Iklan

Iklan

Pertanyaan

Pada suhu dan tekanan tertentu kelarutan Ba ( OH ) 2 ​ ( Ar Ba=137, O=16, H=1), dalam tiap liter larutan jenuhnya sebanyak 171 mg, maka kelarutan dalam BaCl 2 ​ 0,001 M adalah….

Pada suhu dan tekanan tertentu kelarutan  ( Ar Ba=137, O=16, H=1), dalam tiap liter larutan jenuhnya sebanyak 171 mg, maka kelarutan begin mathsize 14px style Ba open parentheses O H close parentheses subscript 2 end styledalam  0,001 M    adalah….

  1. begin mathsize 14px style 4 cross times 10 to the power of negative sign 4 end exponent space mol forward slash L end style 

  2. begin mathsize 14px style 2 cross times 10 to the power of negative sign 4 end exponent space mol forward slash L end style 

  3. begin mathsize 14px style 1 cross times 10 to the power of negative sign 4 end exponent space mol forward slash L end style 

  4. begin mathsize 14px style 1 cross times 10 to the power of negative sign 3 end exponent space mol forward slash L end style 

  5. begin mathsize 14px style 4 cross times 10 to the power of negative sign 3 end exponent space mol forward slash L end style 

Iklan

Q. 'Ainillana

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

jawaban yang benar adalah D.

Iklan

Pembahasan

Penambahan ion senam akan mengakibatkan kelarutannya menurun. Sebelumnya kita cari terlebih dahulu konsentrasi larutan jenuh . Selanjutnya menentukan . Berikutnya menentukan kelarutan dalam 0,001 M. Maka, Jadi, jawaban yang benar adalah D.

Penambahan ion senam akan mengakibatkan kelarutannya menurun. Sebelumnya kita cari terlebih dahulu konsentrasi larutan jenuh begin mathsize 14px style Ba open parentheses O H close parentheses subscript 2 end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row M equals cell massa over M subscript r cross times 1000 over V end cell row blank equals cell fraction numerator 171 cross times 10 to the power of negative sign 3 end exponent over denominator 171 end fraction cross times 1000 over 1000 end cell row blank equals cell 10 to the power of negative sign 3 end exponent space M end cell end table end style 

begin mathsize 14px style Ba open parentheses O H close parentheses subscript 2 equilibrium Ba to the power of 2 plus sign and 2 O H to the power of minus sign space space 10 to the power of negative sign 3 end exponent space space space space space space space space 10 to the power of negative sign 3 end exponent space space space 2 cross times 10 to the power of negative sign 3 end exponent end style 

Selanjutnya menentukan begin mathsize 14px style K subscript sp end style.

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 Ba to the power of 2 plus sign close square brackets left square bracket 2 O H to the power of minus sign right square bracket squared space end cell row blank equals cell left parenthesis 10 to the power of negative sign 3 end exponent right parenthesis left parenthesis 2 cross times 10 to the power of negative sign 3 end exponent right parenthesis squared space end cell row blank equals cell 4 cross times 10 to the power of negative sign 9 end exponent end cell end table end style 

Berikutnya menentukan kelarutan undefined dalam begin mathsize 14px style Ba Cl subscript 2 end style 0,001 M.

begin mathsize 14px style space space Ba Cl subscript 2 equilibrium Ba to the power of 2 plus sign and 2 Cl to the power of minus sign 0 comma 001 M space space 0 comma 001 M space space 0 comma 002 M end style 

Maka,

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 Ba to the power of 2 plus sign close square brackets left square bracket 2 O H to the power of minus sign right square bracket squared space end cell row cell 4 cross times 10 blank to the power of negative sign 9 end exponent end cell equals cell left parenthesis 0 comma 001 right parenthesis left parenthesis 2 s right parenthesis squared space end cell row cell 2 s end cell equals cell square root of fraction numerator 4 cross times 10 blank to the power of negative sign 9 end exponent over denominator 10 to the power of negative sign 3 end exponent end fraction end root end cell row s equals cell fraction numerator 2 cross times 10 to the power of negative sign 3 end exponent over denominator 2 end fraction end cell row s equals cell 10 to the power of negative sign 3 end exponent end cell end table end style 

Jadi, jawaban yang benar adalah D.

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

31

Iklan

Iklan

Pertanyaan serupa

Bila diketahui K sp ​ AgI = 1 , 8 × 1 0 − 14 . Hitunglah kelarutan AgI dalam larutan HI 0,1 M.

12

0.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia