Roboguru

Menurut reaksi: harga tetapan kesetimbangan ionisasi air adalah....

Menurut reaksi: begin mathsize 14px style H subscript 2 O open parentheses italic l close parentheses equilibrium H to the power of plus sign open parentheses italic a italic q close parentheses plus O H to the power of minus sign open parentheses italic a italic q close parentheses end style harga tetapan kesetimbangan ionisasi air adalah....

  1. ...

  2. ...undefined

Jawaban:

Pada suhu begin mathsize 14px style 25 to the power of o C end style harga derajat ionisasi begin mathsize 14px style open parentheses alpha close parentheses end style air adalah begin mathsize 14px style 1 comma 8 cross times 10 to the power of negative sign 9 end exponent end style sehingga :

Persamaan stoikiometri dari begin mathsize 14px style H subscript 2 O end style :
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space terurai end cell equals cell alpha cross times mol space mula bond mula end cell row blank equals cell alpha cross times y space mol end cell row blank equals cell αy space mol end cell end table end style 


Misalkan untuk 1 L (1000 mL) air, berarti :
massa jenis air begin mathsize 14px style open parentheses rho close parentheses end style = 1 g/mL

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell massa space air end cell equals cell rho cross times V end cell row blank equals cell 1 g forward slash mL cross times 1000 space mL end cell row blank equals cell 1000 space g end cell end table end style  

Maka untuk mol air :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space air end cell equals cell massa over M subscript r end cell row blank equals cell 1000 over 18 space mol end cell end table end style 

Volume air = 1 L, maka M = mol (n)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell αy space M space end cell row blank equals cell 1 comma 8 cross times 10 to the power of negative sign 9 end exponent cross times 1000 over 18 end cell row blank equals cell 10 to the power of negative sign 7 space end exponent M end cell end table end style   

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets double bond open square brackets O H to the power of minus sign close square brackets equals 10 to the power of negative sign 7 end exponent space M end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript W end cell equals 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 row blank equals cell left parenthesis 10 to the power of negative sign 7 end exponent right parenthesis left parenthesis 10 to the power of negative sign 7 end exponent right parenthesis end cell row blank equals cell 10 to the power of negative sign 14 end exponent end cell end table end style  

Dengan demikian, maka jawaban yang tepat adalah  begin mathsize 14px style K subscript w space end style=begin mathsize 14px style 10 to the power of negative sign 14 end exponent end style.

0

Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved