Roboguru

Ke dalam 1 dm³ larutan basa lemah MOH 0,2 M ditambahkan 1 dm³ larutan HCI 0,2 M. Besarnya pH larutan adalah ....

Ke dalam 1 dm³ larutan basa lemah MOH 0,2 M begin mathsize 14px style open parentheses K subscript b equals 10 to the power of negative sign 7 end exponent close parentheses end style ditambahkan 1 dm³ larutan HCI 0,2 M. Besarnya pH larutan adalah ....

  1. 4,0space 

  2. 5,0space 

  3. 7,0space 

  4. 8,0space 

  5. 9,0space 

Jawaban:

begin mathsize 14px style n subscript MOH equals 0 comma 2 cross times 1 equals 0 comma 2 space mol n subscript HCl equals 0 comma 2 cross times 1 equals 0 comma 2 space mol  MOH open parentheses aq close parentheses and H Cl open parentheses aq close parentheses yields MCl open parentheses aq close parentheses and H subscript 2 O open parentheses l close parentheses 0 comma 2 space space space space space space space space space space space space space space 0 comma 2 bottom enclose negative sign 0 comma 2 space space space space space space space space space space minus sign 0 comma 2 space space space space space space space space space plus 0 comma 2 space space space space space space space space plus 0 comma 2 end enclose 0 space space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space 0 comma 2 space space space space space space space space space space space space space 0 comma 2  open square brackets MCl close square brackets equals bevelled fraction numerator 0 comma 2 over denominator 2 end fraction equals 0 comma 1 space M M to the power of plus sign open parentheses aq close parentheses and H subscript 2 O open parentheses l close parentheses equilibrium MOH open parentheses aq close parentheses and H to the power of plus sign open parentheses aq close parentheses open square brackets H to the power of plus sign close square brackets equals square root of K subscript w over K subscript b open square brackets M to the power of plus sign close square brackets end root open square brackets H to the power of plus sign close square brackets equals square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 7 end exponent open parentheses 0 comma 1 close parentheses end root open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 4 end exponent space M  pH equals minus sign log open square brackets H to the power of plus sign close square brackets pH equals minus sign log space open parentheses 10 to the power of negative sign 4 end exponent close parentheses pH equals 4 end style

Dengan demikian, maka jawaban yang tepat adalah A.

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