Roboguru

Sebanyak 100 mL larutan  0,05 M direaksikan dengan 100 mL  yang kemolarannya 0,05 M. Reaksi yang terjadi: Jika  air  dan , nilai pH campuran tersebut adalah ....

Pertanyaan

Sebanyak 100 mL larutan begin mathsize 14px style H N O subscript 2 end style 0,05 M direaksikan dengan 100 mL begin mathsize 14px style K O H end style yang kemolarannya 0,05 M. Reaksi yang terjadi:


begin mathsize 14px style K O H left parenthesis italic a italic q right parenthesis plus H N O subscript 2 left parenthesis italic a italic q right parenthesis yields K N O subscript 2 left parenthesis italic a italic q right parenthesis plus H 2 O open parentheses italic l close parentheses space end style


Jika undefined air begin mathsize 14px style equals 10 to the power of negative sign 14 end exponent end style dan begin mathsize 14px style K subscript a space H N O subscript 2 equals 5 space x space 10 to the power of negative sign 4 end exponent end style, nilai pH campuran tersebut adalah .... 

  1. 1undefined 

  2. 5undefined 

  3. 6undefined 

  4. 7undefined 

  5. 8undefined 

Pembahasan Video:

Pembahasan Soal:

Menghitung mol KOH

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell M cross times V end cell row blank equals cell 0 comma 05 space M cross times 100 space mL end cell row blank equals cell 5 space mmol end cell end table end style

Menghitung mol begin mathsize 14px style H N O subscript 2 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell M cross times V end cell row blank equals cell 0 comma 05 space M cross times 100 space mL end cell row blank equals cell 5 space mmol end cell end table end style

Dari perhitungan tersebut, maka reaksi yang terjadi adalah sebagai berikut:

 

Persamaan reaksi tersebut menunjukkan asam dan basa habis bereaksi, maka larutan yang terbentuk adalah garam basa yang terhidrolisis.

Menghitung konsentrasi garam:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets K N O subscript 2 close square brackets end cell equals cell n over V end cell row blank equals cell fraction numerator 5 space mmol over denominator 200 space mL end fraction end cell row blank equals cell 25 cross times 10 to the power of negative sign 3 space end exponent M end cell end table end style 

Menghitung pH

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 italic w over K subscript italic a cross times open square brackets G close square brackets 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 5 cross times 10 to the power of negative sign 4 end exponent end fraction cross times 25 cross times 10 to the power of negative sign 3 end exponent end root end cell row blank equals cell square root of 0 comma 5 cross times 10 to the power of negative sign 12 end exponent end root end cell row blank equals cell 10 to the power of negative sign 6 end exponent cross times square root of 0 comma 5 end root end cell 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 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 10 to the power of negative sign 6 end exponent cross times square root of 0 comma 5 end root right parenthesis end cell row blank equals cell 6 minus sign log square root of 0 comma 5 end root end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH and pOH end cell equals 14 row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign left parenthesis 6 minus sign log square root of 0 comma 5 end root right parenthesis end cell row blank equals cell 8 plus log square root of 0 comma 5 end root end cell row blank equals cell 8 plus left parenthesis minus sign 0 comma 15 right parenthesis end cell row blank equals cell 7 comma 85 almost equal to 8 end cell end table end style 


Jadi, nilai pH campuran tersebut adalah 8, dan jawaban yang tepat adalah E. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Hidayati

Mahasiswa/Alumni Universitas Indonesia

Terakhir diupdate 30 Agustus 2021

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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

Sebanyak 100 mL  0,2 M dicampur dengan 100 mL larutan  0,2 M. Jika  , maka pH larutan setelah dicampur adalah ....

Pembahasan Soal:

Langkah 1: Menghitung mol begin mathsize 14px style C H subscript bold 3 C O O H end style dan begin mathsize 14px style Na O H 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 100 space mL cross times 0 comma 2 space M end cell row blank equals cell 20 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 NaOH end cell equals cell V subscript NaOH cross times M subscript NaOH end cell row blank equals cell 100 space mL cross times 0 comma 2 space M end cell row blank equals cell 20 space mmol end cell end table end style 


Langkah 2: membuat persamaan reaksi RMS


Langkah 3: menghitung konsentrasi garam 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 n over V end cell row blank equals cell fraction numerator 20 space mmol over denominator 200 space mL end fraction end cell row blank equals cell 0 comma 1 space M end cell end table end style


Langkah 4: menghitung konsentrasi begin mathsize 14px style O H to the power of bold minus sign 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 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 10 to the power of negative sign 1 end exponent right parenthesis 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 space M end cell end table end style 


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 10 to the power of negative sign 5 end exponent end cell row blank equals 5 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 5 end cell row blank equals 9 end table end style 


Dengan demikian, jawaban yang tepat adalah E.

0

Roboguru

Tentukan pH dari larutan berikut! 100 mL larutan Na2S 0,1 M (Ka H2S =  )

Pembahasan Soal:

Na2S merupakan garam yang terbentuk dari basa kuat (NaOH) dan asam lemah (H2S) sehingga bersifat basa

menentukan pH garam basa Na2S :

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of Kw over Ka cross times open square brackets garam close square brackets end root open square brackets O H to the power of minus sign close square brackets equals square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 1 cross times 10 to the power of negative sign 7 end exponent end fraction 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 fraction numerator 10 to the power of up diagonal strike negative sign 15 end strike end exponent over denominator 1 cross times 10 to the power of up diagonal strike negative sign 7 end strike end exponent end fraction 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 8 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 4 end exponent  pOH space equals minus sign log space open square brackets O H to the power of minus sign close square brackets pOH space equals minus sign log space left square bracket 10 to the power of negative sign 4 end exponent right square bracket pOH space equals 4 pH space space space equals 14 minus sign pOH pH space space space equals 14 minus sign 4 pH space space space equals 10 end style 

Jadi, pH 100 mL larutan Na2S 0,1 M adalah 10

1

Roboguru

Sejumlah  padat dimasukkan ke dalam  mL larutan M . Jika larutan yang dihasilkan mempunyai pH = , massa  yang dimasukkan adalah ....

Pembahasan Soal:

Reaksi antara NaOH dan begin mathsize 14px style C H subscript 3 C O O H end style akan menghasilkan garam begin mathsize 14px style C H subscript 3 C O O Na end style dan begin mathsize 14px style H subscript 2 O end style. Garam undefined akan mengalami hidrolisis parsial dan menghasilkan ion begin mathsize 14px style O H to the power of minus sign end style yang akan mempengaruhi pH larutan 

begin mathsize 14px style Na O H and C H subscript 3 C O O H yields C H subscript 3 C O O Na and H subscript 2 O 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 yields C H subscript 3 C O O H and O H to the power of minus sign end style 

Tahap 1. Menentukan begin mathsize 14px style open square brackets NaOH close square brackets end style

begin mathsize 14px style space space space space space space pH equals 9 space space space pOH equals 14 minus sign 9 equals 5 space space space pOH equals minus sign log space open square brackets O H to the power of minus sign close square brackets space space space space space space space space 5 equals minus sign log space open square brackets O H to the power of minus sign close square brackets open square brackets O H to the power of minus sign close square brackets equals 1 cross times 10 to the power of negative sign 5 end exponent space space space space space space space space space space space open square brackets O H to the power of minus sign close square brackets equals square root of Kw over Ka middle dot open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end root space space space space space space space 1 cross times 10 to the power of negative sign 5 end exponent equals 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 middle dot open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end root space space left parenthesis space 1 cross times 10 to the power of negative sign 5 end exponent right parenthesis squared equals left parenthesis square root of 10 to the power of negative sign 9 end exponent middle dot open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end root right parenthesis squared space space space space space space space space space space space 1 cross times 10 to the power of negative sign 10 end exponent equals 10 to the power of negative sign 9 end exponent cross times open square brackets C H subscript 3 C O O to the power of minus sign close square brackets space open square brackets C H subscript 3 C O O to the power of minus sign close square brackets equals fraction numerator 1 cross times 10 to the power of negative sign 10 end exponent over denominator 10 to the power of negative sign 9 end exponent end fraction space space open square brackets C H subscript 3 C O O to the power of minus sign close square brackets equals 0 comma 1 space M end style        

begin mathsize 14px style open square brackets Na O H close square brackets equals open square brackets C H subscript 3 C O O Na close square brackets equals open square brackets C H subscript 3 C O O to the power of minus sign close square brackets equals 0 comma 1 space M end style

Tahap 2. Menentukan massa NaOH

begin mathsize 14px style n space Na O H equals 0 comma 1 space M cross times 100 space mL n space Na O H equals 0 comma 1 space mol space L to the power of negative sign 1 end exponent cross times 0 comma 1 space L n space Na O H equals 0 comma 01 space mol end style 

begin mathsize 14px style massa space NaOH equals 0 comma 01 space mol cross times 40 space straight g divided by mol massa space NaOH equals 0 comma 4 space straight g massa space NaOH equals 0 comma 4 cross times space 1000 space mg massa space NaOH equals 400 space miligram end style 

Jadi, jawaban yang benar tidak ada di option jawaban.space

 

 

 

0

Roboguru

Diketahui   =  dan    = . Manakah yang mempunyai pH lebih tinggi,  0,1 M atau  0,1 M? Jelaskan.

Pembahasan Soal:

pH begin mathsize 14px style Na C H subscript 3 C O O end style 0,1 M

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 cross times open square brackets G close square brackets 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 1 comma 8 cross times 10 to the power of negative sign 5 end exponent end fraction cross times left parenthesis 0 comma 1 right parenthesis end root end cell row blank equals cell square root of fraction numerator 10 to the power of negative sign 10 end exponent over denominator 1 comma 8 end fraction end root end cell row blank equals cell square root of 5 comma 56 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell 7 comma 45 cross times 10 to the power of negative sign 6 end exponent end cell row pOH equals cell negative sign log left parenthesis 7 comma 45 cross times 10 to the power of negative sign 6 end exponent right parenthesis end cell row blank equals cell 5 comma 13 end cell row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign 5 comma 13 end cell row blank equals cell 8.87 end cell end table end style  


pH begin mathsize 14px style Na C N end style 0,1 M

Error converting from MathML to accessible text. 


Nilai pH yang lebih tinggi dimiliki oleh NaCN. Hal ini terjadi karena HCN adalah asam yang lebih lemah daripada begin mathsize 14px style C H subscript 3 C O O H end style, dilihat dari nilai begin mathsize 14px style K subscript a end style HCN yang jauh lebih kecil. Semakin lemah kekuatan suatu asam, maka basa konjugasinya justru akan semakin kuat. Oleh karena itu, pH garam yang dihasilkan oleh NaCN akan lebih tinggi dibandingkan pH garam begin mathsize 14px style Na C H subscript 3 C O O end style dengan konsentrasi yang sama.

Jadi, pH garam yang lebih tinggi adalah NaCN.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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