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Tentukan pH NH4Cl 0,1 M ()!

Pertanyaan

Tentukan pH NH4Cl 0,1 M (K subscript b equals 10 to the power of negative sign 5 end exponent)!

Pembahasan Soal:

menentukan nilai OH- :

open square brackets H to the power of plus sign close square brackets equals square root of K subscript W over K subscript b end root cross times M open square brackets H to the power of plus sign close square brackets 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 end root cross times 0 comma 1 open square brackets H to the power of plus sign close square brackets 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 end root cross times 10 to the power of negative sign 1 end exponent open square brackets H to the power of plus sign close square brackets equals square root of 10 to the power of negative sign 15 end exponent over 10 to the power of negative sign 5 end exponent 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 10 end exponent end root open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 5 end exponent

menentukan pH :

pH equals minus sign log space open square brackets H to the power of plus sign close square brackets pH equals minus sign log space left square bracket 10 to the power of negative sign 5 end exponent right square bracket pH equals 5 

Jadi, pH dari garam NH₄Cl 0,1 M adalah 5.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Nurul

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 04 Juni 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.

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 

Roboguru

Jika 200 mL  0,8 M direaksikan dengan 200 mL larutan  0,8 M, Kb , pH campuran setelah bereaksi adalah...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mmol space N H subscript 4 O H end cell equals cell M middle dot V open parentheses mL close parentheses end cell row blank equals cell 0 comma 8 space M middle dot 200 space mL end cell row blank equals cell 160 space mmol end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mmol space H Cl end cell equals cell M middle dot V open parentheses mL close parentheses end cell row blank equals cell 0 comma 8 space M middle dot 200 space mL end cell row blank equals cell 160 space mmol end cell end table end style


Reaksi antara begin mathsize 14px style N H subscript 4 O H end style dan begin mathsize 14px style H Cl end style adalah:



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 begin inline style fraction numerator 160 space mmol over denominator V space total end fraction end style end cell row blank equals cell begin inline style fraction numerator 160 space mmol over denominator 400 space mL end fraction end style end cell row blank equals cell 0 comma 4 space M end cell end table end style


Konsentrasi begin mathsize 14px style N H subscript 4 to the power of plus end style pada begin mathsize 14px style N H subscript 4 Cl end style adalah



Maka, pH begin mathsize 14px style N H subscript 4 Cl end style adalah:


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 begin inline style Kw over Kb end style middle dot open square brackets G close square brackets end root end cell row blank equals cell square root of begin inline style 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent end style middle dot 4 middle dot 10 to the power of negative sign 1 end exponent end root end cell row blank equals cell square root of 4 middle dot 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 2 middle dot 10 to the power of negative sign 5 end exponent space M end cell end table end style


begin mathsize 14px style 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 2 middle dot 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 2 end cell end table end style


Jadi, pH campuran adalah begin mathsize 14px style 5 minus sign log space 2 end style.

Roboguru

Perhatikan gambar di bawah ini!     Jika diketahui  dan . Tentukanlah pH dari masing-masing larutan garam tersebut!

Pembahasan Soal:

Garam merupakan hasil reaksi antara asam dan basa. Garam terdiri dari garam netral, garam asam, dan garam basa.

  1. Larutan begin mathsize 14px style Na Cl end style merupakan garam netral karena terbentuk dari reaksi asam kuat dan basa kuat. Nilai pH garam netral adalah 7.
  2. Larutan begin mathsize 14px style Na subscript 2 C O subscript 3 end style merupakan garam basa karena terbentuk dari basa kuat dan asam lemah. NIlai pH larutan tersebut ditentukan dengan perhitungan berikut.

    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 end root open square brackets garam close square brackets end cell row blank equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 4 comma 5 cross times 10 to the power of negative sign 7 end exponent end fraction cross times end root 10 to the power of negative sign 1 end exponent end cell row blank equals cell 4 comma 7 cross times 10 to the power of negative sign 5 end exponent end cell row blank blank blank row pOH equals cell negative sign log space 4 comma 7 cross times 10 to the power of negative sign 5 end exponent end cell row blank equals cell 5 minus sign log space 4 comma 7 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 5 minus sign log space 4 comma 7 right parenthesis end cell row blank equals cell 9 plus log space 4 comma 7 end cell end table end style
     
  3. Larutan begin mathsize 14px style N H subscript 4 Cl end style merupakan garam asam karena terbentuk dari reaksi asam kuat dengan basa lemah. Nilai pH larutan tersebut ditentukan melalui perhitungan 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 italic w over K subscript italic b cross times open square brackets garam 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 10 to the power of negative sign 1 end exponent end root end cell row blank equals cell 7 comma 4 cross times 10 to the power of negative sign 6 end exponent end cell row blank blank blank row pH equals cell negative sign log space 7 comma 4 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 6 minus sign log space 7 comma 4 end cell end table end style 
     

Jadi, jawaban yang benar adalah sesuai perhitungan di atas.

Roboguru

Pasangan larutan garam berikut yang dapat memerahkan kertas lakmus biru adalah . . . .

Pembahasan Soal:

Garam yang dapat memerahkan lakmus biru, merupakan garam yang bersifat asam. Berdasarkan pilihan jawaban yang merupakan pasangan garam asam adalah begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end style karena bersal dari asam kuat dan basa lemah dan begin mathsize 14px style N H subscript 4 C H subscript 3 C O O end style karena berasal dari asam dan basa lemah sehingga sifat garam tergantung besaran nilai begin mathsize 14px style K subscript a end style dan begin mathsize 14px style K subscript b end style, jika begin mathsize 14px style K subscript a greater than K subscript b end style maka garam bersifat asam. 

Jadi, jawaban yang tepat adalah C.

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