Roboguru

Ke dalam 100 mL larutan netral yang merupakan campuran dari larutan garam  0,001 M,  0,001 M, dan 0,001 M ditambah 100 mL larutan  0,002 M. Campuran diaduk sampai tercampur rata.; ;. Endapan yang terjadi adalah garam-garam ....

Pertanyaan

Ke dalam 100 mL larutan netral yang merupakan campuran dari larutan garam begin mathsize 14px style K Cl end style 0,001 M, begin mathsize 14px style Na subscript 2 Cr O subscript 4 end style 0,001 M, dan begin mathsize 14px style K subscript 2 S O subscript 4 end style0,001 M ditambah 100 mL larutan begin mathsize 14px style Pb open parentheses N O subscript 3 close parentheses subscript 2 end style 0,002 M. Campuran diaduk sampai tercampur rata.begin mathsize 14px style 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 end stylebegin mathsize 14px style 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 end style;begin mathsize 14px style 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 end style. Endapan yang terjadi adalah garam-garam ....space 

  1. begin mathsize 14px style Pb Cl subscript 2 end style sajaundefined 

  2. begin mathsize 14px style Pb S O subscript 4 end style sajaundefined 

  3. begin mathsize 14px style Pb Cr O subscript 4 end style sajaundefined 

  4. undefined dan undefined undefined 

  5. undefined dan undefined undefined 

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 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 Maret 2021

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