Roboguru

Hitung pH dari larutan campuran 100 ml HCI 0,5 M dengan 400 ml  0,125 M. ()

Pertanyaan

Hitung pH dari larutan campuran 100 ml HCI 0,5 M dengan 400 ml N H subscript 4 O H 0,125 M. (K subscript b equals 10 to the power of negative sign 5 end exponent)

Pembahasan Soal:

Untuk mengetahui jenis campuran yang terbentuk, kita tuliskan persamaan reaksi mrs terlebih dahulu.

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H Cl end cell equals cell M cross times V end cell row blank equals cell 0 comma 5 space M cross times 100 space mL end cell row blank equals cell 50 space mmol end cell row cell mol space N H subscript 4 O H end cell equals cell M cross times V end cell row blank equals cell 0 comma 125 space M cross times 400 space mL end cell row blank equals cell 50 space mmol end cell end table 

Karena di akhir reaksi hanya tersisa garam N H subscript 4 Cl, maka pH campuran ditentukan dari pH garam terhidrolisis. Garam N H subscript 4 Cl merupakan garam terhidrolisis parsial karena dibentuk dari asam kuat dan basa lemah. Dimana sisa basa lemah akan bereaksi dengan air melepas ion H to the power of plus sign sehingga sifat garamnya adalah asam. Untuk mencari pH menggunakan 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 square root of K subscript w over K subscript b cross times left square bracket N H subscript 4 to the power of plus right square bracket 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 50 space mmol over denominator left parenthesis 100 plus 400 right parenthesis 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 10 to the power of negative sign 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 10 to the power of negative sign 5 end exponent end cell row blank equals 5 end table 

Jadi, pH campuran tersebut adalah 5.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Q. 'Ainillana

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 07 Juni 2021

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Pertanyaan yang serupa

Larutan100 ml  0,1 M direaksikan dengan 400 ml  0,025 M. pH larutan yang terjadi sebesar .... ()

Pembahasan Soal:

Untuk mengetahui pH campuran, terlebih dahulu kita tuliskan reaksi mrs.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space N H subscript 3 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 cell mol space H Cl end cell equals cell M cross times V end cell row blank equals cell 0 comma 025 cross times 400 end cell row blank equals cell 10 space mmol end cell end table end style 

 

Karena yang terbentuk larutan garam, maka kita tentukan terlebih dahulu sifat garam ini. Karena disusun oleh basa lemah dan asam kuat, artinya sisa basa lemah mengalami hidrolisis dan melepaskan ion begin mathsize 14px style H to the power of plus sign end style (hidrolisis parsial), maka untuk mencari pH:

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 left square bracket N H subscript 4 to the power of plus right square bracket 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 02 end root end cell row blank equals cell square root of 2 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell square root of 2 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 2 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 2 end cell end table end style 

Jadi, pH campuran tersebut adalah begin mathsize 14px style bottom enclose bold 5 bold comma bold 5 bold minus sign bold log square root of bold 2 end enclose end style.

 

0

Roboguru

Diketahui larutan 0,1 M  dan . Hitung harga , , , dan pH larutan!

Pembahasan Soal:

begin mathsize 14px style N H subscript 4 C I end style adalah suatu larutan garam yang terbentuk dari asam kuat HCl dan basa lemah NH3. Garam begin mathsize 14px style N H subscript 4 C I end style yang terbentuk adalah garam yang bersifat asam.
 

  • Penentuan tetapan hidrolisis (begin mathsize 14px style K subscript h end style).

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript h end cell equals cell K subscript w over K subscript b end cell row cell K subscript h 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 end cell row cell K subscript h end cell equals cell 10 to the power of negative sign 9 end exponent end cell end table end style 
 

  • Penentuan begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left square bracket H to the power of plus right square bracket end cell equals cell square root of K subscript h cross times M end root end cell row cell left square bracket H to the power of plus right square bracket end cell equals cell square root of K subscript h cross times open square brackets N H subscript 4 Cl close square brackets end root end cell row cell left square bracket H to the power of plus right square bracket end cell 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 cell left square bracket H to the power of plus right square bracket end cell equals cell square root of 10 to the power of negative sign 10 end exponent end root end cell row cell left square bracket H to the power of plus right square bracket end cell equals cell 10 to the power of negative sign 5 end exponent end cell end table end style  
 

  • Penentuan begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets 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 fraction numerator K subscript w over denominator open square brackets H to the power of plus close square brackets end fraction 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 14 end exponent over 10 to the power of negative sign 5 end exponent 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 9 end exponent end cell end table end style   
 

  • Penentuan pH.

begin mathsize 14px style pH =- log space open square brackets H to the power of plus close square brackets pH equals minus sign log space 10 to the power of negative sign 5 end exponent pH equals 5 end style 
 


Jadi, nilai begin mathsize 14px style K subscript h end stylebegin mathsize 14px style open square brackets H to the power of plus sign close square brackets end stylebegin mathsize 14px style open square brackets O H to the power of minus sign close square brackets end style, pH berturut-turut adalah begin mathsize 14px style 10 to the power of negative sign 9 end exponent comma space 10 to the power of negative sign 5 end exponent comma space 10 to the power of negative sign 9 end exponent comma space 5 end style.

0

Roboguru

Konsentrasi  dalam larutan  0,036 M adalah...

