Roboguru

Pertanyaan

Bila terdapat 125 mL larutan begin mathsize 14px style N H subscript 4 Cl end style 0,1 M dah harga begin mathsize 14px style K subscript b equals 1 middle dot 10 to the power of negative sign 5 end exponent end style. Tentukan pH !

S. Lubis

Master Teacher

Jawaban terverifikasi

Jawaban

pH larutan adalah 5.

Pembahasan

Diketahui:
Larutan begin mathsize 14px style N H subscript 4 Cl end style 0,1 M, 125 mL 
begin mathsize 14px style K subscript b equals 1 middle dot 10 to the power of negative sign 5 end exponent end style.

Ditanya:
pH larutan begin mathsize 14px style N H subscript 4 Cl end style

Jawab:

1. Menentukan sifat larutan

begin mathsize 14px style N H subscript 4 Cl yields N H subscript 4 to the power of plus sign and Cl to the power of minus sign end style

begin mathsize 14px style N H subscript 4 Cl end style merupakan senyawa garam, yaitu garam yang bersifat asam karena begin mathsize 14px style N H subscript 4 Cl end style merupakan garam yang terdiri dari kation dari basa lemah dan anion dari asam kuat. 

2. Menghitung konsentrasi H+

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of fraction numerator K italic w over denominator K italic b end fraction cross times open square brackets G subscript open parentheses kation close parentheses end subscript close square brackets end root end style  

karena koefisien garam dan kation itu sama maka nilai konsentrasinya juga sama,

begin mathsize 14px style open square brackets N H subscript 4 to the power of plus sign close square brackets double bond open square brackets N H subscript 4 Cl close square brackets end style 

sehingga :

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of fraction numerator K italic w over denominator K italic b end fraction cross times open square brackets N H subscript 4 Cl close square brackets end root end style  

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of fraction numerator 1 middle dot 10 to the power of negative sign 14 end exponent over denominator 1 middle dot 10 to the power of negative sign 5 end exponent end fraction cross times 0 comma 1 end root open square brackets H to the power of plus sign close square brackets equals square root of fraction numerator 1 middle dot 10 to the power of negative sign 14 end exponent over denominator 1 middle dot 10 to the power of negative sign 5 end exponent end fraction cross times 1 middle dot 10 to the power of negative sign 1 end exponent end root open square brackets H to the power of plus sign close square brackets equals square root of fraction numerator 1 middle dot 10 to the power of negative sign 15 end exponent over denominator 1 middle dot 10 to the power of negative sign 5 end exponent end fraction end root open square brackets H to the power of plus sign close square brackets equals square root of 1 middle dot 10 to the power of negative sign 10 end exponent end root open square brackets H to the power of plus sign close square brackets equals 1 middle dot 10 to the power of negative sign 5 end exponent end style 

3. Menentukan pH larutan:

 begin mathsize 14px style pH equals minus sign log space open square brackets H to the power of plus sign close square brackets pH equals minus sign log space left square bracket 1 middle dot 10 to the power of negative sign 5 end exponent right square bracket pH equals 5 end style  


Jadi, pH larutan adalah 5.

7

0.0 (0 rating)

Pertanyaan serupa

Sebanyak 50 mL larutan NH3​ 0,1 M dicampur dengan 50 mL larutan HCI 0,1 M. tentukan pH larutan setelah dicampur!

16

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 Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia