Roboguru

Sebanyak 200 mL larutan SrCl2​ 0,05 M ditambahkan ke dalam 200 mL larutan K2​CrO4​ 0,05 M. Berapakah massa zat yang mengendap? (Ksp​SrCrO4​=3,6×10−5; Ar​:Sr=88gmol−1, Cl=35,5gmol−1, K=39gmol−1, Cr=52gmol−1, dan O=16gmol−1).

Pertanyaan

Sebanyak 200 mL larutan begin mathsize 14px style Sr Cl subscript 2 end style 0,05 M ditambahkan ke dalam 200 mL larutan begin mathsize 14px style K subscript 2 Cr O subscript 4 end style 0,05 M. Berapakah massa zat yang mengendap? begin mathsize 14px style left parenthesis K subscript sp space Sr Cr O subscript 4 equals 3 comma 6 cross times 10 to the power of negative sign 5 end exponent end stylebegin mathsize 14px style italic A subscript r colon space Sr equals 88 space g space mol to the power of negative sign 1 end exponent end stylebegin mathsize 14px style Cl equals 35 comma 5 space g space mol to the power of negative sign 1 end exponent end stylebegin mathsize 14px style K equals 39 space g space mol to the power of negative sign 1 end exponent end stylebegin mathsize 14px style Cr equals 52 space g space mol to the power of negative sign 1 end exponent end style, dan begin mathsize 14px style O equals 16 space g space mol to the power of negative sign 1 end exponent right parenthesis end style.undefined 

  1. ...undefined 

  2. ...

Pembahasan Video:

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.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Rochmawatie

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Jika diketahui Ksp​Hg2​Cl2​=1×10−18, kelarutan Hg2​Cl2​ dalam larutan CaCl2​ 0,01 M adalah ....

Pembahasan Soal:

Adanya ion senama di dalam larutan akan memperkecil atau mengurangi kelarutan suatu zat tersebut.

Langkah pertama, menentukan reaksi ionisasi Ca Cl subscript 2.


Ca Cl subscript 2 open parentheses italic s close parentheses yields Ca to the power of 2 plus sign left parenthesis italic a italic q right parenthesis plus bold 2 Cl to the power of bold minus sign bold left parenthesis italic a italic q bold right parenthesis 0 comma 01 space M space space space space space space space space space 0 comma 01 space M space space space space space bold 0 bold comma bold 02 bold space italic M 


langkah kedua, menentukan reaksi ionisasi Hg subscript 2 Cl subscript 2.


Hg subscript 2 Cl subscript 2 open parentheses italic s close parentheses yields 2 space Hg to the power of 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 space s space M space space space space space space space space space space space space space space 2 s space M space space space space space space space space space space space 0 comma 02 space M


Langkah ketiga, menentukan nilai kelarutan dengan mengunakan rumus K subscript sp.


table attributes columnalign right center left columnspacing 0px end attributes row cell Ksp space Hg subscript 2 Cl subscript 2 end cell equals cell open square brackets Hg to the power of plus sign close square brackets to the power of 2 space end exponent open square brackets Cl to the power of minus sign close square brackets squared space space space space space end cell row cell 1 cross times 10 to the power of negative sign 18 end exponent end cell equals cell left parenthesis 2 s right parenthesis squared left parenthesis 0 comma 02 right parenthesis squared end cell row cell 1 cross times 10 to the power of negative sign 18 end exponent end cell equals cell left parenthesis 4 s squared right parenthesis left parenthesis 0 comma 0004 right parenthesis end cell row cell 4 s squared space end cell equals cell fraction numerator 1 cross times 10 to the power of negative sign 18 end exponent over denominator 4 cross times 10 to the power of negative sign 4 end exponent end fraction end cell row cell 4 s squared end cell equals cell 0 comma 25 cross times 10 to the power of negative sign 14 end exponent end cell row cell s squared end cell equals cell fraction numerator 0 comma 25 cross times 10 to the power of negative sign 14 end exponent over denominator 4 end fraction end cell row blank equals cell 0 comma 0625 cross times 10 to the power of negative sign 14 end exponent end cell row s equals cell square root of 0 comma 0625 cross times 10 to the power of negative sign 14 end exponent end root end cell row blank equals cell 0 comma 25 cross times 10 to the power of negative sign 7 end exponent end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 8 end exponent end cell end table 


Jadi, kelarutannya adalah 2 comma 5 cross times 10 to the power of negative sign 8 end exponent.

0

Roboguru

Jika diketaui Ksp PbCl2​=1,7×10−5. Hitunglah kelarutan PbCl2​ dalam larutan CaCl2​ 0,1 M!

Pembahasan Soal:

Jika diasumsikan larutan begin mathsize 14px style Ca Cl subscript 2 end style sebanyak 1 L, maka

begin mathsize 14px style n subscript Ca Cl subscript 2 end subscript equals 0 comma 1 cross times 1 equals 0 comma 1 space mol Ca Cl subscript 2 open parentheses aq close parentheses yields Ca to the power of 2 plus sign open parentheses aq close parentheses and 2 Cl to the power of minus sign open parentheses aq close parentheses n subscript Cl to the power of minus sign end subscript equals 2 cross times n subscript Ca Cl subscript 2 end subscript equals 0 comma 2 space mol open square brackets Cl to the power of minus sign close square brackets equals begin inline style bevelled fraction numerator 0 comma 2 over denominator 1 end fraction end style equals 0 comma 2 space M  Pb Cl subscript 2 open parentheses s close parentheses equilibrium Pb to the power of 2 plus sign open parentheses aq close parentheses and 2 Cl to the power of minus sign open parentheses aq close parentheses s space space space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space space space 2 s K subscript sp equals 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 1 comma 7 cross times 10 to the power of negative sign 5 end exponent equals open parentheses s close parentheses open parentheses 0 comma 2 plus 2 s close parentheses squared s space diabaikan space karena space perbandingan space K subscript sp space end subscript dan space open square brackets Cl to the power of minus sign close square brackets space almost equal to 10 to the power of negative sign 4 end exponent 1 comma 7 cross times 10 to the power of negative sign 5 end exponent equals open parentheses s close parentheses open parentheses 0 comma 04 close parentheses s equals 4 comma 25 cross times 10 to the power of negative sign 4 end exponent space M end style

Dengan demikian, maka jawaban yang tepat adalah sesuai penjelasan di atas.

0

Roboguru

Kelarutan L(OH)2​ pada suhu tertentu adalah 2×10−5mol/L. Hitunglah kelarutan zat tersebut di dalam 500 mL larutan yang mempunyai pH = 4 - log 2.

Pembahasan Soal:

Jika ke dalam sistem kesetimbangan kelarutan ditambahkan ion yang senama (ion sejenis), kelarutan senyawa tersebut akan berkurang.

Tahapan yang dapat dilakukan adalah:

  • Menentukan Ksp begin mathsize 14px style L open parentheses O H close parentheses subscript 2 end style

  • Menentukan kelarutan begin mathsize 14px style L open parentheses O H close parentheses subscript 2 end styledalam 500 mL
     

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 4 minus sign log space 2 end cell row cell H to the power of plus sign end cell equals cell 2 cross times 10 to the power of negative sign 4 end exponent end cell row cell open square brackets H to the power of plus sign close square brackets open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 14 end exponent end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 5 cross times 10 to the power of negative sign 11 end exponent end cell row blank blank blank end table end style

Sehingga, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell L open parentheses O H close parentheses subscript 2 end cell rightwards harpoon over leftwards harpoon cell L 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 s space space space space space space space 5 cross times 10 to the power of negative sign 11 end exponent end cell row Ksp equals cell open square brackets L 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 3 comma 2 cross times 10 to the power of negative sign 14 end exponent end cell equals cell s cross times open parentheses 2 s plus up diagonal strike 5 cross times 10 to the power of negative sign 11 end exponent end strike close parentheses squared end cell row s equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell end table end style
 

undefined dari pH = 4 - log 2 sangat kecil, maka diabaikan. 

Jadi, kelarutan zat tersebut adalah begin mathsize 14px style 2 cross times 10 to the power of negative sign 5 end exponent end style mol/L.

