Roboguru

Pertanyaan

Larutan N H subscript 4 Cl yang mempunyai pH = 5. Jika K subscript b space N H subscript 4 O H equals 10 to the power of negative sign 5 end exponent, hitunglah kemolaran N H subscript 4 Cl.

D. Santika

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Pembahasan

Kemolaran N H subscript 4 Cl adalah 0,1 M.

Larutan N H subscript 4 Cl merupakan larutan garam yang bersifat asam dan dapat terhidrolisis sebagian. Garam tersebut terionisasi sebagai berikut:

N H subscript 4 Cl yields N H subscript 4 to the power of plus plus Cl to the power of minus sign 

Untuk menghitung kemolaran N H subscript 4 Cl melalui tahapan berikut:

menghitung nilai Kh

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript h end cell equals cell K subscript w over K subscript b end cell row blank equals cell 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent end cell row blank equals cell 10 to the power of negative sign 9 end exponent end cell end table 

mengubah pH ke dalam open square brackets H to the power of plus sign close square brackets

table attributes columnalign right center left columnspacing 0px end attributes row pH equals 5 row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign 5 end exponent end cell end table 

menghitung kemolaran N H subscript 4 Cl

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 square root of K subscript h cross times open square brackets N H subscript 4 to the power of plus close square brackets end root end cell row cell 10 to the power of negative sign 5 end exponent end cell equals cell square root of 10 to the power of negative sign 9 end exponent cross times open square brackets N H subscript 4 to the power of plus close square brackets end root space space space left parenthesis kuadratkan space kedua space ruas right parenthesis end cell row cell open parentheses 10 to the power of negative sign 5 end exponent close parentheses squared end cell equals cell open parentheses square root of 10 to the power of negative sign 9 end exponent cross times open square brackets N H subscript 4 to the power of plus close square brackets end root close parentheses squared end cell row cell 10 to the power of negative sign 10 end exponent end cell equals cell 10 to the power of negative sign 9 end exponent cross times open square brackets N H subscript 4 to the power of plus close square brackets end cell row cell open square brackets N H subscript 4 to the power of plus close square brackets end cell equals cell 10 to the power of negative sign 10 end exponent over 10 to the power of negative sign 9 end exponent end cell row blank equals cell 0 comma 1 space M end cell end table 

berdasarkan reaksi ionisasi, open square brackets N H subscript 4 Cl close square brackets equals open square brackets N H subscript 4 to the power of plus close square brackets equals 0 comma 1 space M.

4

0.0 (0 rating)

Pertanyaan serupa

Di dalam 200 mL larutan terlarut 5,35 gram NH4​Cl. Jika Kb​=10−5 (ArN=14;H=1;Cl=35,5), hitunglah pH larutan tersebut.

3

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