Roboguru

Dalam 200 mL air dapat larut 0,233 mg BaSO4​. Hasil kali kelarutan  adalah.... (Mr = 233)

Pertanyaan

Dalam 200 mL air dapat larut 0,233 mg begin mathsize 14px style Ba S O subscript 4 end style. Hasil kali kelarutan begin mathsize 14px style Ba S O subscript 4 end style adalah.... (Mr begin mathsize 14px style Ba S O subscript 4 end style= 233)

  1. begin mathsize 14px style 1 cross times 10 to the power of negative sign 10 end exponent end style

  2. begin mathsize 14px style 3 comma 6 cross times 10 to the power of negative sign 9 end exponent end style

  3. begin mathsize 14px style 3 cross times 10 to the power of negative sign 7 end exponent end style 

  4. begin mathsize 14px style 2 comma 5 cross times 10 to the power of negative sign 11 end exponent end style

  5. begin mathsize 14px style 1 cross times 10 to the power of negative sign 3 end exponent end style

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space Ba S O subscript 4 end cell equals cell space m over Mr cross times 1000 over V end cell row blank equals cell fraction numerator 0 comma 233 space mg over denominator 233 end fraction cross times fraction numerator 1 over denominator 200 mL end fraction end cell row blank equals cell left parenthesis 1 cross times 10 to the power of negative sign 3 end exponent right parenthesis cross times left parenthesis 5 cross times 10 to the power of negative sign 3 end exponent right parenthesis end cell row blank equals cell 5 cross times 10 to the power of negative sign 6 end exponent end cell end table end style


Persamaan reaksi kesetimbangan kelarutan undefined adalah:

undefined    

 

begin mathsize 14px style 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 left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket end cell row blank equals cell left parenthesis 5 cross times 10 to the power of negative sign 6 end exponent right parenthesis left parenthesis 5 cross times 10 to the power of negative sign 6 end exponent right parenthesis end cell row blank equals cell 25 cross times 10 to the power of negative sign 12 end exponent end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 11 end exponent end cell row blank blank blank end table end style 

Dengan demikian, maka jawaban yang tepat adalah D

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

V. Seftari

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Penulisan persamaan tetapan hasil kali kelarutan dari larutan jenuh BaCl2​ adalah ....

Pembahasan Soal:

Tetapan hasil kali kelarutan (Ksp) merupakan perkalian konsentrasi ion-ion elektrolit yang sukar larut dalam larutan jenuhnya dipangkatkan koefisien masing-masing. Satuan tetapan hasil kali kelarutan bergantung pada nilai kelarutan (s) nya, dimana satuan kelarutan = mol/L.

Reaksi ionisasi larutan jenuh begin mathsize 14px style Ba Cl subscript 2 end style yaitu :


begin mathsize 14px style Ba Cl subscript 2 space rightwards harpoon over leftwards harpoon space Ba to the power of 2 plus sign space plus space 2 Cl to the power of minus sign space space space space s space space space space space space space space space space space s space space space space space space space space space space space 2 s end style


sehingga persamaan tetapan hasil kali kelarutan (Ksp) begin mathsize 14px style Ba Cl subscript 2 end style dapat ditentukan dengan rumus berikut.


table attributes columnalign right center left columnspacing 0px end attributes row cell Ksp space Ba Cl subscript 2 end cell equals cell open square brackets Ba to the power of 2 plus sign close square brackets. open square brackets Cl to the power of minus sign close square brackets squared end cell row blank equals cell open parentheses s close parentheses begin inline style bevelled mol over L end style space. space left parenthesis 2 s space begin inline style bevelled mol over L end style right parenthesis squared end cell row blank equals cell space 4 s cubed space begin inline style bevelled mol cubed over L cubed end style end cell end table


Dengan demikian, tetapan hasil kali kelarutan begin mathsize 14px style Ba Cl subscript 2 end style yaitu begin mathsize 14px style 4 s cubed space begin inline style bevelled mol cubed over L cubed end style end style.

Jadi, jawaban yang benar adalah C.undefined

0

Roboguru

Berapa pH maksimum penambahan larutan NaOH agar kadar Pb maksimum 3 ppm, jika Ksp Pb(OH)2=1×10−12 (Ar Pb=120)?

Pembahasan Soal:

Soal ini menggabungkan ion senama dengan pH dan Ksp. Penggabungan ini dapat dilihat dari larutan NaOH dengan data Ksp yang diketahui, yaitu Pb(OH)2. Pada kedua larutan memiliki ion senama yaitu OH-.

Untuk menentukan pH maksimum pada larutan NaOH, dengan menentukan [Pb2+] dari Kadar Pb yang diketahui 3 ppm, artinya terdapat 3 miligram Pb dalam satu liter larutan.

open square brackets Pb to the power of 2 plus sign close square brackets equals fraction numerator 3 cross times 10 to the power of negative sign 3 end exponent over denominator 120 end fraction cross times 1000 over 1000 open square brackets Pb to the power of 2 plus sign close square brackets equals 2 comma 5 cross times 10 to the power of negative sign 5 end exponent space M

Pb open parentheses O H close parentheses subscript 2 subscript open parentheses italic s close parentheses end subscript rightwards harpoon over leftwards harpoon Pb to the power of 2 plus sign subscript open parentheses italic a italic q close parentheses end subscript plus 2 O H to the power of minus sign subscript open parentheses italic a italic q close parentheses end subscript space space space space s space space space space space space space space space space space 2 comma 5 cross times 10 to the power of negative sign 5 end exponent M 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 space open square brackets O H to the power of minus sign close square brackets squared 1 cross times 10 to the power of negative sign 12 end exponent equals open parentheses 25 cross times 10 to the power of negative sign 6 end exponent close parentheses space open square brackets O H to the power of minus sign close square brackets squared open square brackets O H to the power of minus sign close square brackets equals 2 cross times 10 to the power of negative sign 4 end exponent space M

Na O H yields Na to the power of plus sign and O H to the power of minus sign space space space space space space space space space space space space space space space space space space space space space space 2 cross times 10 to the power of negative sign 4 end exponent space M pOH equals minus sign log space open square brackets O H to the power of minus sign close square brackets space space space space space space space equals minus sign log space open parentheses space 2 cross times 10 to the power of negative sign 4 end exponent close parentheses pOH equals 4 minus sign log space 2 pH and pOH equals 14 pH equals 14 minus sign open parentheses 4 minus sign log space 2 close parentheses pH equals 10 plus log space 2

Maka, pH maksimum penambahan larutan NaOH adalah 10+ log 2.space

0

Roboguru

Hitunglah KSp​ dari senyawa-senyawa berikut: b. Ce(OH)3​,s=1×10−6molL−1

Pembahasan Soal:

Untuk menentukan begin mathsize 14px style K subscript italic Sp end style maka diperlukan ditentukan rumus begin mathsize 14px style K subscript italic Sp end style nya terlebih dahulu


begin mathsize 14px style Ce open parentheses O H close parentheses subscript 3 yields Ce to the power of 3 plus sign and 3 O H to the power of minus sign 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 3 italic s end style

begin mathsize 14px style 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 Ce to the power of 3 plus sign close square brackets space open square brackets O H to the power of minus sign close square brackets cubed space equals left parenthesis italic s right parenthesis left parenthesis 3 italic s right parenthesis cubed space equals 27 italic s to the power of italic 4 end cell row blank equals cell 27 cross times left parenthesis 1 cross times 10 to the power of negative sign 6 end exponent right parenthesis to the power of 4 end cell row blank equals cell 27 cross times 10 to the power of negative sign 24 end exponent end cell end table end style 


Jadi, Ksp senyawa tersebut adalah begin mathsize 14px style 27 cross times 10 to the power of negative sign 24 end exponent end style.space 

0

Roboguru

Hasil kali kelarutan Ag2​SO4​  yang memiliki kelarutan sebesar 5,4gL−1  adalah ... (Ar​Ag=108gmol−1,S=32gmol−1,danO=16gmol−1)

Pembahasan Soal:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell kelarutan space Ag subscript 2 S O subscript 4 space end cell equals cell space 5 comma 4 space g forward slash L end cell row cell italic M subscript italic r italic space Ag subscript 2 S O subscript 4 space end cell equals cell space 312 space g forward slash mol end cell end table end style 

Error converting from MathML to accessible text.

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold Persamaan bold space bold reaksinya end cell row cell Ag subscript bold 2 S O subscript bold 4 bold italic open parentheses italic s bold italic close parentheses end cell bold rightwards harpoon over leftwards harpoon cell bold 2 Ag to the power of bold plus sign bold italic left parenthesis italic a italic q bold italic right parenthesis bold plus bold space S O subscript bold 4 to the power of bold 2 bold minus sign end exponent bold italic left parenthesis italic a italic q bold italic right parenthesis end cell row blank blank cell italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic 2 italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic s end cell row cell K subscript italic s italic p end subscript end cell equals cell left parenthesis 2 s right parenthesis squared open parentheses s close parentheses space end cell row cell K subscript italic s italic p end subscript end cell equals cell 4 s cubed end cell row cell K subscript italic s italic p end subscript end cell equals cell 4 left parenthesis 0 comma 0173 space mol forward slash L bold right parenthesis cubed space end cell row cell K subscript bold sp end cell bold equals cell bold 2 bold comma bold 07 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end cell end table end style

 

Jadi, begin mathsize 14px style K subscript bold sp Ag subscript bold 2 S O subscript bold 4 bold equals bold 2 bold comma bold 07 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end style dan karena pilihan jawaban tidak ada maka pilih yang paling mendekati yaitu B.

 

0

Roboguru

Jika konsentrasi Ca2+ dalam larutan jenuh CaF2​=2×10−4mol/L, maka hasil kali kelarutan CaF2 adalah ....

Pembahasan Soal:

begin mathsize 14px style Ca F subscript 2 yields Ca to the power of 2 plus sign and 2 F to the power of minus sign space space s space space space space space space space space space space s space space space space space space space space space 2 s end style 

s adalah konsentrasi jenuh maka nilai Ksp CaF2 yaitu:

begin mathsize 14px style K subscript sp space equals space open square brackets Ca to the power of 2 plus sign close square brackets left square bracket 2 F to the power of minus sign right square bracket squared space K subscript sp space equals space open square brackets s close square brackets left square bracket 2 s right square bracket squared space K subscript sp space equals space 4 s cubed Ksp space equals space 4 cross times left parenthesis 2 cross times 10 to the power of negative sign 4 end exponent right parenthesis cubed space Ksp space equals space 4 cross times 8 cross times 10 to the power of negative sign 12 end exponent Ksp space equals space 32 cross times 10 to the power of negative sign 12 end exponent Ksp space equals space 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end style   


Jadi, nilai Ksp dari CaF2 yaitu begin mathsize 14px style 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end style

1

Roboguru

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