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Larutan natrium asetat 0,1 M dengan tetapan hidrolisis  mempunyai pH sebesar ...

Pertanyaan

Larutan natrium asetat 0,1 M dengan tetapan hidrolisis 1 cross times 10 to the power of negative sign 9 end exponent mempunyai pH sebesar ...space 

  1. 5  

  2. 7  

  3. 9  

  4. 11 

Pembahasan Soal:

Larutan Na C H subscript 3 C O O adalah garam yang bersifat basa karena berasal dari asam lemah dan basa kuat yang habis bereaksi menghasilkan garam basa yang terhidrolisis dalam air menghasilkan ion O H to the power of minus sign.

C H subscript 3 C O O Na yields with plus air on top C H subscript 3 C O O to the power of minus sign and Na to the power of plus sign C H subscript 3 C O O to the power of minus sign and H subscript 2 O equilibrium C H subscript 3 C O O H and O H to the power of minus sign 

Menentukan konsentrasi O H to the power of bold minus sign 

open square brackets O H to the power of minus sign close square brackets equals square root of K subscript h cross times open square brackets C H subscript 3 C O O Na close square brackets end root open square brackets O H to the power of minus sign close square brackets equals square root of 10 to the power of negative sign 9 end exponent cross times left square bracket 10 to the power of negative sign 1 end exponent right square bracket end root open square brackets O H to the power of minus sign close square brackets equals square root of 10 to the power of negative sign 10 end exponent end root open square brackets O H to the power of minus sign close square brackets equals 10 to the power of negative sign 5 end exponent space M 
 

Menentukkan pH 

table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space open square brackets 10 to the power of negative sign 5 end exponent space M close square brackets end cell row pOH equals 5 row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 5 end cell row pH equals 9 end table 

Jadi pH garam tersebut adalah 9.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Yassa

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Hitunglah pH larutan :

Pembahasan Soal:

begin mathsize 14px style C H subscript 3 C O O Na end style merupakan senyawa garam yang terbentuk dari anion dari asam lemah dan kation dari basa kuat, sehingga dalam air mengalami hidrolisis parsial dan bersifat basa. Reaksi ionisasi garamnya adalah sebagai berikut.

C H subscript 3 C O O Na space rightwards arrow space C H subscript 3 C O O to the power of minus sign space plus space Na to the power of plus sign 


Untuk mengetahui pH larutan begin mathsize 14px style C H subscript 3 C O O Na end style 1 M begin mathsize 14px style left parenthesis K subscript a equals 1 cross times 10 to the power of negative sign 5 end exponent right parenthesis end style dapat ditentukan dengan cara :

  • Menghitung konsentrasi begin mathsize 14px style O H to the power of minus sign end style

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times M subscript anion end root end cell row blank equals cell square root of fraction numerator 1 cross times 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 5 end exponent end fraction cross times 1 space M end root end cell row blank equals cell square root of 1 cross times 10 to the power of negative sign 9 end exponent space M end root end cell row blank equals cell 1 cross times 10 to the power of negative sign 4 comma 5 end exponent space M end cell end table 

  • Menghitung pOH

table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log left parenthesis 1 cross times 10 to the power of negative sign 4 comma 5 end exponent right parenthesis end cell row blank equals cell 4 comma 5 end cell end table

  • Menghitung pH

table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign 4 comma 5 end cell row blank equals cell 9 comma 5 end cell end table  


Jadi, pH larutan C H subscript bold 3 C O O Na adalah 9,5.undefined

0

Roboguru

Besarnya pH larutan dari 100 mL  0,4 M  adalah ....

Pembahasan Soal:

Larutan C H subscript 3 C O O K merupakan garam yang disusun oleh sisa asam lemah dan sisa basa kuat. Garam ini mengalami hidrolisis parsial dimana ion sisa asam lemah membentuk reaksi kesetimbangan dengan air sehingga melepas ion O H to the power of minus sign dengan reaksi sebagai berikut:

C H subscript 3 C O O to the power of minus sign and H subscript 2 O equilibrium C H subscript 3 C O O H and O H to the power of minus sign 

Berdasarkan reaksi diatas, larutan garam ini bersifat basa, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end root end cell row blank equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent cross times 0 comma 4 end root end cell row blank equals cell square root of 4 cross times 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell row pOH equals cell negative sign log space open square brackets O H to the power of minus 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 cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign left parenthesis 5 minus sign log space 2 right parenthesis end cell row blank equals cell 9 plus log space 2 end cell end table   

Jadi, jawaban yang benar adalah D.

