Roboguru

Suatu larutan penyangga terdiri atas asam lemah HA dan garam natriumnya (NaA). Jika konsentrasi asam HA dalam larutan tersebut adalah 0,2 M; konsentrasi larutan NaA sehingga pH larutan sama dengan pKa asam HA adalah...

Pertanyaan

Suatu larutan penyangga terdiri atas asam lemah HA dan garam natriumnya (NaA). Jika konsentrasi asam HA dalam larutan tersebut adalah 0,2 M; konsentrasi larutan NaA sehingga pH larutan sama dengan pKa asam HA adalah...space 

  1. 0,02 Mspace 

  2. 0,1 Mspace 

  3. 0,2 Mspace 

  4. 1,0 Mspace 

  5. 2,0 Mspace 

Pembahasan Soal:

Untuk menentukan konsentrasi H to the power of plus sign dalam larutan penyangga digunakan rumus berikut


open square brackets H to the power of plus sign close square brackets double bond K subscript a cross times fraction numerator mol space asam over denominator mol space garam end fraction


pH larutan penyangga sama seperti harga pKa maka jumlah mol asam dan mol garam sama sehingga menghasilkan perbandingan sehingga harga pH sama dengan harga pKa

Jika dianggap volume kedua larutan sama maka konsentrasi garam adalah 0,2M

Jadi, jawaban yang benar adalah C.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Muhammad

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 30 April 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Jika kedalam 50 mL larutan penyangga dengan pH=5 ditambahkan 50 mL akuades, maka...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank Volume end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank awal end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank V end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank pH end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 10 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank to the power of negative sign 5 end exponent end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank M end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank a end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator fraction numerator mol space asam space lemah over denominator V 1 end fraction over denominator fraction numerator mol space garam over denominator V 1 end fraction end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank a end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator fraction numerator mol space asam space lemah over denominator V 1 end fraction over denominator fraction numerator mol space garam over denominator V 1 end fraction end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank a end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator V 1 over denominator mol space garam end fraction end cell end table bold left square bracket H to the power of bold plus sign table attributes columnalign right center left columnspacing 0px end attributes row right square bracket equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript italic a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator bold mol bold space bold asam bold space bold lemah over denominator bold mol bold space bold garam end fraction end cell end table end style    

Jika ditambahkan 50 mL akuades maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank V end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank V end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 50 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank mL end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator fraction numerator mol space asam space lemah over denominator V 2 end fraction over denominator fraction numerator mol space garam over denominator V 2 end fraction end fraction end cell end table left square bracket H to the power of plus sign table attributes columnalign right center left columnspacing 0px end attributes row right square bracket equals K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator fraction numerator mol space asam space lemah over denominator V 2 end fraction over denominator fraction numerator mol space garam over denominator V 2 end fraction end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator mol space asam space lemah over denominator V 2 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator V 2 over denominator mol space garam end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold left square bracket end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank H end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank to the power of bold plus end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold right square bracket end table table attributes columnalign right center left columnspacing 0px end attributes row blank bold equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank K end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript italic a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator bold mol bold space bold asam bold space bold lemah over denominator bold mol bold space bold garam end fraction end cell end table end style  

Jadi dapat disimpulkan bahwa penambahan 50 mL akuades tidak akan mengubah pH.

Oleh karena itu, jawaban yang benar adalah C.undefined 

Roboguru

Tentukan keasaman larutan penyangga berikut ini ! Campuran antara campuran dari larutan  berlebih dengan

Pembahasan Soal:

Basa kuat yang berlebih dengan asam lemah tidak akan membentuk larutan penyangga, tetapi terbentuk larutan yang bersifat basa.

Jadi, jawaban yang benar adalah bersifat basa.undefined

Roboguru

Perbandingan volume larutan CH3COOH 0,1 M dan NaOH 0,1 M yang harus dicampurkan untuk membuat larutan penyangga dengan pH = 6 adalah ... (Ka = 10-5)

Pembahasan Soal:

menentukan mol masing - masing larutan :

n space C H subscript 3 C O O H double bond M cross times V subscript 1 space space space space space space space space space space space space space space space space equals 0 comma 1 cross times V subscript 1 space space space space space space space space space space space space space space space space equals 0 comma 1 space V subscript 1 n space Na O H equals space M cross times V subscript 2 space space space space space space space space space space space equals 0 comma 1 cross times V subscript 2 space space space space space space space space space space space equals 0 comma 1 V subscript 2 

menentukan reaksi MRS:

