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pH campuran dari 20 mL larutan CH3​COOH 0,3 M (Ka​=10−5) dan 40 mL larutan NaOH 0,15 M adalah ....

Pertanyaan

pH campuran dari 20 mL larutan C H subscript 3 C O O H 0,3 M left parenthesis K subscript a equals 10 to the power of negative sign 5 end exponent right parenthesis dan 40 mL larutan Na O H 0,15 M adalah ....

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Pembahasan Soal:

puran dari 20 mL larutan C H subscript 3 C O O H 0,3 M left parenthesis K subscript a equals 10 to the power of negative sign 5 end exponent right parenthesis dan 40 mL larutan Na O H 0,15 M

Mol C H subscript 3 C O O H

table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell M cross times V end cell row blank equals cell 0 comma 3 space M cross times 20 space mL end cell row blank equals cell 6 space mmol end cell end table 

Mol Na O H

table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell M cross times V end cell row blank equals cell 0 comma 15 space M cross times 40 space mL end cell row blank equals cell 6 space mmol end cell end table 

Persamaan reaksinya:

Karena pada akhir reaksi hanya dihasilkan garam C H subscript 3 C O O Na, pH ditentukan dari pH garam terhidrolisis. Larutan C H subscript 3 C O O Na merupakan larutan garam yang disusun oleh sisa asam lemah dan sisa basa kuat. Ion sisa asam lemah akan membentuk reaksi kesetimbangan dengan air sehingga melepas ion O H to the power of minus sign dengan reaksi 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 

Sehingga garam ini bersifat basa, persamaan O H to the power of minus sign-nya sebagai berikut:

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 fraction numerator 6 space mmol over denominator 60 space mL end fraction end root end cell row blank equals cell square root of 10 to the power of negative sign 9 end exponent cross times 10 to the power of negative sign 1 end exponent end root end cell row blank equals cell square root of 10 to the power of negative sign 10 end exponent end root end cell row blank 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 blank equals cell negative sign log space 10 to the power of negative sign 5 end exponent end cell row blank equals 5 row cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign 5 end cell row blank equals 9 end table 

Jadi, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Q. 'Ainillana

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Besarnya [OH−] dalam 200 mL larutan yang mengandung 16,2 gram kalium sianat (KOCN) adalah ....(Ka​HOCN=10−6;Ar​:K=39gmol−1,N=14gmol−1,O=16gmol−1,C=12gmol−1,H=1gmol−1)

Pembahasan Soal:

Kalium sianat (begin mathsize 14px style K O C N end style) merupakan garam basa yang mengalami hidrolisis parsial dengan reaksi sebagai berikut.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K O C N space end cell rightwards arrow cell space K to the power of plus sign space plus space O C N to the power of minus sign end cell row cell K to the power of plus sign space plus space H subscript 2 O space end cell rightwards arrow cell space tidak space bereaksi end cell row cell O C N to the power of minus sign space plus space H subscript 2 O space end cell rightwards harpoon over leftwards harpoon cell space H O C N space plus space O H to the power of minus sign end cell end table end style


Besarnya begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets end style dalam 200 mL larutan yang mengandung 16,2 gram begin mathsize 14px style K O C N end style dapat ditentukan dengan rumus yaitu :


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Mr space K O C N end cell equals cell 1 point Ar space K and 1 point Ar space O and 1 point Ar space C and 1 point Ar space N end cell row blank equals cell 1.39 plus 1.16 plus 1.12 plus 1.14 end cell row blank equals 81 row blank blank blank row cell M subscript anion end cell equals cell M subscript O C N to the power of minus sign end subscript double bond M subscript KOCN end cell row blank equals cell g over Mr cross times 1000 over mL end cell row blank equals cell fraction numerator 16 comma 2 over denominator 81 end fraction cross times 1000 over 200 end cell row blank equals 1 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 w over K subscript a cross times M subscript anion 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 6 end exponent cross times 1 end root end cell row blank equals cell square root of 10 to the power of negative sign 8 end exponent end root end cell row blank equals cell 10 to the power of negative sign 4 end exponent end cell end table end style


Jadi, besarnya begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets end style dalam larutan tersebut adalah begin mathsize 14px style 10 to the power of negative sign 4 end exponent end style M.space 

0

Roboguru

Sebanyak 4,1 g padatan natrium asetat (Mr​=82) dilarutkan dalam 0,5 L air. Jika ditambahkan 1,5 L air lagi, perubahan pH yang terjadi adalah .... (Ka​ asam asetat=10−5)

Pembahasan Soal:

Garam natrium asetat open parentheses C H subscript 3 C O O Na close parentheses berasal dari asam lemah C H subscript 3 C O O H dan basa kuat Na O H. Reaksi ionisasi dan hidrolisis garamnya, yaitu sebagai berikut.


C H subscript 3 C O O Na yields 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
 

Oleh karena yang bersisa adalah ion O H to the power of minus sign, garam tersebut bersifat basa. Persamaan matematis yang digunakan untuk menghitung pH garam basa, yaitu sebagai berikut.
 

begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets equals 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 style
 

pH larutan garam pada volume air 0,5 L dapat dihitung sebagai berikut.
 

Error converting from MathML to accessible text.
 

Penambahan 1,5 L air mengubah volume menjadi 2 L, artinya pH larutan akan berubah. Dengan menggunakan persamaan yang sama, pH larutannya menjadi seperti berikut.
 

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 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent cross times fraction numerator 4 comma 1 over denominator 82 end fraction cross times 1 half 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 025 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 25 cross times 10 to the power of negative sign 12 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 5 cross times 10 to the power of negative sign 6 end exponent end cell row blank blank blank 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 5 cross times 10 to the power of negative sign 6 end exponent close square brackets end cell row pOH equals cell 6 minus sign log space 5 end cell row blank blank blank row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign open parentheses 6 minus sign log space 5 close parentheses end cell row pH equals cell 8 plus log space 5 end cell end table


Perubahan pH yang terjadi adalah 9 menjadi 8 + log 5.

Jadi, jawaban yang benar adalah C.

0

Roboguru

Larutan NaOH 0,2 M sebanyak 10 ml ditambahkan ke dalam 20 mL larutan CH3​COOH 0,1 M (Ka​=10−5) maka tentukan harga pH larutan campuran tersebut!

Pembahasan Soal:

Penyelesaian:

Untuk mengetahui pH campuran, kita periksa terlebih dahulu reaksi mrs-nya untuk mengetahui sifat larutan.

1. Menghitung mol masing-masing larutan 

begin mathsize 14px style 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 V end cell row blank equals cell 0 comma 1 cross times 20 end cell row blank equals cell 2 space mmol end cell row cell mol space Na O H end cell equals cell M cross times V end cell row blank equals cell 0 comma 2 cross times 10 end cell row blank equals cell 2 space mmol end cell end table end style 

2. Menuliskan reaksi MRS untuk mengetahui sifat larutan

Reaksinya adalah sebagai berikut:

 

Berdasarkan reaksi diatas, dapat diketahui campuran tersebut membentuk garam yang bersifat basa karena terbentuk dari reaksi asam lemah dan basa kuat.

3. Menghitung konsentrasi larutan garam basa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row M equals cell mol over V subscript total end cell row blank equals cell fraction numerator 2 space mmol over denominator 30 space mL end fraction end cell row blank equals cell 0 comma 067 space M end cell end table end style 

4. Menghitung konsentrasi OH- larutan garam

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 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 0 comma 067 end root end cell row blank equals cell square root of 0 comma 67 cross times 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 0 comma 81 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 8 comma 1 cross times 10 to the power of negative sign 6 end exponent space M end cell end table  

5. Menghitung pH larutan

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 1 cross times 10 blank to the power of negative sign 6 end exponent end cell row blank equals cell 6 minus sign log space 8 comma 1 end cell end table 

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 left parenthesis 6 minus sign log space 8 comma 1 right parenthesis end cell row blank equals cell 8 plus log space 8 comma 1 end cell end table  

Jadi, pH campuran tersebut adalah 8 + log 8,1. 

0

Roboguru

Sebanyak 100 mL larutan CH3​COONa mempunyai pH = 10. Jika Ka​CH3​COOH=10−5 dan Mr  = 82, tentukan massa garam CH3​COONa yang terlarut dalam 100 mL larutan !

Pembahasan Soal:

Langkah 1: menghitung begin mathsize 14px style begin bold style open square brackets O H to the power of minus sign close square brackets end style end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell pKw bond pH end cell row blank equals cell 14 minus sign 10 end cell row blank equals 4 end table end style

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 4 equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets 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 4 end exponent end cell end table end style 


Langkah 2: menghitung begin mathsize 14px style begin bold style open square brackets C H subscript 3 C O O Na close square brackets end style end style 

begin mathsize 14px 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 open square brackets C H subscript 3 C O O Na close square brackets end root end cell row cell 10 to the power of negative sign 4 end exponent 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 open square brackets C H subscript 3 C O O Na close square brackets end root end cell row cell open parentheses 10 to the power of negative sign 4 end exponent close parentheses squared end cell equals cell open parentheses square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent open square brackets C H subscript 3 C O O Na close square brackets end root close parentheses squared end cell row cell 10 to the power of negative sign 8 end exponent end cell equals cell 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent open square brackets C H subscript 3 C O O Na close square brackets end cell row cell open square brackets C H subscript 3 C O O Na close square brackets end cell equals cell fraction numerator 10 to the power of negative sign 8 end exponent cross times 10 to the power of negative sign 5 end exponent over denominator 10 to the power of negative sign 14 end exponent end fraction end cell row blank equals cell 10 space M end cell end table end style


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

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 Na close square brackets end cell equals cell m over Mr cross times 1000 over V end cell row cell 10 space M end cell equals cell m over 82 cross times fraction numerator 1000 over denominator 100 space mL end fraction end cell row m equals cell 82 space gram end cell row blank blank blank end table end style

Jadi, massa garam begin mathsize 14px style C H subscript 3 C O O Na end style yang terlarut adalah 82 gram.

0

Roboguru

pH hasil hidrolisis larutan NaCN 0,01 M adalah .... (Ka​HCN=10−10)

Pembahasan Soal:

Larutan Na C N merupakan garam yang disusun oleh sisa basa kuat dan sisa asam lemah. 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 N to the power of minus sign and H subscript 2 O equilibrium H C N 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 10 end exponent cross times 0 comma 01 end root end cell row blank equals cell square root of 10 to the power of negative sign 6 end exponent end root end cell row blank equals cell 10 to the power of negative sign 3 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 10 to the power of negative sign 3 end exponent end cell row blank equals 3 row cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign 3 end cell row blank equals 11 end table  

Jadi, jawaban yang benar adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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