Roboguru

Diketahui  adalah . Untuk mengendapkan  dari larutan  0,01 M; diperlukan konsentrasi ion  sebesar ...

Pertanyaan

Diketahui K subscript sp space Ag subscript 2 Cr O subscript 4 adalah 1 cross times 10 to the power of negative sign 12 end exponent. Untuk mengendapkan Ag subscript 2 Cr O subscript 4 dari larutan K subscript 2 Cr O subscript 4 0,01 M; diperlukan konsentrasi ion Ag to the power of plus sign sebesar ...space 

  1. 1 cross times 10 to the power of negative sign 7 space end exponent Mspace  

  2. 1 cross times 10 to the power of negative sign 6 space end exponent Mspace 

  3. 1 cross times 10 to the power of negative sign 5 space end exponent Mspace 

  4. 1 cross times 10 to the power of negative sign 4 space end exponent Mspace 

  5. 1 cross times 10 to the power of negative sign 2 space end exponent Mspace 

Pembahasan Soal:

Larutan K subscript 2 Cr O subscript 4 0,01 M mengandung ion sebagai berikut:

K subscript 2 Cr O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 K to the power of plus sign left parenthesis italic a italic q right parenthesis plus Cr O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis space space space space 0 comma 01 space M space space space space space space space 0 comma 02 space M space space space space space space space 0 comma 01 space M  

Apabila kedalam larutan ditambahkan Ag subscript 2 Cr O subscript 4 padat, kristal tersebut akan larut hingga larutan jenuh. Misal kelarutan Ag subscript 2 Cr O subscript 4 double bond s space mol space L to the power of negative sign 1 end exponent maka:

Ag subscript 2 Cr O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 Ag to the power of plus sign left parenthesis italic a italic q right parenthesis plus Cr O subscript 4 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis space 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 space italic space italic space italic space italic space italic space 2 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  

Konsentrasi total ion Cr O subscript 4 to the power of 2 minus sign dalam larutan adalah 0 comma 01 space M and italic s space mol space L to the power of negative sign 1 end exponent space, nilai s sangat kecil maka kosentrasi ion total Cr O subscript 4 to the power of 2 minus sign dianggap 0,01 M.

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ag subscript 2 Cr O subscript 4 end cell equals cell open square brackets Ag to the power of plus sign close square brackets squared open square brackets Cr O subscript 4 to the power of 2 minus sign end exponent close square brackets end cell row cell 1 cross times 10 to the power of negative sign 12 end exponent end cell equals cell open square brackets 2 s close square brackets squared open square brackets 10 to the power of negative sign 2 end exponent close square brackets end cell row cell 4 s squared end cell equals cell fraction numerator 1 cross times 10 to the power of negative sign 12 end exponent over denominator 10 to the power of negative sign 2 end exponent end fraction end cell row cell 4 s squared end cell equals cell 10 to the power of negative sign 10 end exponent end cell row cell 2 s end cell equals cell 10 to the power of negative sign 5 end exponent space M end cell row cell 2 s end cell equals cell open square brackets Ag to the power of plus sign close square brackets equals 10 to the power of negative sign 5 end exponent space M end cell end table  

Jadi konsentrasi Ag to the power of bold plus sign yang diperlukan adalah bold 1 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent bold space italic M.

Oleh karena itu, jawaban yang benar adalah C.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Rahayu

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 01 Mei 2021

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Sebanyak 5 gelas kimia berisi larutan dengan volume yang sama. Jika ke dalam kelima gelas kimia terebut dilarutkan sejumlah perak klorida padat, perak klorida padat akan paling mudah larut dalam gelas...

Pembahasan Soal:

Campuran tersebut memiliki ion senama yaitu ion Cl to the power of minus sign. Semakin tinggi konsentrasi ion senama maka kelarutan padatan perak klorida tersebut akan semakin kecil dan sebaliknya.

Jadi perak klorida akan semakin mudah larut di larutan asam klorida dengan konsentrasi terkecil yaitu 0,01 M.

Oleh karena itu, jawaban yang benar adalah A.space 

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Pembahasan Soal:

Harga begin mathsize 14px style K subscript sp end style suatu zat dapat digunakan untuk memperkirakan zat tersebut larut atau mengendap. Apabila harga undefined suatu zat dibandingkan dengan hasil kali konsentrasi ion-ion zat tersebut dipangkatkan masing-masing koefisien reaksi (begin mathsize 14px style italic Q subscript sp end style), akan ada tiga kemungkinan seperti berikut.

  1. begin mathsize 14px style italic Q subscript sp space less than space K subscript sp end style, belum mengendap
  2. begin mathsize 14px style italic Q subscript sp space equals space K subscript sp end style, mulai terjadi endapan
  3. begin mathsize 14px style italic Q subscript sp space greater than space K subscript sp end style, terjadi endapan

Guna mengetahui endapan garam mana saja yang terbentuk, maka perlu diketahui konsentrasi ion-ion pembentuk garam.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space K Cl space end cell equals cell space mol space Na subscript 2 Cr O subscript 4 space equals space mol space K subscript 2 S O subscript 4 end cell row cell mol space K Cl space end cell equals cell space M space K Cl cross times V space K Cl end cell row blank equals cell space 0 comma 001 space M cross times 100 space mL end cell row blank equals cell space 0 comma 1 space mmol end cell row cell mol space Pb open parentheses N O subscript 3 close parentheses subscript 2 space end cell equals cell space M space Pb open parentheses N O subscript 3 close parentheses subscript 2 cross times V space Pb open parentheses N O subscript 3 close parentheses subscript 2 end cell row blank equals cell space 0 comma 002 thin space M cross times 100 space mL end cell row blank equals cell space 0 comma 2 space mmol end cell end table end style


Adapun dari reaksi ionisasi larutan-larutan tersebut, dapat diperoleh mol ion-ion pembentuk garamnya.


begin mathsize 14px style K Cl left parenthesis italic a italic q right parenthesis yields K 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 0 comma 1 space mmol space space space space space space space space space space space space space space space space 0 comma 1 space mmol Na subscript 2 Cr O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 Na to the power of plus sign left parenthesis italic a italic q right parenthesis plus Cr O subscript 4 to the power of 2 minus sign left parenthesis italic a italic q right parenthesis 0 comma 1 space mmol 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 1 space mmol K subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 K 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 left parenthesis italic a italic q right parenthesis 0 comma 1 space mmol 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 1 space mmol Pb open parentheses N O subscript 3 close parentheses subscript 2 left parenthesis italic a italic q right parenthesis yields Pb to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis 0 comma 2 space mmol space space space space space space space space space space 0 comma 2 space mmol end style


Setelah diperoleh nilai mol tiap-tiap ion, maka dapat dihitung konsentrasi ion-ion tersebut.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Cl to the power of minus sign close square brackets space end cell equals cell space open square brackets Cr subscript 2 O subscript 4 to the power of minus sign close square brackets space equals space open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell open square brackets Cl to the power of minus sign close square brackets space end cell equals cell space fraction numerator mol space open square brackets Cl to the power of minus sign close square brackets space space over denominator V space total end fraction end cell row blank equals cell space fraction numerator 0 comma 1 space mmol over denominator 200 space mL end fraction end cell row blank equals cell space 5 cross times 10 to the power of negative sign 4 end exponent space M end cell row cell open square brackets Pb to the power of 2 plus sign close square brackets space end cell equals cell space fraction numerator mol space open square brackets Pb to the power of 2 plus sign close square brackets space over denominator V space total end fraction end cell row blank equals cell space fraction numerator 0 comma 2 space mmol over denominator 200 space mL end fraction end cell row blank equals cell space 10 to the power of negative sign 3 end exponent space M end cell end table end style


Nilai undefined tiap-tiap garam dapat dihitung menggunakan data konsentrasi ion-ion penyusunnya. Perhitungan nilai begin mathsize 14px style italic Q subscript sp italic space Pb Cl subscript 2 end style adalah sebagai berikut.


begin mathsize 14px style Pb Cl subscript 2 open parentheses italic s close parentheses rightwards harpoon over leftwards harpoon Pb 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 italic Q subscript sp italic space Pb Cl subscript 2 space equals space open square brackets Pb to the power of 2 plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets squared equals space left parenthesis 10 to the power of negative sign 3 end exponent right parenthesis left parenthesis 5 cross times 10 to the power of negative sign 4 end exponent right parenthesis squared equals space 2 comma 5 cross times 10 to the power of negative sign 10 end exponent K subscript sp italic space Pb Cl subscript 2 space equals space 1 comma 7 cross times 10 to the power of negative sign 5 end exponent italic Q subscript sp space Pb Cl subscript 2 space less than space K subscript sp space Pb Cl subscript 2 table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table end style


Berdasarkan perhitungan di atas, dapat disimpulkan bahwa garam begin mathsize 14px style Pb Cl subscript 2 end style belum mengendap.

Hal yang sama dilakukan untuk garam begin mathsize 14px style Pb Cr O subscript 4 end style dan begin mathsize 14px style Pb S O subscript 4 end style.


begin mathsize 14px style Pb Cr O subscript 4 open parentheses italic s close parentheses rightwards harpoon over leftwards harpoon Pb to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus Cr O subscript 4 to the power of 2 minus sign left parenthesis italic a italic q right parenthesis italic Q subscript sp italic space Pb Cr O subscript 4 space equals space open square brackets Pb to the power of 2 plus sign close square brackets space open square brackets Cr O subscript 4 to the power of 2 minus sign close square brackets equals space left parenthesis 10 to the power of negative sign 3 end exponent right parenthesis left parenthesis 5 cross times 10 to the power of negative sign 4 end exponent right parenthesis equals space 5 cross times 10 to the power of negative sign 7 end exponent K subscript sp italic space Pb Cr O subscript 4 space equals space 1 comma 8 cross times 10 to the power of negative sign 14 end exponent italic Q subscript sp space Pb Cr O subscript 4 space greater than space K subscript sp space Pb Cr O subscript 4  Pb S O subscript 4 open parentheses italic s close parentheses rightwards harpoon over leftwards harpoon 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 left parenthesis italic a italic q right parenthesis italic Q subscript sp italic space Pb S O subscript 4 space equals space open square brackets Pb to the power of 2 plus sign close square brackets space open square brackets S O subscript 4 to the power of 2 minus sign close square brackets equals space left parenthesis 10 to the power of negative sign 3 end exponent right parenthesis left parenthesis 5 cross times 10 to the power of negative sign 4 end exponent right parenthesis equals space 5 cross times 10 to the power of negative sign 7 end exponent K subscript sp italic space Pb S O subscript 4 space equals space 1 comma 8 cross times 10 to the power of negative sign 8 end exponent italic Q subscript sp space Pb S O subscript 4 space greater than space K subscript sp space Pb S O subscript 4 table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table end style


Berdasarkan perhitungan di atas, garam undefined dan undefinedmembentuk endapan.


Jadi, jawaban yang benar adalah D.space 

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Roboguru

Sebanyak 100 mL Ag­NO­3 0,01 M dicampur dengan 100 mL H2SO4 0,01 M. Diketahui Ksp  Ag2SO4 = 3,2.10-5. Pernyataan yang benar mengenai campuran tersebut adalah

Pembahasan Soal:

A g subscript 2 S O subscript 4 space end subscript leftwards arrow over rightwards arrow space 2 A g to the power of plus space plus space S O subscript 4 to the power of negative 2 end exponent  Q c space equals space left square bracket A g to the power of plus right square bracket squared subscript c a m p u r a n end subscript space left square bracket S O subscript 4 to the power of negative 2 end exponent right square bracket subscript c a m p u r a n end subscript  space space space space space space space space space space space space equals space open square brackets fraction numerator M subscript space a w a l end subscript space cross times space V subscript space a w a l end subscript over denominator V subscript space t o t a l end subscript end fraction close square brackets squared space A g to the power of plus cross times space open square brackets fraction numerator space M subscript a w a l end subscript space cross times space V subscript a w a l end subscript over denominator V subscript t o t a l end subscript end fraction close square brackets space S O subscript 4 to the power of negative 2 end exponent    space space space space space space space space space space space space equals open square brackets fraction numerator 100 space x space 0 comma 01 over denominator 200 space end fraction close square brackets squared space space x space space open square brackets fraction numerator 100 space x space 0 comma 01 over denominator 200 space end fraction close square brackets  space space space space space space space space space space space space equals space left parenthesis 2 comma 5 space x space 10 to the power of negative 5 end exponent space right parenthesis space x space left parenthesis 5 space x space 10 to the power of negative 2 end exponent right parenthesis  space space space space space space space space space space space space equals space 1 comma 25 space x space 10 to the power of negative 6 end exponent    Q c space less than space K s p space left parenthesis e n d a p a n space b e l u m space t e r b e n t u k right parenthesis.

