Roboguru

Sebanyak 690 mg HCOOH dan 566 mg HCOONa dilarutkan dalam air hingga 100 ml larutan.   b. Jika ke dalam larutan ditambahkan 1 mL larutan  0,25 M; berapakah pH larutan sekarang?

Pertanyaan

Sebanyak 690 mg HCOOH dan 566 mg HCOONa dilarutkan dalam air hingga 100 ml larutan. left parenthesis K subscript a space H C O O H equals 1 comma 8 cross times 10 to the power of negative sign 4 end exponent right parenthesis 

b. Jika ke dalam larutan ditambahkan 1 mL larutan Ba open parentheses O H close parentheses subscript 2 0,25 M; berapakah pH larutan sekarang?space 

Pembahasan Soal:

Campuran tersebut adalah campuran yang membentuk penyangga maka pH campuran setelah ditambahkan asam menjadi


table attributes columnalign right center left columnspacing 0px end attributes row cell mol subscript HCCOH end cell equals cell fraction numerator 0 comma 69 g over denominator 47 space g forward slash mol end fraction equals 0 comma 014 space mol end cell row cell mol subscript HCOONa end cell equals cell fraction numerator 0 comma 566 g over denominator 68 end fraction equals 0 comma 008 space mol end cell row cell mol subscript open parentheses Ba O H close parentheses subscript 2 end subscript end cell equals cell 0 comma 25 M cross times 0 comma 001 L equals 0 comma 00025 end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell K subscript a cross times fraction numerator mol space asam bond mol space basa over denominator mol space garam and mol space basa end fraction end cell row blank equals cell 1 comma 8 cross times 10 to the power of negative sign 4 end exponent cross times fraction numerator 0 comma 014 minus sign 0 comma 00025 over denominator 0 comma 008 plus 0 comma 00025 end fraction end cell row blank equals cell 1 comma 8 cross times 10 to the power of negative sign 4 end exponent cross times fraction numerator 0 comma 013 over denominator 0 comma 0083 end fraction end cell row blank equals cell 2 comma 8 cross times 10 to the power of negative sign 4 end exponent end cell row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell 4 minus sign log space 2 comma 8 end cell end table  


Jadi, pH larutan tersebut menjadi 4-log 2,8.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Muhammad

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

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

0

Roboguru

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 

0

Roboguru

Tentukan pH dari larutan penyangga yg dibuat dengan mencampurkan 50 mL larutan  0,3 M dengan 50 mL larutan  0,1M! ()

Pembahasan Soal:


begin mathsize 14px style open square brackets H to the power of plus sign close square brackets double bond K subscript a point space fraction numerator n subscript asam over denominator n subscript garam point space valensi end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 5 end exponent. space fraction numerator 10 over denominator 5. space 1 end fraction open square brackets H to the power of plus sign close square brackets equals 2 cross times 10 to the power of negative sign 5 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 2 cross times 10 to the power of negative sign 5 end exponent pH equals 5 minus sign log space 2 end style 


Jadi, pHnya adalah 5-log 2.space  

0

Roboguru

Tentukan berapa gram  yang harus ditambahkan kedalam 100 mL  0,1 M () agar dihasilkan larutan penyangga dengan pH 5 - log 2 ... ()

Pembahasan Soal:

Untuk mengetahui massa garam open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca, kita cari terlebih dahulu konsentrasi ion H to the power of plus sign:

table attributes columnalign right center left columnspacing 0px end attributes row cell italic p H 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 5 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 negative sign log space 2 cross times 10 to the power of negative sign 5 end exponent 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 5 end exponent space M end cell end table 

Nilai open square brackets H to the power of plus sign close square brackets larutan penyangga asam diperoleh dari persamaan berikut:

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 Ka cross times n subscript asam space lemah end subscript over n subscript basa space konjugasi end subscript end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell Ka cross times n subscript C H subscript 3 C O O H end subscript over n subscript C H subscript 3 C O O to the power of minus sign end subscript end cell end table  

maka dapat diperoleh mol garam open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca:

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 Ka cross times n subscript C H subscript 3 C O O H end subscript over n subscript C H subscript 3 C O O to the power of minus sign end subscript end cell row cell 2 cross times 10 to the power of negative sign 5 end exponent end cell equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator left parenthesis 0 comma 1 M cross times 0 comma 1 L right parenthesis over denominator n subscript C H subscript 3 C O O to the power of minus sign end subscript end fraction end cell row cell n subscript C H subscript 3 C O O to the power of minus sign end subscript end cell equals cell fraction numerator 0 comma 01 over denominator 2 end fraction end cell row blank equals cell 5 cross times 10 to the power of negative sign 3 end exponent space mol end cell end table  

dengan demikian, mol garam open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca:

open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca space space rightwards arrow space space 2 C H subscript 3 C O O to the power of minus sign space space plus space space Ca to the power of 2 plus sign bold 2 bold comma bold 5 bold cross times bold 10 to the power of bold minus sign bold 3 end exponent bold space bold mol bold space space space space space space space space 5 cross times 10 to the power of negative sign 3 end exponent space mol space space space space space 2 comma 5 cross times 10 to the power of negative sign 3 end exponent space mol 

dengan demikian, massa garam open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca:

table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca end subscript end cell equals cell n cross times italic M subscript r end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 3 end exponent cross times 158 end cell row blank equals cell 0 comma 395 space gram end cell end table  

Oleh karena itu, massa open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca adalah 0,395 gram.

Jadi, tidak ada jawaban yang benar.

0

Roboguru

400 mL larutan  0,1 M direaksikan dengan 100 mL larutan  0,1 M. Tentukan: () a. pH larutan b. pH larutan jika ditambahkan 0,001 mol HCl c. pH larutan jika ditambahkan 0,001 mol NaOH d. pH larutan ...

Pembahasan Soal:

a. pH larutan

Menentukan mol larutan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 3 C O O H end subscript end cell equals cell M cross times V end cell row blank equals cell 0 comma 1 cross times 400 end cell row blank equals cell 40 space mmol end cell row blank equals cell 0 comma 04 space mol end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript Ca open parentheses O H close parentheses subscript 2 end subscript end cell equals cell M cross times V end cell row blank equals cell 0 comma 1 cross times 100 end cell row blank equals cell 10 space mmol end cell row blank equals cell 0 comma 01 space mol end cell end table end style 

Reaksi yang terjadi:

 

Menentukan pH larutan:

begin mathsize 14px style 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 K subscript a cross times n subscript asam over n subscript garam end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator 0 comma 02 over denominator 0 comma 01 end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic p H end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 2 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 2 end cell row blank equals cell 4 comma 7 end cell end table end style 

Maka, pH larutan adalah 4,7.

b. pH larutan jika ditambahkan 0,001 mol HCl

Reaksi yang terjadi:

 

Menentukan pH larutan:

begin mathsize 14px style 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 K subscript a cross times n subscript asam over n subscript garam end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator 0 comma 021 over denominator 0 comma 009 end fraction end cell row blank equals cell 2 comma 3 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic p H end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 2 comma 3 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 2 comma 3 end cell row blank equals cell 4 comma 6 end cell end table end style 

Maka, pH larutan setelah ditambahkan HCl adalah 4,7.

c. pH larutan jika ditambahkan 0,001 mol NaOH

Reaksi yang terjadi:

 

Menentukan pH larutan:

begin mathsize 14px style 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 K subscript a cross times n subscript asam over n subscript garam end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator 0 comma 019 over denominator 0 comma 011 end fraction end cell row blank equals cell 1 comma 7 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic p H end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 1 comma 7 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 1 comma 7 end cell row blank equals cell 4 comma 8 end cell end table end style 

Maka, pH larutan setelah ditambahkan NaOH adalah 4,8.

d. pH larutan jika diencerkan dengan menambahkan 100 mL air

Menentukan konsentrasi larutan setelah penambahan 100 mL air:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets C H subscript 3 C O O H close square brackets end cell equals cell n over V subscript total end cell row blank equals cell fraction numerator 0 comma 02 over denominator 500 plus 100 end fraction end cell row blank equals cell fraction numerator 0 comma 02 over denominator 600 end fraction end cell row blank equals cell 3 comma 3 cross times 10 to the power of negative sign 5 end exponent space mmol end cell row blank equals cell 3 comma 3 cross times 10 to the power of negative sign 8 end exponent space mol end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left square bracket open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca right square bracket end cell equals cell n over V subscript total end cell row blank equals cell fraction numerator 0 comma 01 over denominator 500 plus 100 end fraction end cell row blank equals cell fraction numerator 0 comma 01 over denominator 600 end fraction end cell row blank equals cell 1 comma 7 cross times 10 to the power of negative sign 5 end exponent space mmol end cell row blank equals cell 1 comma 7 cross times 10 to the power of negative sign 8 end exponent space mol end cell end table end style 

Menentukan pH larutan:

begin mathsize 14px style 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 K subscript a cross times fraction numerator open square brackets C H subscript 3 C O O H close square brackets over denominator open square brackets open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca close square brackets end fraction end cell row blank equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator 3 comma 3 cross times 10 to the power of negative sign 8 end exponent over denominator 1 comma 7 cross times 10 to the power of negative sign 8 end exponent end fraction end cell row blank equals cell 1 comma 9 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic p H end cell equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign log space 1 comma 9 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 1 comma 9 end cell row blank equals cell 4 comma 7 end cell end table end style 

Maka, pH larutan setelah pengenceran adalah 4,7.

Jadi, jawaban yang benar sesuai penjelasan diatas.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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