0

Roboguru

Larutan  0,2 M sebanyak 10 mL ditambahkan ke dalam 20 mL larutan  0,1 M (), maka tentukan harga pH larutan campuran tersebut!

Pembahasan Soal:

Langkah 1: menghitung mol begin mathsize 14px style Na O H bold space bold dan bold space C H subscript bold 3 C O O H end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol subscript NaOH end cell equals cell V subscript NaOH cross times M subscript NaOH space end cell row blank equals cell 10 space mL cross times 0 comma 2 space M end cell row blank equals cell 2 space mmol end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol subscript C H subscript 3 C O O H end subscript end cell equals cell V subscript C H subscript 3 C O O H end subscript cross times M subscript C H subscript 3 C O O H end subscript end cell row blank equals cell 20 space mL cross times 0 comma 1 space M end cell row blank equals cell 2 space mmol end cell end table end style


Langkah 2: membuat persamaan reaksi MRS

Langkah 3: menghitung konsentrasi garam begin mathsize 14px style C H subscript bold 3 C O O Na end style

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Langkah 4: menghitung konsentrasi begin mathsize 14px style O H to the power of bold minus sign end style

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H close square brackets end cell equals cell square root of K subscript w over K subscript a open square brackets C H subscript 3 C O O Na close square brackets cross times Valensi space Garam end root end cell row blank equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent left parenthesis 6 comma 7 cross times 10 to the power of negative sign 2 end exponent right parenthesis end root cross times 1 end cell row blank equals cell square root of 6 comma 7 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell 8 comma 18 cross times 10 to the power of negative sign 6 end exponent space M end cell end table  


Langkah 5: menghitung pH

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log space 8 comma 18 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 6 minus sign log space 8 comma 18 end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell pKw bond pOH end cell row blank equals cell 14 minus sign left parenthesis 6 minus sign log space 8 comma 18 right parenthesis end cell row blank equals cell 8 plus log space 8 comma 18 end cell end table end style


Jadi, pH larutan campuran tersebut adalah 8 + log 8,18.

0

Roboguru

100 mL larutan 0,3 M bereaksi dengan 200 mL larutan 0,15 M. Tentukan pH campuran zat tersebut! ()

Pembahasan Soal:

Persamaan reaksi kimia yang terjadi adalah

Karena asam dan basa habis bereaksi, maka terbentuk garam terhidrolisis basa.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times open square brackets C H subscript 3 C O O K close square brackets end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent cross times open parentheses bevelled fraction numerator 30 space mmol over denominator 300 space mL end fraction close parentheses end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 9 end exponent cross times 0 comma 1 end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 10 end exponent end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 5 end exponent end cell row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space 10 to the power of negative sign 5 end exponent end cell row pOH equals 5 row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 5 end cell row bold pH bold equals bold 9 end table


Jadi, pH campuran zat tersebut adalah 9.space

0

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Hitunglah pH dari 3,528 gram NaCN yang dilarutkan ke dalam air sehingga volume larutan menjadi 2 liter. ( Na=23, C=12, N=14)  HCN=

Pembahasan Soal:

Garam NaCN adalah garam yang bersifat basa karena berasal dari campuran antara basa kuat dan asam lemah sehingga pH dari garam tersebut akan memiliki nilai lebih dari 7.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Na C N close square brackets end cell equals cell fraction numerator mol over denominator V open parentheses L close parentheses end fraction end cell row blank equals cell fraction numerator massa forward slash Mr over denominator V open parentheses L close parentheses end fraction end cell row blank equals cell fraction numerator 3 comma 528 forward slash 49 over denominator 2 end fraction end cell row blank equals cell 0 comma 036 space M end cell row blank blank blank row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript italic w over K subscript italic a cross times open square brackets Na C N close parentheses end root end cell row blank equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 7 cross times 10 to the power of negative sign 10 end exponent end fraction cross times 360 cross times 10 to the power of negative sign 2 end exponent end root end cell row blank equals cell square root of 51 comma 42 cross times 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 7 comma 17 cross times 10 to the power of negative sign 3 end exponent end cell row blank blank blank row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell negative sign log space 7 comma 17 cross times 10 to the power of negative sign 3 end exponent end cell row blank equals cell 3 minus sign log space 7 comma 17 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign left parenthesis 3 minus sign log space 7 comma 17 right parenthesis end cell row blank equals cell 11 plus log space 7 comma 17 end cell end table 
 

Jadi, pH dari larutan tersebut adalah 11+log 7,17.

0

Roboguru

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