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Tentukan nilai pH dari larutan 0,1 M  !

Pertanyaan

Tentukan nilai pH dari larutan 0,1 M begin mathsize 14px style N H subscript 4 N O subscript 3 end style ! begin mathsize 14px style left parenthesis Kb space N H subscript 3 equals 10 to the power of negative sign 5 end exponent right parenthesis end style 

Pembahasan Soal:

Larutan di atas terbentuk dari asam kuat begin mathsize 14px style H N O subscript 3 end style dan basa lemah begin mathsize 14px style N H subscript 4 O H end style, sehingga terbentuk larutan hidrolisis dengan konsentrasi garam begin mathsize 14px style N H subscript 4 N O subscript 3 end style 0,1 M.


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 square root of K subscript w over K subscript b cross times open square brackets N H subscript 4 N O subscript 3 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 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 pH equals cell negative sign log open parentheses 10 to the power of negative sign 5 end exponent close parentheses end cell row blank equals 5 end table end style 


Sehingga pH larutan yang terbentuk adalah 5.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Zharva

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 01 April 2021

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Pertanyaan yang serupa

Larutan  mempunyai pH = 5. Jika , hitunglah kemolaran .

Pembahasan Soal:

N H subscript 4 Cl merupakan garam yang mengalami hidrolisis sebagian dan bersifat asam. Penentuan kemolaran garam N H subscript 4 Cl dapat dihitung berdasarkan nilai pH larutan.

table attributes columnalign right center left columnspacing 0px end attributes row cell N H subscript 4 Cl end cell rightwards arrow cell N H subscript 4 to the power of plus sign and Cl to the power of minus sign end cell row cell N H subscript 4 to the power of plus sign and H subscript 2 O end cell rightwards harpoon over leftwards harpoon cell N H subscript 4 O H and H to the power of plus sign end cell row blank blank blank row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of K subscript w over K subscript b open square brackets N H subscript 4 Cl close square brackets end root end cell row cell 10 to the power of negative sign pH end exponent end cell equals cell square root of K subscript w over K subscript b open square brackets N H subscript 4 Cl close square brackets end root end cell row cell 10 to the power of negative sign 5 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 N H subscript 4 Cl close square brackets end root end cell row cell open square brackets N H subscript 4 Cl close square brackets end cell equals cell fraction numerator left parenthesis 10 to the power of negative sign 5 end exponent right parenthesis squared 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 to the power of negative sign 15 end exponent over 10 to the power of negative sign 14 end exponent end cell row blank equals cell 0 comma 1 space M end cell end table 

Jadi, kemolaran N H subscript 4 Cl adalah 0,1 M.

0

Roboguru

Jika tetapan hidrolisis , tentukan pH larutan 0,5 M dengan volume 100 mL!

Pembahasan Soal:

Garam yang berasal dari basa lemah dan asam kuat akan terionisasi sempurna dalam air dan akan menghasilkan kation anion. Kation berasal dari basa lemah dan anion dari asam kuat seperti reaksi berikut.

begin mathsize 14px style N H subscript 4 Cl left parenthesis italic a italic q right parenthesis rightwards arrow with plus air on top N H subscript 4 to the power of plus left parenthesis italic a italic q right parenthesis plus Cl to the power of minus sign left parenthesis italic a italic q right parenthesis end style 

Kation dari basa lemah begin mathsize 14px style N H subscript 4 to the power of plus end style akan terhidrolisis dengan reaksi berikut.

begin mathsize 14px style N H subscript 4 to the power of plus open parentheses aq close parentheses and H subscript 2 O open parentheses italic l close parentheses equilibrium N H subscript 3 left parenthesis italic a italic q right parenthesis plus H subscript 3 O to the power of plus sign left parenthesis italic a italic q right parenthesis N H subscript 4 to the power of plus left parenthesis italic a italic q right parenthesis equilibrium N H subscript 3 open parentheses aq close parentheses and H to the power of plus sign left parenthesis italic a italic q right parenthesis end style  

Adanya ion begin mathsize 14px style H to the power of plus sign end style dalam hasil reaksi  menunjukkan bahwa larutan garam bersifat asam. Ion begin mathsize 14px style Cl to the power of minus sign end style berasal dari asam kuat tidak bereaksi dengan air (tidak terhidrolisis) sehingga reaksinya adalah hidrolisis parsial. Konsentrasi kation sama dengan konsentrasi begin mathsize 14px style N H subscript 4 Cl end style.

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K subscript italic h cross times open square brackets kation close square brackets subscript garam end root open square brackets H to the power of plus sign close square brackets equals square root of K subscript italic h cross times open square brackets N H subscript 4 Cl close square brackets end root open square brackets H to the power of plus sign close square brackets equals square root of 10 to the power of negative sign 8 end exponent cross times 5 cross times 10 to the power of negative sign 1 end exponent end root open square brackets H to the power of plus sign close square brackets equals 7 comma 07 cross times 10 to the power of negative sign 5 end exponent space M  pH equals minus sign log open square brackets H to the power of plus sign close square brackets pH equals minus sign log left square bracket 7 comma 07 cross times 10 to the power of negative sign 5 end exponent right square bracket pH equals 5 minus sign log space 7 comma 07 end style 

Jadi pH begin mathsize 14px style N H subscript bold 4 Cl bold space bold adalah bold space bold 5 bold minus sign bold log bold space bold 7 bold comma bold 07 end style .undefined 

0

Roboguru

Berapa gram  harus dilarutkan dalam 500 mL larutan agar memiliki nilai pH = 5 - log 2 ? ( = ,  = 14,  = 1,  = 80)

Pembahasan Soal:

pH equals minus sign log space open square brackets H to the power of plus sign close square brackets 5 minus sign log space 2 equals minus sign log space open square brackets H to the power of plus sign close square brackets 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 
 

open square brackets H to the power of plus sign close square brackets equals square root of K subscript w over K subscript b cross times open square brackets G close square brackets end root 2 cross times 10 to the power of negative sign 5 end exponent equals 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 m over italic M subscript r cross times fraction numerator 1000 over denominator V open parentheses mL close parentheses end fraction end root 4 cross times 10 to the power of negative sign 10 end exponent equals 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 m over 98 cross times 1000 over 500 fraction numerator 7 comma 2 cross times 10 to the power of negative sign 15 end exponent over denominator 10 to the power of negative sign 14 end exponent end fraction equals m over 49 m equals 35 comma 28 space gram 

 

Jadi, jawaban sesuai dengan uraian di atas.space  

0

Roboguru

Hitunglah pH larutan:  0,1 M (Kb = )

Pembahasan Soal:

N H subscript 4 Cl adalah garam asam yang mengalami hidrolisis parsial. Ion yang mengalami hidrolisis pada N H subscript 4 Cl adalah ion N H subscript 4 to the power of plus yang berasal dari basa  lemah.


N H subscript 4 Cl yields N H subscript 4 and Cl to the power of minus sign


Koefisien Cl to the power of minus sign sama dengan koefisien N H subscript 4 Cl maka konsentrasi Cl to the power of minus sign adalah 0,1 M.


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 square root of begin inline style Kw over Kb end style cross times M end root end cell row blank equals cell square root of begin inline style fraction numerator 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 end style cross times 0 comma 1 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 end table


table attributes columnalign right center left columnspacing 0px end attributes 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 negative sign log space 10 to the power of negative sign 5 end exponent end cell row blank equals 5 end table


Jadi, pH larutan adalah 5.

0

Roboguru

Sebanyak 30 mL suatu larutan amonia tepat bereaksi dengan 20 mL asam sulfat 0,15 M. Jika diketahui  , tentukan: b. pH larutan setelah penambahan asam sulfat

Pembahasan Soal:

Campuran dari asam kuat dan basa lemah akan membentuk larutan garam yang bersifat asam sehingga untuk menentukan pH nya maka ditentukan berapakah mol garam yang terbentuk



table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Garam close square brackets end cell equals cell fraction numerator 3 mmol over denominator left parenthesis 30 mL and 20 mL right parenthesis end fraction equals 0 comma 06 M end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of K subscript w over K subscript a cross times open square brackets Garam close square brackets cross times open square brackets Valensi 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 06 cross times 2 end root end cell row blank equals cell square root of 12 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell square root of 12 cross times 10 to the power of negative sign 5 comma 5 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 negative sign log space square root of 12 cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row blank equals cell 5 comma 5 minus sign log space square root of 12 end cell end table


Jadi, pH larutan setelah dicampur menjadi 5 comma 5 minus sign log space square root of 12.space 

0

Roboguru

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