Roboguru

250 mL larutan HCI 0,02 M direaksikan dengan 250 m...

250 mL larutan HCI 0,02 M direaksikan dengan 250 mL larutan NaOH 0,03 M akan dihasilkan larutan dengan pH sebesar ... space

Jawaban:

menentukan mol masing-masing :

mol space H Cl space equals space V space cross times space M space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 250 space mL space cross times space 0 comma 02 space M space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 space mmol space space 

mol space Na O H space equals space V space cross times space M space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 250 space mL space cross times space 0 comma 03 space M space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 7 comma 5 space mmol space space 

menentukan mol yang tersisa :

 

Karena yang tersisa basa , maka menentukan nilai [OH-] :

V total = 500 ml

open square brackets O H to the power of minus sign close square brackets equals fraction numerator mol over denominator V space total end fraction open square brackets O H to the power of minus sign close square brackets equals fraction numerator 2 comma 5 space mmol over denominator 500 space ml end fraction open square brackets O H to the power of minus sign close square brackets equals 0 comma 005 M 

menentukan nilai pH yang terbentuk :

pOH space equals space minus sign space log space open square brackets O H to the power of minus sign close square brackets space space space space space space space space space space space space space space equals space minus sign space log space left square bracket 5 cross times 10 to the power of negative sign 3 end exponent right square bracket space space space space space space space equals space 3 minus sign Log space 5  pH space space space equals 14 minus sign pOH space space space space space space equals 14 minus sign left parenthesis 3 minus sign log space 5 right parenthesis space space space space space space equals 11 plus log space 5 space 

Jadi, pH yang larutan yang tersisa adalah 11+log 5.

 

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