Pembahasan Soal:

Garam begin mathsize 14px style N H subscript 4 Cl end style berasal dari basa lemah dan asam kuat yang terionisasi sempurna dalam air menghasilkan kation dan anion. Kation dari basa lemah dan anion dari asam kuat. Reaksinya sebagai berikut:

begin mathsize 14px style N H subscript 4 Cl yields with plus air on top N H subscript 4 to the power of plus sign and Cl to the power of minus sign N H subscript 4 to the power of plus sign and H subscript 2 O equilibrium N H subscript 3 and H subscript 3 O to the power of plus sign N H subscript 4 to the power of plus sign equilibrium N H subscript 3 and H to the power of plus sign end style 

Adanya ion begin mathsize 14px style H to the power of plus sign end style dalam hasil reaksi menunjukan bahwa larutan garam tersebut bersifat asam dan kita dapat mencari konsentrasi ion tersebut sebagai berikut:

begin mathsize 14px style H to the power of plus sign equals square root of K subscript w over K subscript b x open square brackets N H subscript 4 Cl close square brackets end root H to the power of plus sign equals square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent x left square bracket 36 cross times 10 to the power of negative sign 3 end exponent right square bracket end root H to the power of bold plus sign bold equals bold 6 bold cross times bold 10 to the power of bold minus sign bold 6 end exponent bold space italic M end style 

Oleh karena itu, jawaban yang benar adalah C.space 

0

Roboguru

21,4 gram  dilarutkan dalam air hingga 800 mL, jika , tentukan: pH larutan Derajat hidrolisis Berapa persen dari garam yang mengalami hidrolisis

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.

Mencari pH larutan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets N H subscript 4 Cl close square brackets end cell equals cell fraction numerator italic g space N H subscript 4 Cl over denominator M subscript r space N H subscript 4 Cl end fraction cross times fraction numerator 1000 over denominator V space larutan end fraction end cell row cell open square brackets N H subscript 4 Cl close square brackets end cell equals cell fraction numerator 21 comma 4 space gram over denominator 53 comma 5 space gram space mol to the power of negative sign 1 end exponent end fraction cross times fraction numerator 1000 over denominator 800 space mL end fraction end cell row cell open square brackets N H subscript 4 Cl close square brackets end cell equals cell 5 cross times 10 to the power of negative sign 1 end exponent space M 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 italic w over K subscript italic b cross times open square brackets kation close square brackets subscript garam end root 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 italic w over K subscript italic b cross times open square brackets N H subscript 4 Cl close square brackets end root end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 cross times 10 to the power of negative sign 5 end exponent end fraction cross times 5 cross times 10 to the power of negative sign 1 end exponent end root end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 1 comma 58 cross times 10 to the power of negative sign 5 end exponent space M end cell row blank blank blank row pH equals cell negative sign log open square brackets H to the power of plus sign close square brackets end cell row pH equals cell negative sign log left square bracket 1 comma 58 cross times 10 to the power of negative sign 5 end exponent right square bracket end cell row pH equals cell 5 minus sign log space 1 comma 58 end cell end table end style 

Jadi pH larutan adalah begin mathsize 14px style bold 5 bold minus sign bold log bold space bold 1 bold comma bold 58 end style.
 

Mencari Derajat Hidrolisis

begin mathsize 14px style alpha equals square root of K subscript h over open square brackets kation close square brackets subscript garam end root alpha equals square root of fraction numerator begin display style bevelled K subscript italic w over K subscript italic b end style over denominator open square brackets N H subscript 4 Cl close square brackets end fraction end root alpha equals square root of fraction numerator begin display style bevelled fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 cross times 10 to the power of negative sign 5 end exponent end fraction end style over denominator 5 cross times 10 to the power of negative sign 1 end exponent end fraction end root alpha equals 3 comma 16 cross times 10 to the power of negative sign 5 end exponent end style 

Jadi derajat hidrolisisnya adalah begin mathsize 14px style bold 3 bold comma bold 16 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end style.
 

Mencari % hidrolisis

begin mathsize 14px style percent sign space hidrolisis equals space 3 comma 16 cross times 10 to the power of negative sign 5 end exponent cross times 100 percent sign percent sign space hidrolisis equals 0 comma 00316 percent sign end style 

Jadi % hidrolisisnya adalah 0,00316%undefined 

0

Roboguru

pH dari 100 mL  0,9 M apabila  adalah ....

Pembahasan Soal:

Larutan begin mathsize 14px style N H subscript 4 Br end style merupakan larutan garam terhidrolisis parsial yang terdiri dari sisa basa lemah dan sisa asam kuat, dimana dalam air ion sisa basa lemah akan bereaksi dengan air melepaskan ion H+ (bersifat asam). untuk menentukan pH garam terhidrolisis dapat menggunakan rumus berikut:

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 left square bracket N H subscript 4 to the power of plus right square bracket 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 9 end root end cell row blank equals cell square root of 10 to the power of negative sign 9 end exponent cross times 9 cross times 10 to the power of negative sign 1 end exponent end root end cell row blank equals cell square root of 9 cross times 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 3 cross times 10 to the power of negative sign 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 3 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 3 end cell end table end style 

Jadi, pH larutan tersebut adalah 5 - log 3.

0

Roboguru

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