Iklan

Iklan

Pertanyaan

Di dalam suatu larutan terdapat ion ion X 2 + ; Y 2 + ; dan Z 2 + dengan konsentrasi masing masing 0,1M. ke dalam larutan ini ditambahkan NaOH padat, sehingga pH larutan menjadi 8. Berdasarkan data berikut: Ksp X ( OH ) 2 ​ = 2 , 8 × 1 0 − 10 Ksp Y ( OH ) 2 ​ = 4 , 5 × 1 0 − 11 Ksp Z ( OH ) 2 ​ = 1 , 6 × 1 0 − 14 Maka hidroksida yang larut adalah…

Di dalam suatu larutan terdapat ion ion ; ; dan dengan konsentrasi masing masing 0,1M. ke dalam larutan ini ditambahkan  padat, sehingga pH larutan menjadi 8.

Berdasarkan data berikut:
 

Ksp  =  

Ksp  =   

Ksp  =   


Maka hidroksida yang larut adalah…
 

  1. undefined 

  2. undefined 

  3. undefined 

  4. undefined dan undefined 

  5. undefined dan undefined 

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

pH larutan = 8, maka pOH = 6. Jadi, jawaban yang benar adalah D.

pH larutan = 8, maka pOH = 6.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row 6 equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row cell 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 6 end exponent end cell end table end style 

begin mathsize 14px style X open parentheses O H close parentheses subscript 2 equilibrium X to the power of 2 plus sign and 2 O H to the power of minus sign space space space space space space space space space space space space space space space 10 to the power of minus sign space space space space 10 to the power of negative sign 6 end exponent end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript sp space X open parentheses O H close parentheses subscript 2 end cell equals cell open square brackets X 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 1 end exponent right parenthesis left parenthesis 10 to the power of negative sign 6 end exponent right parenthesis squared space end cell row blank equals cell 10 to the power of negative sign 13 end exponent end cell row cell K subscript sp space end subscript X open parentheses O H close parentheses subscript 2 end cell equals cell 2 comma 8 cross times 10 to the power of negative sign 10 end exponent end cell row cell Q subscript sp end cell less than cell K subscript sp space end subscript left parenthesis belum space mengendap forward slash larut right parenthesis end cell end table end style   

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript sp space Y open parentheses O H close parentheses subscript 2 end cell equals cell open square brackets Y 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 1 end exponent right parenthesis left parenthesis 10 to the power of negative sign 6 end exponent right parenthesis squared space end cell row blank equals cell 10 to the power of negative sign 13 end exponent end cell row cell K subscript sp space end subscript Y open parentheses O H close parentheses subscript 2 end cell equals cell 4 comma 5 cross times 10 to the power of negative sign 11 end exponent end cell row cell Q subscript sp end cell less than cell K subscript sp space end subscript left parenthesis belum space mengendap forward slash larut right parenthesis end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript sp space Z open parentheses O H close parentheses subscript 2 end cell equals cell open square brackets Z 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 1 end exponent right parenthesis left parenthesis 10 to the power of negative sign 6 end exponent right parenthesis squared space end cell row blank equals cell 10 to the power of negative sign 13 end exponent end cell row cell K subscript sp space end subscript Z open parentheses O H close parentheses subscript 2 end cell equals cell 1 comma 6 cross times 10 to the power of negative sign 14 end exponent end cell row cell Q subscript sp end cell greater than cell K subscript sp space end subscript open parentheses mengendap close parentheses end cell end table end style 

Jadi, jawaban yang benar adalah D.

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

3

Ica Tengku

Jawaban tidak sesuai

Iklan

Iklan

Pertanyaan serupa

Jika K sp ​ CaC 2 ​ O 4 ​ = 2 , 6 × 1 0 − 9 , konsentrasi ion oksalat yang diperlukan untuk membentuk endapan dalam larutan yang mengandung 0,02 M ion kalsium adalah …

35

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