RoboguruRoboguru
SD

Padatan Ca(OH)2​, (Mr=74 g mol−1) ditambahkan ke dalam 500 mL larutan HCl 0,8 M berlebih. Jika larutan yang terbentuk mempunyai pH= 1 - log 2, massa   yang ditambahkan sebanyak ....

Pertanyaan

Padatan Ca open parentheses O H close parentheses subscript 2, left parenthesis Mr equals 74 space g space mol to the power of negative sign 1 end exponent right parenthesis spaceditambahkan ke dalam 500 mL larutan H Cl 0,8 M berlebih. Jika larutan yang terbentuk mempunyai pH= 1 - log 2, massa Ca open parentheses O H close parentheses subscript 2  yang ditambahkan sebanyak ....

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

massa Ca open parentheses O H close parentheses subscript 2  yang ditambahkan sebanyak 11,1 gram.

Pembahasan

Massa padatan Ca open parentheses O H close parentheses subscript 2 dapat dihitung berdasarkan reaksinya dengan larutan H Cl. Adapun langkah pengerjaannya sebagai berikut.

1) Tentukan open square brackets H Cl close square brackets berdasarkan nilai pH larutan

pH larutan yang terbentuk adalah 1 - log 2 artinya mol H Cl masih tersisa sedangkan mol Ca open parentheses O H close parentheses subscript 2 habis bereaksi maka,

table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row cell 1 minus sign log space 2 end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 1 end exponent end cell row blank blank blank row cell open square brackets H to the power of plus sign close square brackets end cell equals cell open square brackets H Cl close square brackets space. space valensi space asam end cell row cell 2 cross times 10 to the power of negative sign 1 end exponent end cell equals cell open square brackets H Cl close square brackets space. space 1 end cell row cell open square brackets H Cl close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 1 end exponent space M end cell row cell open square brackets H Cl close square brackets end cell equals cell 0 comma 2 end cell end table

2) Hitung mol H Cl yang tersisa dan mula-mula.

mol H Cl sisa

mol space H Cl double bond open square brackets H Cl close square brackets space. space volume open parentheses L close parentheses mol space H Cl equals 0 comma 2 space M space. space 0 comma 5 space L mol space H Cl equals 0 comma 1 space mol

mol HCl mula-mula

mol space H Cl equals 0 comma 8 space M space. space 0 comma 5 space L mol space H Cl equals 0 comma 4 space mol

3) Tentukan mol Ca open parentheses O H close parentheses subscript 2 berdasarkan reaksi MRS.

reaksi colon space space 2 H Cl space plus space Ca open parentheses O H close parentheses subscript 2 space rightwards arrow space Ca Cl subscript 2 space plus space 2 H subscript 2 O M space space space space space space space space 0 comma 4 space mol space space space space space x space mol space space space space space space space space space space space space minus sign space space space space space space space space space space space minus sign R space space space space space space space space 0 comma 3 space mol space space space 0 comma 15 space mol space space space space space space 0 comma 15 space mol space space 0 comma 3 space mol S space space space space space space space space space 0 comma 1 space mol space space space space space space space space minus sign space space space space space space space space space space space space 0 comma 15 space mol space space 0 comma 3 space mol

karena Ca open parentheses O H close parentheses subscript 2 habis bereaksi maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Ca open parentheses O H close parentheses subscript 2 space mula bond mula end cell equals cell mol space Ca open parentheses O H close parentheses subscript 2 space yang space bereaksi end cell row cell mol space Ca open parentheses O H close parentheses subscript 2 space mula bond mula end cell equals cell 0 comma 15 space mol end cell end table

4) Hitung massa Ca open parentheses O H close parentheses subscript 2 yang ditambahkan.

table attributes columnalign right center left columnspacing 0px end attributes row cell massa space Ca open parentheses O H close parentheses subscript 2 end cell equals cell mol space Ca open parentheses O H close parentheses subscript 2 space. space Mr space Ca open parentheses O H close parentheses subscript 2 space end cell row cell massa space Ca open parentheses O H close parentheses subscript 2 end cell equals cell 0 comma 15 space mol space. space 74 space g space mol to the power of negative sign 1 end exponent end cell row cell massa space Ca open parentheses O H close parentheses subscript 2 end cell equals cell 11 comma 1 space gram end cell end table


Jadi, massa Ca open parentheses O H close parentheses subscript 2  yang ditambahkan sebanyak 11,1 gram.

55

0.0 (0 rating)

Pertanyaan serupa

Tentukan pH sebelum dan sesudah dicampurkan dari campuran berikut 25 mL Ca(OH)2​ 0,1 M dan 50 mL HNO3​ 0,1 M

208

1.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

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