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Roboguru

10 mL  0,03 M dicampur dengan 20 mL  0,06 M. Maka, pernyataan yang benar adalah .... ()

Pembahasan Soal:

Harga begin mathsize 14px style K subscript sp end style suatu zat dapat digunakan untuk memperkirakan zat tersebut larut atau mengendap. Apabila harga undefined suatu zat dibandingkan dengan hasil kali konsentrasi ion-ion zat tersebut dipangkatkan masing-masing koefisien reaksi (begin mathsize 14px style italic Q subscript sp end style), akan ada tiga kemungkinan seperti berikut.

  1. begin mathsize 14px style italic Q subscript italic sp italic space less than italic space K subscript sp end style, belum mengendap
  2. begin mathsize 14px style italic Q subscript italic s italic p end subscript italic space equals italic space K subscript sp end style, mulai terjadi endapan
  3. begin mathsize 14px style italic Q subscript italic s italic p end subscript italic space greater than italic space K subscript sp end style, terjadi endapan

Guna mengetahui endapan garam mana saja yang terbentuk, maka perlu diketahui konsentrasi ion-ion pembentuk garam.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na Cl space end cell equals cell space M space Na Cl cross times V space Na Cl end cell row blank equals cell space 0 comma 06 space M cross times 20 space mL end cell row blank equals cell space 1 comma 2 space mmol end cell row cell mol space Pb open parentheses N O subscript 3 close parentheses subscript 2 space end cell equals cell space M space Pb open parentheses N O subscript 3 close parentheses subscript 2 cross times V space Pb open parentheses N O subscript 3 close parentheses subscript 2 end cell row blank equals cell space 0 comma 03 thin space M cross times 10 space mL end cell row blank equals cell space 0 comma 3 space mmol end cell end table end style


Reaksi yang terjadi antara kedua larutan adalah sebagai berikut.


begin mathsize 14px style Pb open parentheses N O subscript 3 close parentheses subscript 2 left parenthesis italic a italic q right parenthesis plus 2 Na Cl left parenthesis italic a italic q right parenthesis yields Pb Cl subscript 2 open parentheses italic s close parentheses and 2 Na N O subscript 3 left parenthesis italic a italic q right parenthesis end style


Garam yang akan diamati pembentukkan endapannya adalah garam begin mathsize 14px style Pb Cl subscript 2 end style. Berdasarkan reaksi ionisasi kedua larutan, dapat diperoleh mol tiap-tiap ion pembentuk garam sebagai berikut.


begin mathsize 14px style Pb open parentheses N O subscript 3 close parentheses subscript 2 left parenthesis italic a italic q right parenthesis yields Pb to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis 0 comma 3 space mmol space space space space space space space space space space space space 0 comma 3 space mmol  Na Cl 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 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis 1 comma 2 space mmol space space space space space space space space space space space space space space space space space space space space 1 comma 2 space mmol space end style


Setelah diperoleh nilai mol tiap-tiap ion, maka dapat dihitung konsentrasi ion-ion tersebut.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Cl to the power of minus sign close square brackets space end cell equals cell space fraction numerator mol space open square brackets Cl to the power of minus sign close square brackets space space over denominator V space total end fraction end cell row blank equals cell space fraction numerator 1 comma 2 space mmol over denominator 30 space mL end fraction end cell row blank equals cell space 0 comma 04 space M end cell row cell open square brackets Pb to the power of 2 plus sign close square brackets space end cell equals cell space fraction numerator mol space open square brackets Pb to the power of 2 plus sign close square brackets space over denominator V space total end fraction end cell row blank equals cell space fraction numerator 0 comma 3 space mmol over denominator 30 space mL end fraction end cell row blank equals cell space 0 comma 01 space M end cell end table end style


Kemudian, perhitungan nilai begin mathsize 14px style italic Q subscript sp space Pb Cl subscript 2 end style adalah sebagai berikut.


begin mathsize 14px style Pb Cl subscript 2 open parentheses italic s close parentheses rightwards harpoon over leftwards harpoon Pb 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 italic Q subscript sp italic space Pb Cl subscript 2 space equals space open square brackets Pb to the power of 2 plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets squared equals space left parenthesis 0 comma 01 right parenthesis left parenthesis 0 comma 04 right parenthesis squared equals space 1 comma 6 cross times 10 to the power of negative sign 5 end exponent K subscript sp italic space Pb Cl subscript 2 space equals space 1 comma 6 cross times 10 to the power of negative sign 6 end exponent italic Q subscript sp space Pb Cl subscript 2 space greater than space K subscript sp space Pb Cl subscript 2 table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table end style


Berdasarkan perhitungan di atas, dapat disimpulkan bahwa garam undefined membentuk endapan.

Jadi, jawaban yang benar adalah A.undefined 

0

Roboguru

Sebanyak 100 ml larutan  0,05 M ditambahkan kedalam larutan ,  dan  dengan konsentrasi masing-masing sebesar 0,05 M dan volume 100 ml. Garam yang akan mengendap adalah .... (; ;)

Pembahasan Soal:

Harga begin mathsize 14px style K subscript sp end style suatu zat dapat digunakan untuk memperkirakan zat tersebut larut atau mengendap. Apabila harga undefined suatu zat dibandingkan dengan hasil kali konsentrasi ion-ion zat tersebut dipangkatkan masing-masing koefisien reaksi (begin mathsize 14px style italic Q subscript sp end style), akan ada tiga kemungkinan seperti berikut.

  1. begin mathsize 14px style italic Q subscript sp space less than space K subscript sp end style, belum mengendap
  2. begin mathsize 14px style italic Q subscript sp space equals space K subscript sp end style, mulai terjadi endapan
  3. begin mathsize 14px style italic Q subscript sp space greater than space K subscript sp end style, terjadi endapan

Guna mengetahui endapan garam mana saja yang terbentuk, maka perlu diketahui konsentrasi ion-ion pembentuk garam.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na Cl space end cell equals cell space mol space Na subscript 2 Cr O subscript 4 space equals space mol space Na subscript 2 S O subscript 4 end cell row cell mol space Na Cl space end cell equals cell space M space Na Cl cross times V space Na Cl end cell row blank equals cell space 0 comma 05 space M cross times 100 space mL end cell row blank equals cell space 5 space mmol end cell row cell mol space Pb open parentheses N O subscript 3 close parentheses subscript 2 space end cell equals cell space M space Pb open parentheses N O subscript 3 close parentheses subscript 2 cross times V space Pb open parentheses N O subscript 3 close parentheses subscript 2 end cell row blank equals cell space 0 comma 05 thin space M cross times 100 space mL end cell row blank equals cell space 5 space mmol end cell end table end style


Adapun dari reaksi ionisasi larutan-larutan tersebut, dapat diperoleh mol ion-ion pembentuk garamnya.


begin mathsize 14px style Na Cl 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 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis 5 space mmol space space space space space space space space space space space space space space space space space space space space space space space 5 space mmol Na subscript 2 Cr O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 Na to the power of plus sign left parenthesis italic a italic q right parenthesis plus Cr O subscript 4 to the power of 2 minus sign left parenthesis italic a italic q right parenthesis 5 space mmol 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 space 5 space mmol Na subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis yields 2 Na 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 left parenthesis italic a italic q right parenthesis 5 space mmol 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 5 space mmol Pb open parentheses N O subscript 3 close parentheses subscript 2 left parenthesis italic a italic q right parenthesis yields Pb to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus 2 N O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis 5 space mmol space space space space space space space space space space space space space space 5 space mmol end style


Setelah diperoleh nilai mol tiap-tiap ion, maka dapat dihitung konsentrasi ion-ion tersebut.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Cl to the power of minus sign close square brackets space end cell equals cell space open square brackets Cr subscript 2 O subscript 4 to the power of minus sign close square brackets space equals space open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell open square brackets Cl to the power of minus sign close square brackets space end cell equals cell space fraction numerator mol space open square brackets Cl to the power of minus sign close square brackets space space over denominator V space total end fraction end cell row blank equals cell space fraction numerator 5 space mmol over denominator 200 space mL end fraction end cell row blank equals cell space 0 comma 025 space M end cell row cell open square brackets Pb to the power of 2 plus sign close square brackets space end cell equals cell space fraction numerator mol space open square brackets Pb to the power of 2 plus sign close square brackets space over denominator V space total end fraction end cell row blank equals cell space fraction numerator 5 space mmol over denominator 200 space mL end fraction end cell row blank equals cell space 0 comma 025 space M end cell end table end style


Nilai undefined tiap-tiap garam dapat dihitung menggunakan data konsentrasi ion-ion penyusunnya. Perhitungan nilai begin mathsize 14px style italic Q subscript sp space Pb Cl subscript 2 end style adalah sebagai berikut.


begin mathsize 14px style Pb Cl subscript 2 open parentheses italic s close parentheses rightwards harpoon over leftwards harpoon Pb 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 italic Q subscript sp italic space Pb Cl subscript 2 space equals space open square brackets Pb to the power of 2 plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets squared equals space left parenthesis 0 comma 025 right parenthesis left parenthesis 0 comma 025 right parenthesis squared equals space 1 comma 56 cross times 10 to the power of negative sign 5 end exponent K subscript sp italic space Pb Cl subscript 2 space equals space 1 comma 70 cross times 10 to the power of negative sign 5 end exponent italic Q subscript sp space Pb Cl subscript 2 space less than space K subscript sp space Pb Cl subscript 2 table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table end style


Berdasarkan perhitungan di atas, dapat disimpulkan bahwa garam begin mathsize 14px style Pb Cl subscript 2 end style belum mengendap.

begin mathsize 14px style Pb Cr O subscript 4 open parentheses italic s close parentheses rightwards harpoon over leftwards harpoon Pb to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus Cr O subscript 4 to the power of 2 minus sign left parenthesis italic a italic q right parenthesis italic Q subscript sp italic space Pb Cr O subscript 4 space equals space open square brackets Pb to the power of 2 plus sign close square brackets space open square brackets Cr O subscript 4 to the power of 2 minus sign close square brackets equals space left parenthesis 0 comma 025 right parenthesis left parenthesis 0 comma 025 right parenthesis equals space 6 comma 25 cross times 10 to the power of negative sign 4 end exponent K subscript sp italic space Pb Cr O subscript 4 space equals space 2 comma 0 cross times 10 to the power of negative sign 14 end exponent italic Q subscript sp space Pb Cr O subscript 4 space greater than space K subscript sp space Pb Cr O subscript 4  Pb S O subscript 4 open parentheses italic s close parentheses rightwards harpoon over leftwards harpoon 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 left parenthesis italic a italic q right parenthesis italic Q subscript sp italic space Pb S O subscript 4 space equals space open square brackets Pb to the power of 2 plus sign close square brackets space open square brackets S O subscript 4 to the power of 2 minus sign close square brackets equals space left parenthesis 0 comma 025 right parenthesis left parenthesis 0 comma 025 right parenthesis equals space 6 comma 25 cross times 10 to the power of negative sign 4 end exponent K subscript sp italic space Pb S O subscript 4 space equals space 2 cross times 10 to the power of negative sign 6 end exponent italic Q subscript sp space Pb S O subscript 4 space greater than space K subscript sp space Pb S O subscript 4 table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row blank blank blank end table end style


Berdasarkan perhitungan di atas, garam begin mathsize 14px style Pb Cr O subscript 4 end style dan begin mathsize 14px style Pb S O subscript 4 end stylemembentuk endapan.


Jadi, jawaban yang benar adalah D.space 

0

Roboguru

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