1

Roboguru

Dalam suatu larutan terdapat ion Ag+, Ca2+, Ba2+, Sr2+, dan Pb2+ dengan konsentrasi yang sama. Jika larutan tersebut ditetesi dengan larutan K2​SO4​, zat yang mula-mula mengendap adalah ....

Pembahasan Soal:

Konstanta hasil kali kelarutan atau Ksp adalah hasil kali konsentrasi ion-ion tersebut dipangkatkan koefisien reaksi. Tidak semua garam bersifat larut dalam air, apabila garam yang sukar larut itu dikocok dalam air maka garam itu sebagian akan larut dalam air dan bagian yang larut ini akan terurai menjadi ion - ionnya. Suatu larutan dikatakan jenuh, tepat jenuh atau mengendap dapat dilakukan dengan membandingkan Qsp dan Ksp.

size 14px Qsp size 14px space size 14px less than size 14px space size 14px Ksp size 14px comma size 14px space size 14px tidak size 14px space size 14px terjadi size 14px space size 14px pengendapan size 14px Qsp size 14px space size 14px equals size 14px space size 14px Ksp size 14px comma size 14px space size 14px tepat size 14px space size 14px jenuh size 14px Qsp size 14px space size 14px greater than size 14px space size 14px Ksp size 14px comma size 14px space size 14px terjadi size 14px space size 14px endapan
 

begin mathsize 14px style italic a bold right parenthesis bold space bold Ksp bold space Ag subscript bold 2 S O subscript bold 4 bold space bold equals bold space bold 1 bold space italic x bold space bold 10 to the power of bold minus sign bold 5 end exponent Ag subscript 2 S O subscript 4 open parentheses italic s close parentheses equilibrium 2 Ag 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 space end exponent left parenthesis italic a italic q right parenthesis s space space space space space space space space space space space space space space space space space space space space space 2 s space space space space space space space space space space space space space space space space space space s K subscript italic s italic p end subscript italic space Ag subscript 2 S O subscript 4 equals space italic Q subscript italic s italic p end subscript Ag subscript italic 2 S O subscript italic 4 K subscript italic s italic p end subscript Ag subscript 2 S O subscript 4 equals space open square brackets Ag to the power of plus sign close square brackets squared left square bracket S O subscript 4 to the power of 2 minus sign space end exponent right square bracket K subscript italic s italic p end subscript space Ag subscript 2 S O subscript 4 equals left parenthesis 2 s right parenthesis squared open parentheses s close parentheses 1 space x space 10 to the power of negative sign 5 end exponent space space space space equals 4 s cubed space space space space space space space space space space space space space space space space space space s space equals 0 comma 0135 space mol forward slash L  italic b bold right parenthesis bold space bold Ksp bold space Sr S O subscript bold 4 bold space bold equals bold space bold 2 bold comma bold 5 bold space italic x bold space bold 10 to the power of bold minus sign bold 7 end exponent Sr S O subscript 4 bold space left parenthesis italic s right parenthesis equilibrium Sr 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 end exponent left parenthesis italic a italic q right parenthesis s space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space space s K subscript italic s italic p end subscript space Sr S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space Sr S O subscript 4 K subscript italic s italic p end subscript space Sr S O subscript 4 equals space open square brackets Sr to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space Sr S O subscript 4 equals s squared 2 comma 5 space x space 10 to the power of negative sign 7 end exponent space equals s squared space space space space space space space space space space space space space s space equals space 5 x 10 to the power of negative sign 4 end exponent space mol forward slash L  italic c bold right parenthesis bold space bold Ksp bold space Pb S O subscript bold 4 bold space bold equals bold space bold 1 bold comma bold 7 bold space italic x bold space bold 10 to the power of bold minus sign bold 8 end exponent bold space Pb S O subscript 4 open parentheses italic s close parentheses equilibrium 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 end exponent left parenthesis italic a italic q right parenthesis s space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space space s K subscript italic s italic p end subscript space space Pb S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space space Pb S O subscript 4 K subscript italic s italic p end subscript space Pb S O subscript 4 equals space open square brackets Pb to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space space Pb S O subscript 4 equals s squared 1 comma 7 space x space 10 to the power of negative sign 8 end exponent space equals s squared bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space italic s bold space bold equals bold space bold 1 bold comma bold 3 italic x bold 10 to the power of bold minus sign bold 4 end exponent bold space bold mol bold forward slash italic L  italic d bold right parenthesis bold space bold Ksp bold space Ba S O subscript bold 4 bold space bold equals bold space bold 1 bold comma bold 1 bold space italic x bold space bold 10 to the power of bold minus sign bold 10 end exponent space Ba S O subscript 4 open parentheses italic s close parentheses equilibrium Ba 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 end exponent left parenthesis italic a italic q right parenthesis s space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space space s K subscript italic s italic p end subscript space space Ba S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space space Ba S O subscript 4 K subscript italic s italic p end subscript space Ba S O subscript 4 equals space open square brackets Ba to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space space Ba S O subscript 4 equals s squared 1 comma 1 space x space 10 to the power of negative sign 10 end exponent space equals s squared space space space space space space space space space space space space space italic s bold space bold equals bold space bold 1 bold comma bold 0 italic x bold 10 to the power of bold minus sign bold 5 end exponent bold space bold mol bold forward slash italic L  italic e bold right parenthesis bold space bold Ksp bold space Ca S O subscript bold 4 bold space bold equals bold space bold 9 bold space italic x bold space bold 10 to the power of bold minus sign bold 6 end exponent space Ca S O subscript 4 open parentheses italic s close parentheses equilibrium Ca 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 end exponent left parenthesis italic a italic q right parenthesis s space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space space s K subscript italic s italic p end subscript space space Ca S O subscript 4 equals space italic Q subscript italic s italic p end subscript italic space space Ca S O subscript 4 K subscript italic s italic p end subscript space Ca S O subscript 4 equals space open square brackets Ca to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket K subscript italic s italic p end subscript space space Ca S O subscript 4 equals s squared 9 space x space 10 to the power of negative sign 6 end exponent space equals s squared bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space italic s bold space bold equals bold space bold 3 bold comma bold 0 italic x bold 10 to the power of bold minus sign bold 3 end exponent bold space bold mol bold forward slash italic L  bold Semakin bold space bold kecil bold comma bold space bold kelarutannya bold space bold maka bold space bold akan bold space bold semakin bold space bold mudah bold space bold pengendap    space end style

