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 .... (; ;)

Pertanyaan

Sebanyak 100 ml larutan undefined 0,05 M ditambahkan kedalam larutan begin mathsize 14px style Na Cl end style, undefined dan begin mathsize 14px style Na subscript 2 S O subscript 4 end style dengan konsentrasi masing-masing sebesar 0,05 M dan volume 100 ml. Garam yang akan mengendap adalah ....

(begin mathsize 14px style K subscript sp space Pb Cl subscript 2 space equals space 1 comma 70 cross times 10 to the power of negative sign 5 end exponent end style; begin mathsize 14px style K subscript sp space Pb Cr O subscript 4 space equals space 2 comma 0 cross times 10 to the power of negative sign 14 end exponent end style;begin mathsize 14px style K subscript sp space Pb S O subscript 4 space equals space 2 comma 0 cross times 10 to the power of negative sign 6 end exponent end style)undefined 

  1. undefined undefined 

  2. undefinedundefined 

  3. undefined undefined 

  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 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 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 Maret 2021

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