Roboguru

Jika , maka kelarutan  dalam larutan penyangga dengan pH=12 adalah ....

Pertanyaan

Jika K subscript italic s italic p end subscript space Mg open parentheses O H close parentheses subscript 2 equals 3 comma 4 cross times 10 to the power of negative sign 14 end exponent, maka kelarutan Mg open parentheses O H close parentheses subscript 2 dalam larutan penyangga dengan pH=12 adalah ....

  1. 3 comma 4 cross times 10 to the power of negative sign 5 end exponent mol/L

  2. 2 cross times 10 to the power of negative sign 6 end exponent mol/L

  3. 3 comma 4 cross times 10 to the power of negative sign 6 end exponent mol/L

  4. 2 cross times 10 to the power of negative sign 10 end exponent mol/L

  5. 3 comma 4 cross times 10 to the power of negative sign 10 end exponent mol/L

Pembahasan Soal:

Penambahan ion senama bisa mengurangi kelarutan suatu zat. Artinya, semakin banyak ion senama di dalam larutan, zat-zat terlarut semakin sulit untuk larut. Kelarutan dalam larutan penyangga dengan pH=12 adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row pH equals 12 row pOH equals cell 14 minus sign pH end cell row blank equals cell 14 minus sign 12 end cell row blank equals 2 row blank blank blank row cell left square bracket O H minus sign right square bracket end cell equals cell 10 to the power of negative sign 2 space end exponent M end cell row blank blank blank row cell Mg open parentheses O H close parentheses 2 end cell rightwards harpoon over leftwards harpoon cell Mg to the power of 2 plus sign and 2 O H to the power of minus sign end cell row blank blank cell space space space space space space s space space space space space space space space space space space space space s space space space space space space space space 2 s plus 10 to the power of negative sign 2 end exponent end cell row blank blank blank row blank blank blank end table 

Nilai s pada open square brackets O H to the power of minus sign close square brackets diabaikan karena nilainya sangat kecil dibandingan konsentrasi open square brackets O H to the power of minus sign close square brackets tersebut.

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript italic s italic p end subscript end cell equals cell open square brackets Mg 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 3 comma 4 cross times 10 to the power of negative sign 14 end exponent equals open parentheses s close parentheses left parenthesis 10 to the power of negative sign 2 end exponent right parenthesis squared space comma space s equals fraction numerator 3 comma 4 cross times 10 to the power of negative sign 14 end exponent over denominator 10 to the power of negative sign 4 end exponent end fraction end cell row blank equals cell 3 comma 4 cross times 10 to the power of negative sign 10 space end exponent M end cell end table   

 

Jadi, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Krisna

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Hitunglah perbandingan kelarutan  dalam air murni dan dalam larutan

Pembahasan Soal:

Kelarutan adalah kemampuan zat terlarut untuk larut dalam pelarut.
Kelarutan dalam air

Error converting from MathML to accessible text.

Kelarutan dalam begin mathsize 14px style Na subscript 2 S O subscript 4 space 0 comma 01 M end style

begin mathsize 14px style Na subscript 2 S O subscript 4 space space space rightwards arrow space space 2 Na to the power of plus space end exponent plus space S O subscript 4 to the power of 2 minus sign end exponent 0 comma 01 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 0 comma 01 Pb S O subscript 4 space space space space space rightwards arrow space Pb to the power of 2 plus sign space space plus space S O subscript 4 to the power of 2 minus sign end exponent space space space space space space space space space space space space space space space space space space space space space space space space italic s space space space space space space space space space space space space space space K subscript sp space space space equals space open square brackets s close square brackets space left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket 2 comma 25 space cross times space 10 to the power of negative sign 8 end exponent space space equals space italic s italic space italic cross times italic space 0 comma 01 space space space space space space space space space space space space space space space space space space italic s italic space italic space italic space italic space equals italic space fraction numerator 2 comma 25 space cross times space 10 to the power of negative sign 8 end exponent over denominator 0 comma 01 end fraction space space space space space space space space space space space space space space space space space space italic s italic space italic space italic space italic space equals italic space 2 comma 25 space cross times space 10 to the power of negative sign 6 end exponent end style 

Perbandingan kelarutan

Error converting from MathML to accessible text.   

0

Roboguru

Sebanyak 200 mL larutan  0,05 M ditambahkan ke dalam 200 mL larutan  0,05 M. Berapakah massa zat yang mengendap? ; , , , , dan .

Pembahasan Soal:

Reaksi antara begin mathsize 14px style Sr Cl subscript 2 end style dengan begin mathsize 14px style K subscript 2 Cr O subscript 4 end style akan terjadi sesuai reaksi berikut ini.

begin mathsize 14px style Sr Cl subscript 2 and K subscript 2 Cr O subscript 4 yields Sr Cr O subscript 4 and 2 K Cl end style 

begin mathsize 14px style Sr Cr O subscript 4 end style akan mengendap jika begin mathsize 14px style Q subscript sp space Sr Cr O subscript 4 greater than K subscript sp space Sr Cr O subscript 4 end style

Tahap 1. Menentukan terbentuk atau tidaknya endapan undefined 

Diketahui :

begin mathsize 14px style space space space space space open square brackets Sr Cl subscript 2 close square brackets equals 0 comma 05 space M space space space space V space Sr Cl subscript 2 equals 200 space mL space open square brackets K subscript 2 Cr O subscript 4 close square brackets equals 0 comma 05 space M V space K subscript 2 Cr O subscript 4 equals 200 space mL end style 

begin mathsize 14px style straight V subscript 2 equals straight V space total space space space space equals straight V space SrCl subscript 2 plus straight V space straight K subscript 2 CrO subscript 4 space space space space equals 200 space mL plus 200 space mL space space space space equals 400 space mL end style 

begin mathsize 14px style open square brackets Sr Cl subscript 2 close square brackets end style setelah pencampuran

begin mathsize 14px style space space space space space space space space space space space space space space space space space space straight V subscript 1 straight M subscript 1 equals straight V subscript 2 straight M subscript 2 200 space mL cross times 0 comma 05 space straight M equals 400 space mL cross times straight M subscript 2 space space space space space space space space space space space space space space space space space space space space space space straight M subscript 2 equals 0 comma 025 thin space straight M end style 

begin mathsize 14px style space space space Sr Cl subscript 2 space space rightwards arrow space Sr to the power of 2 plus sign space space plus space space 2 Cl to the power of minus sign 0 comma 025 space M space space space 0 comma 025 space M space space space 0 comma 05 space M end style 

begin mathsize 14px style open square brackets straight K subscript 2 CrO subscript 4 close square brackets end style setelah pencampuran

begin mathsize 14px style space space space space space space space space space space space space space space space space space space straight V subscript 1 straight M subscript 1 equals straight V subscript 2 straight M subscript 2 200 space mL cross times 0 comma 05 space straight M equals 400 space mL cross times straight M subscript 2 space space space space space space space space space space space space space space space space space space space space space space straight M subscript 2 equals 0 comma 025 thin space straight M end style 

begin mathsize 14px style space K subscript 2 Cr O subscript 4 yields space space 2 K to the power of plus sign space space plus space Cr O subscript 4 to the power of 2 minus sign end exponent 0 comma 025 space M space space space space 0 comma 05 space M space space space space 0 comma 025 space M end style 

begin mathsize 14px style straight Q subscript sp space SrCrO subscript 4 equals open square brackets Sr to the power of 2 plus end exponent close square brackets open square brackets CrO subscript 4 to the power of 2 minus end exponent close square brackets space space space space space space space space space space space space space space space space space space equals left parenthesis 0 comma 025 right parenthesis cross times left parenthesis 0 comma 025 right parenthesis space space space space space space space space space space space space space space space space space space equals 6 comma 25 cross times 10 to the power of negative 4 end exponent straight K subscript sp space SrCrO subscript 4 equals 3 comma 6 cross times 10 to the power of negative 5 end exponent straight Q subscript sp space SrCrO subscript 4 equals 62 comma 5 cross times 10 to the power of negative 5 end exponent end style 

Terbentuk endapan undefined, karena begin mathsize 14px style straight Q subscript sp space SrCrO subscript 4 greater than straight K subscript sp space SrCrO subscript 4 end style.

Tahap 2. Menentukan massa undefined 

Reaksi antara 200 mL larutan begin mathsize 14px style Sr Cl subscript 2 end style 0,05 M dengan 200 mL larutan begin mathsize 14px style K subscript 2 Cr O subscript 4 end style 0,05 M memiliki stoikiometri sebagai berikut.

begin mathsize 14px style open square brackets Sr Cl subscript 2 close square brackets equals fraction numerator n space Sr Cl subscript 2 over denominator V space Sr Cl subscript 2 end fraction n space Sr Cl subscript 2 equals open square brackets Sr Cl subscript 2 close square brackets cross times V space Sr Cl subscript 2 n space Sr Cl subscript 2 equals 0 comma 05 space mol space L to the power of negative sign 1 end exponent cross times 200 space mL n space Sr Cl subscript 2 equals 0 comma 05 space mol space L to the power of negative sign 1 end exponent cross times 0 comma 2 space L n space Sr Cl subscript 2 equals 0 comma 01 space mol end style 

begin mathsize 14px style space space left square bracket straight K subscript 2 CrO subscript 4 right square bracket equals fraction numerator straight n space straight K subscript 2 CrO subscript 4 over denominator straight V space straight K subscript 2 CrO subscript 4 end fraction straight n space straight K subscript 2 CrO subscript 4 equals left square bracket straight K subscript 2 CrO subscript 4 right square bracket cross times straight V space straight K subscript 2 CrO subscript 4 straight n space straight K subscript 2 CrO subscript 4 equals 0 comma 05 space mol space straight L to the power of negative 1 end exponent cross times 200 space mL straight n space straight K subscript 2 CrO subscript 4 equals 0 comma 05 space mol space straight L to the power of negative 1 end exponent cross times 0 comma 2 space straight L straight n space straight K subscript 2 CrO subscript 4 equals 0 comma 01 space mol end style 

 begin mathsize 14px style Mr space SrCrO subscript 4 equals Ar space Sr plus Ar thin space Cr plus space left parenthesis 4 cross times Ar space straight O right parenthesis Mr space SrCrO subscript 4 equals 88 space straight g space mol to the power of negative 1 end exponent plus 52 space straight g space mol to the power of negative 1 end exponent plus left parenthesis 4 cross times 16 space straight g space mol to the power of negative 1 end exponent right parenthesis Mr space SrCrO subscript 4 equals 140 space straight g space mol to the power of negative 1 end exponent plus space 64 space straight g space mol to the power of negative 1 end exponent Mr space SrCrO subscript 4 equals 204 space straight g space mol to the power of negative 1 end exponent  massa space SrCrO subscript 4 space yang space mengendap equals 0 comma 01 space mol cross times 204 space straight g space mol to the power of negative 1 end exponent massa space SrCrO subscript 4 space yang space mengendap equals 2 comma 04 space straight g 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 end style 

Jadi, massa zat yang mengendap adalah 2,04spacegram.

0

Roboguru

Pada , nilai adalah . Hitunglah kelarutan di dalam larutan yang mempunyai pH = 12.

Pembahasan Soal:

Diketahui:

  • begin mathsize 14px style K subscript sp space Mg open parentheses O H close parentheses subscript 2 space equals space 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end style
  • pH = 12

Ditanya:

s?

Jawab:

Menentukan konsentrasi ion OH 

pHpOH[OH]=====1214pH14122102M 

Menentukan kelarutan:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space end subscript Mg open parentheses O H close parentheses subscript 2 space end cell equals cell space open square brackets Mg 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 end cell row cell 3 comma 2 cross times 10 to the power of negative sign 11 end exponent space end cell equals cell space s space cross times space left parenthesis 10 to the power of negative sign 2 end exponent right parenthesis squared space end cell row cell 3 comma 2 cross times 10 to the power of negative sign 11 end exponent space end cell equals cell space s space cross times space 10 to the power of negative sign 4 end exponent end cell row cell s space end cell equals cell space fraction numerator 3 comma 2 cross times 10 to the power of negative sign 11 end exponent over denominator 10 to the power of negative sign 4 end exponent end fraction end cell row cell s space end cell equals cell space 3 comma 2 cross times 10 to the power of negative sign 7 end exponent space M end cell end table end style 


Jadi, jawaban yang benar kelarutan begin mathsize 14px style Mg open parentheses O H close parentheses subscript 2 end style adalah begin mathsize 14px style 3 comma 2 cross times 10 to the power of negative sign 7 end exponent space M end style. 

0

Roboguru

Apakah yang akan terjadi jika ke dalam larutan jenuh  dialirkan gas ? Jelaskan!

Pembahasan Soal:

Reaksi kesetimbangan Na Cl 

Na Cl open parentheses italic s close parentheses equilibrium 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 

Ketika gas H Cl ditambahkan maka akan terjadi penambahan konsentrasi ion Cl to the power of minus sign dan mengakibatkan kesetimbangan bergeser kearah pembentukan endapan Na Cl yang menyebabkan kelarutan Na Cl berkurang.

0

Roboguru

Pada suhu , . Hitunglah kelarutan pada larutan .

Pembahasan Soal:

Reaksi yang terjadi pada larutan begin mathsize 14px style Ni Cl subscript 2 end style adalah: 

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Misalkan s adalah kelarutan molar begin mathsize 14px style Ni open parentheses O H close parentheses subscript 2 end style dalam larutan begin mathsize 14px style Ni Cl subscript 2 end style, maka perhitungan konsentrasi masing-masing ion adalah:

begin mathsize 14px style table row blank cell Ni open parentheses O H close parentheses subscript 2 left parenthesis italic s right parenthesis end cell rightwards arrow over leftwards arrow cell Ni to the power of 2 plus sign left parenthesis italic a italic q right parenthesis end cell plus cell 2 O H to the power of minus sign open parentheses aq close parentheses end cell row cell mula bond mula end cell blank blank cell 0 comma 001 end cell blank minus sign row reaksi blank blank s blank cell 2 s end cell row sisa blank blank cell 0 comma 001 plus s end cell blank cell 2 s end cell end table end style

Karena nilai begin mathsize 14px style K subscript sp end style sangat kecil, nilai sakan sangat kecil dibanding dengan 0,001, dapat kita anggap:

begin mathsize 14px style 0 comma 001 plus s almost equal to 0 comma 001 end style

sehingga:

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 open square brackets O H to the power of minus sign close square brackets squared end cell row cell 6 x 10 to the power of negative sign 18 end exponent end cell equals cell 0 comma 001 middle dot open parentheses 2 s close parentheses squared end cell row cell 6 x 10 to the power of negative sign 18 end exponent end cell equals cell 0 comma 004 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 end cell end table end style

Jadi, kelarutan begin mathsize 14px style Ni bold open parentheses O H bold close parentheses subscript bold 2 end style pada larutan begin mathsize 14px style Ni Cl subscript bold 2 end style 0,001 M adalah begin mathsize 14px style bold 3 bold comma bold 87 bold cross times bold 10 to the power of bold minus sign bold 8 end exponent end style M.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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