Dalam suatu larutan terdapat dan masing-masing 0,01 M. Larutan tersebut ditetesi sedikit demi sedikit dengan larutan . Anggap volume larutan tidak berubah dengan penambahan larutan tersebut, berapakah konsentrasi ion pada saat mulai mengendap?

Pertanyaan

Dalam suatu larutan terdapat $Ca Cl subscript 2$ dan $Ba Cl subscript 2$ masing-masing 0,01 M. Larutan tersebut ditetesi sedikit demi sedikit dengan larutan $Na subscript 2 S O subscript 4$ . Anggap volume larutan tidak berubah dengan penambahan larutan $Na subscript 2 S O subscript 4$ tersebut, berapakah konsentrasi ion $Ba to the power of 2 plus sign$ pada saat $Ca S O subscript 4$ mulai mengendap?

$left parenthesis K subscript sp space Ca S O subscript 4 equals 9 comma 1 cross times 10 to the power of negative sign 6 end exponent space dan space Ba S O subscript 4 equals 1 comma 1 cross times 10 to the power of negative sign 10 end exponent right parenthesis$ $space$

Pembahasan Soal:

Untuk mengendapkan $Ca S O subscript 4$ diperlukan $S O subscript 4 to the power of 2 minus sign$ sebanyak:

$table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ca S O subscript 4 end cell equals cell open square brackets Ca to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell 9 cross times 10 to the power of negative sign 6 end exponent end cell equals cell open square brackets 0 comma 01 space M close square brackets open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell equals cell fraction numerator 9 cross times 10 to the power of negative sign 6 end exponent over denominator 0 comma 01 space mol space L to the power of negative sign 1 end exponent end fraction end cell row cell open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell equals cell 9 cross times 10 to the power of negative sign 4 end exponent space mol space L to the power of negative sign 1 end exponent end cell end table$

Pada saat $Ca S O subscript 4$ mulai mengendap, konsentrasi $S O subscript 4 to the power of 2 minus sign$ dalam larutan adalah $9 comma 1 cross times 10 to the power of negative sign 4 end exponent space mol space L to the power of negative sign 1 end exponent$ .

$table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ba S O subscript 4 end cell equals cell open square brackets Ba to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell 1 comma 1 cross times 10 to the power of negative sign 10 end exponent end cell equals cell open square brackets Ba to the power of 2 plus sign close square brackets open square brackets 9 comma 1 cross times 10 to the power of negative sign 4 end exponent close square brackets end cell row cell open square brackets Ba to the power of 2 plus sign close square brackets end cell equals cell fraction numerator 1 comma 1 cross times 10 to the power of negative sign 10 end exponent over denominator 9 comma 1 cross times 10 to the power of negative sign 4 end exponent end fraction end cell row cell open square brackets Ba to the power of 2 plus sign close square brackets end cell equals cell 1 comma 21 cross times 10 to the power of negative sign 7 end exponent space mol space L to the power of negative sign 1 end exponent end cell end table$

Jadi konsentrasi $Ba to the power of bold 2 bold plus sign$ pada saat $Ca S O subscript bold 4$ mulai mengendap adalah $bold 1 bold comma bold 21 bold cross times bold 10 to the power of bold minus sign bold 7 end exponent bold space bold mol bold space italic L to the power of bold minus sign bold 1 end exponent$ . $space$

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Rahayu

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 01 Mei 2021

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$Error converting from MathML to accessible text.$

Menentukan konsentrasi Cl dalam $Na Cl$ :

$Na Cl yields Na to the power of plus sign and Cl to the power of bold minus sign 0 comma 001 space space 0 comma 001 space space bold 0 bold comma bold 001$

Menghitung nilai $Q subscript c space Pb Cl subscript 2$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell Pb Cl subscript 2 end cell rightwards arrow cell Pb to the power of 2 plus sign and 2 Cl to the power of minus sign end cell row blank blank blank row cell Q subscript C end cell equals cell open square brackets Pb close square brackets open square brackets Cl to the power of minus sign close square brackets squared end cell row blank equals cell open square brackets 0 comma 002 space M close square brackets open square brackets 0 comma 001 space M close square brackets squared end cell row blank equals cell open square brackets 2 cross times 10 to the power of negative sign 3 end exponent close square brackets open square brackets 10 to the power of negative sign 6 end exponent close square brackets end cell row blank equals cell 2 cross times 10 to the power of negative sign 9 end exponent end cell end table$

Nilai $Q subscript c$ yang dihasilkan adalah $2 cross times 10 to the power of negative sign 9 end exponent$ dan nilai $K subscript sp$ larutan adalah $10 to the power of negative sign 8 end exponent$ . Oleh karena itu, $Q subscript c less than K subscript sp$ menandakan tidak terdapat endapan.

Jadi, jawaban yang benar adalah E.

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Dalam suatu larutan terdapat dan masing-masing 0,01 M. Larutan tersebut ditetesi sedikit demi sedikit dengan larutan . Anggap volume larutan tidak berubah dengan penambahan larutan tersebut, berap...

Pembahasan Soal:

Untuk mengendapkan $Ba S O subscript 4$ diperlukan $S O subscript 4 to the power of 2 minus sign$ sebanyak:

$table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ba S O subscript 4 end cell equals cell open square brackets Ba to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell 1 comma 1 cross times 10 to the power of negative sign 10 end exponent end cell equals cell open square brackets 0 comma 01 space M close square brackets open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell equals cell fraction numerator 1 comma 1 cross times 10 to the power of negative sign 10 end exponent over denominator 0 comma 01 space mol space L to the power of negative sign 1 end exponent end fraction end cell row cell open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell equals cell 1 comma 1 cross times 10 to the power of negative sign 8 end exponent space mol space L to the power of negative sign 1 end exponent end cell end table$

Pada saat $Ba S O subscript 4$ mulai mengendap, konsentrasi $S O subscript 4 to the power of 2 minus sign$ dalam larutan adalah $1 comma 1 cross times 10 to the power of negative sign 8 end exponent space mol space L to the power of negative sign 1 end exponent$ .

$K subscript sp space Ca S O subscript 4 equals open square brackets Ca to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign close square brackets 9 comma 1 cross times 10 to the power of negative sign 6 end exponent equals open square brackets Ca to the power of 2 plus sign close square brackets open square brackets 1 comma 1 cross times 10 to the power of negative sign 8 end exponent close square brackets open square brackets Ca to the power of 2 plus sign close square brackets equals fraction numerator 9 comma 1 cross times 10 to the power of negative sign 6 end exponent over denominator 1 comma 1 cross times 10 to the power of negative sign 8 end exponent end fraction open square brackets Ca to the power of 2 plus sign close square brackets equals 827 comma 27 space mol space L to the power of negative sign 1 end exponent$

Jadi konsentrasi $Ca to the power of bold 2 bold plus sign$ pada saat $Ba S O subscript bold 4$ mulai mengendap adalah $bold 827 bold comma bold 27 bold space bold mol bold space italic L to the power of bold minus sign bold 1 end exponent$ . $space$

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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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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.

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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.

$begin mathsize 14px style italic Q subscript sp space less than space K subscript sp end style$ , belum mengendap
$begin mathsize 14px style italic Q subscript sp space equals space K subscript sp end style$ , mulai terjadi endapan
$begin mathsize 14px style italic Q subscript sp space greater than space K subscript sp end style$ , 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 $begin mathsize 14px style Pb S O subscript 4 end style$ membentuk endapan.

Jadi, jawaban yang benar adalah D. $space$

Roboguru