Jadi, jawaban yang benar adalah D.

0

Roboguru

Berapa mol NaF yang harus dilarutkan ke dalam 1,0 liter larutan jenuh PbF2​ pada 25∘C agar konsentrasi ion Pb2+ dalam larutan menjadi 1×10−6molar? (Ksp​PbF2​pada25∘C=4,0×10−8)

Pembahasan Soal:

Adanya ion senama dapat mempengaruhi kelarutan. Ion senama merupakan ion merupakan salah satu bahan terjadinya endapan. Kelarutan suatu endapan akan berkurang, Apabila salah satu ion senama terdapat dalam jumlah yang berlebihan.
 

Pb F subscript 2 yields Pb to the power of 2 plus sign and 2 F to the power of minus sign Ksp space Pb F subscript 2 double bond open square brackets Pb to the power of 2 plus sign close square brackets open square brackets F to the power of minus sign close square brackets squared space 4 cross times 10 to the power of negative sign 8 end exponent equals left parenthesis 10 to the power of negative sign 6 end exponent right parenthesis open square brackets F to the power of minus sign close square brackets squared space 4 cross times 10 to the power of negative sign 2 end exponent equals open square brackets F to the power of minus sign close square brackets squared space 2 cross times 10 to the power of negative sign 1 end exponent equals open square brackets F to the power of minus sign close square brackets  Na F yields Na to the power of plus sign and F to the power of minus sign 2 cross times 10 to the power of negative sign 1 end exponent space space space space 2 cross times 10 to the power of negative sign 1 end exponent   
 

Sehingga mol NaF,


mol space Na F double bond M cross times V mol space Na F equals 2 cross times 10 to the power of negative sign 1 end exponent cross times 1 mol space Na F equals 2 cross times 10 to the power of negative sign 1 end exponent equals 0 comma 2 space mol   


Jadi, jawaban yang paling tepat adalah D.undefined 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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