NaOH harus habis agar larutan bersifat penyangga asam karena memiliki pH =6

menentukan perbandingan mol asam dengan  mol garam :

pH space space space space equals 6 open square brackets H to the power of plus sign close square brackets space equals 10 to the power of negative sign 6 end exponent open square brackets H to the power of plus sign close square brackets double bond Ka fraction numerator mol space asam over denominator mol space garam end fraction left square bracket 10 to the power of negative sign 6 end exponent right square bracket equals 10 to the power of negative sign 5 end exponent fraction numerator mol space C H subscript 3 C O O H over denominator mol space C H subscript 3 C O O Na end fraction left square bracket 10 to the power of negative sign 6 end exponent right square bracket equals 10 to the power of negative sign 5 end exponent fraction numerator 0 comma 1 V subscript 1 minus sign 0 comma 1 V subscript 2 over denominator 0 comma 1 space V subscript 2 end fraction left square bracket up diagonal strike 10 to the power of negative sign 6 end exponent end strike right square bracket equals up diagonal strike 10 to the power of negative sign 5 end exponent end strike fraction numerator up diagonal strike 0 comma 1 end strike open parentheses V subscript 1 bond V subscript 2 close parentheses over denominator up diagonal strike 0 comma 1 end strike space V subscript 2 end fraction V subscript 2 space space space space space equals 10 space open parentheses V subscript 1 bond V subscript 2 close parentheses V subscript 2 space space space space space equals 10 space V subscript 1 minus sign 10 space V subscript 2 V subscript 2 plus 10 space V subscript 2 space equals 10 space V subscript 1 11 space V subscript 2 space equals 10 space V subscript 1 V subscript 2 over V subscript 1 space space space space equals 10 over 11 

Jadi, perbandingan volume NaOH dengan CH3COOH adalah 10:11.

Roboguru

Berapakah pH larutan, jika ke dalam 0,4 L larutan 0,1 M CH3COOH dilarutkan 0,04 mol CH3COONa? Harga .

Pembahasan Soal:

Larutan asam lemah C H subscript 3 C O O H dicampurkan dengan garam C H subscript 3 C O O Na akan membentuk larutan buffer asam. pH larutan buffer tersebut dapat dihitung dengan cara:

open square brackets H to the power of plus sign close square brackets double bond K subscript a cross times fraction numerator n space C H subscript 3 C O O H over denominator n space C H subscript 3 C O O Na end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 7 end exponent cross times fraction numerator 0 comma 4 space L cross times 0 comma 1 space M over denominator 0 comma 04 space mol end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 7 end exponent cross times fraction numerator 0 comma 04 space mol over denominator 0 comma 04 space mol end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 7 end exponent space M   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 parenthesis 10 to the power of negative sign 7 end exponent right parenthesis pH equals 7  


Jadi, pH larutan buffer asam tersebut adalah 7.space space space space

Roboguru

Suatu larutan penyangga terdiri dari 0,030 M  dan 0,020 M  . Jika ke dalam 100 mL larutan penyangga tersebut ditambahkan 5 mL  0,10 M, pH larutan yang terbentuk adalah ...

Pembahasan Soal:

Campuran antara larutan C H subscript 3 C O O Na dan larutan C H subscript 3 C O O H tersebut akan membentuk suatu larutan penyangga asam, dimana C H subscript 3 C O O H sebagai asam lemah dan C H subscript 3 C O O Na sebagai basa konjugasi berupa ion C H subscript 3 C O O to the power of minus sign. Ke dalam larutan tersebut kemudian ditambahkan sedikit asam yaitu H Cl, maka agar dapat mengetahui nilai pH setelah penambahan asam, perlu dihitung terlebih dahulu mol larutan C H subscript 3 C O O H dan larutan C H subscript 3 C O O Na sebagai berikut.space


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space C H subscript 3 C O O H end cell equals cell M cross times volume end cell row cell mol space C H subscript 3 C O O H end cell equals cell 0 comma 02 space M cross times 100 space mL end cell row cell mol space C H subscript 3 C O O H end cell equals cell 2 space mmol end cell row blank blank blank row cell mol space C H subscript 3 C O O to the power of minus sign end cell equals cell M cross times volume end cell row cell mol space C H subscript 3 C O O to the power of minus sign end cell equals cell 0 comma 03 space M cross times 100 space mL end cell row cell mol space C H subscript 3 C O O to the power of minus sign end cell equals cell 3 space mmol end cell end table 


Selanjutnya, larutan H Cl yang ditambahkan sebagai H to the power of plus sign akan bereaksi dengan basa konjugasi yaitu ion C H subscript 3 C O O to the power of minus sign dalam sistem penyangga dan kembali membentuk asam lemah C H subscript 3 C O O H seperti pada persamaan reaksi berikut:space


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H to the power of plus sign end cell equals cell M space H Cl cross times vol point space H Cl end cell row cell mol space H to the power of plus sign end cell equals cell 0 comma 1 space M cross times 5 space mL end cell row cell mol space H to the power of plus sign end cell equals cell 0 comma 5 space mmol end cell end table 


 


Berdasarkan persamaan reaksi di atas, dihasilkan 2,5 mmol asam lemah dan 2,5 mmol basa konjugasinya, maka nilai pH larutan dapat dihitung dengan cara berikut.space


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 fraction numerator mol space C H subscript 3 C O O H over denominator mol space C H subscript 3 C O O to the power of minus sign end fraction cross times K subscript a end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell fraction numerator 2 comma 5 space mmol over denominator 2 comma 5 space mmol end fraction cross times 2 middle dot 10 to the power of negative sign 5 end exponent 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 5 end exponent end cell row blank blank blank row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row pH equals cell negative sign log space 2 cross times 10 to the power of negative sign 5 end exponent end cell row pH equals cell 5 minus sign log space 2 end cell end table 


Jadi, jawaban yang tepat adalah B.space

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

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