parentheses subscript 2 end style dengan menggunakan reaksi yang terjadi pada larutan. Pada larutan, volume yang digunakan dimisalkan 1 L, sehingga mol (n) akan sama dengan molaritas (M).

 begin mathsize 14px style Pb open parentheses O H close parentheses subscript 2 equilibrium Pb to the power of 2 plus sign and 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 10 to the power of negative sign 5 end exponent space space space 2 cross times 10 to the power of negative sign 5 end exponent end style 

Dari persamaan di atas dapat dihitung begin mathsize 14px style Q subscript sp space Pb open parentheses O H close parentheses subscript 2 end style.

Error converting from MathML to accessible text. 

Sebelumnya telah diketahui bahwa larutan undefined adalah larutan jenuh. Dengan demikian, nilai begin mathsize 14px style K subscript sp space Pb open parentheses O H close parentheses subscript 2 end style sama dengan begin mathsize 14px style Q subscript sp space Pb open parentheses O H close parentheses subscript 2 end style yaitu begin mathsize 14px style 4 cross times 10 to the power of negative sign 15 end exponent end style.  

Jadi, jawaban yang benar adalah D. 

 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved