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Sebanyak 6,84 gram aluminium sulfat ditambahkan ke dalam air sehingga volume larutan menjadi 2 liter. Jika diketahui  dan , pH larutan yang terbentuk adalah ...

Pertanyaan

Sebanyak 6,84 gram aluminium sulfat begin mathsize 14px style left parenthesis Al subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 right parenthesis end styleditambahkan ke dalam air sehingga volume larutan menjadi 2 liter. Jika diketahui begin mathsize 14px style A subscript r space Al equals 27 space g space mol to the power of negative sign 1 end exponent comma space S equals 32 space g space mol to the power of negative sign 1 end exponent end style dan begin mathsize 14px style O equals 16 space g space mol to the power of negative sign 1 end exponent end style, pH larutan yang terbentuk adalah ...undefined 

begin mathsize 14px style left parenthesis K subscript w equals 10 to the power of negative sign 14 end exponent comma space K subscript b space Al open parentheses O H close parentheses subscript 3 equals 2 comma 5 cross times 10 to the power of negative sign 9 end exponent right parenthesis end style 

  1. begin mathsize 14px style 12 plus log space 2 comma 5 end style 

  2. begin mathsize 14px style 10 plus log space 2 comma 5 end style 

  3. begin mathsize 14px style 5 minus log space 2 comma 8 end style 

  4. begin mathsize 14px style 4 minus log space 2 comma 8 end style 

  5. begin mathsize 14px style 3 minus log space 2 comma 8 end style 

Pembahasan Video:

Pembahasan Soal:

undefined merupakan garam yang terbentuk dari asam kuat dan basa lemah, sehingga garam ini akan terhidrolisis sbagian. pH begin mathsize 14px style Al subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end style dapat ditentukan dengan persamaan berikut:

 

undefined

keterangan

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space konsentrasi space ion space H to the power of plus sign end style

Kw = tetapan kesetimbangan air

Kb = tetapan kesetimbangan basa

[G] = konsentrasi garam

val = valensi garam


Dari soal diketahui:

Error converting from MathML to accessible text.

V = 2 L = 2000 mL

Error converting from MathML to accessible text.

Error converting from MathML to accessible text.

Error converting from MathML to accessible text.

Error converting from MathML to accessible text.

begin mathsize 14px style Kw space equals space 10 to the power of negative sign 14 end exponent end style

Sehingga

begin mathsize 14px style Mr space Al subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 space equals space left parenthesis 2 cross times A subscript r Al right parenthesis space plus space left parenthesis 3 cross times A subscript r S right parenthesis space plus left parenthesis 12 cross times A subscript r O right parenthesis end style

Error converting from MathML to accessible text.

Error converting from MathML to accessible text.

Tentukan begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style terlebih dahulu, dengan cara berikut:

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space square root of Kw over Kb point open square brackets G close square brackets point val end root end style

Error converting from MathML to accessible text.

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 comma 5.10 to the power of negative sign 9 end exponent end fraction. fraction numerator 6 comma 84 over denominator 342 end fraction 1000 over 2000.2 end root end style 

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 comma 5.10 to the power of negative sign 9 end exponent end fraction. fraction numerator 6 comma 84 over denominator 342 end fraction fraction numerator 1 up diagonal strike 000 over denominator 2 up diagonal strike 000 end fraction.2 end root end style

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 comma 5.10 to the power of negative sign 9 end exponent end fraction.0 comma 02 end root end style

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space square root of fraction numerator 2.10 to the power of negative sign 16 end exponent over denominator 2 comma 5.10 to the power of negative sign 9 end exponent end fraction end root end style

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space square root of 8.10 to the power of negative sign 8 end exponent end root end style

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals space 2 comma 8.10 to the power of negative sign 4 end exponent end style

Lalu tentukan nilai pH dengan cara berikut:

size 14px pH size 14px space size 14px equals size 14px space size 14px minus sign size 14px log size 14px space begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style

begin mathsize 14px style pH space equals space minus sign log space left square bracket 2 comma 8.10 to the power of negative sign 4 end exponent right square bracket end style

begin mathsize 14px style pH space equals space 4 minus sign log space 2 comma 8 end style


Jadi, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Hidayati

Mahasiswa/Alumni Universitas Indonesia

Terakhir diupdate 30 Agustus 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

Hitunglah pH larutan  0,4 M, diketahui .

Pembahasan Soal:

Larutan begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end style merupakan garam yang berasal dari basa lemah dan asam kuat sehingga garam tersebut bersifat asam. pH larutan tersebut sebagai 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 a cross times open square brackets N H subscript 4 to the power of plus sign close square brackets cross times 2 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 2 cross times 10 to the power of negative sign 5 end exponent end fraction cross times left parenthesis 4 cross times 10 to the power of negative sign 1 end exponent right parenthesis cross times 2 end root end cell row blank equals cell square root of 4 cross times 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 end exponent 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 blank equals cell negative sign log left parenthesis 2 cross times 10 to the power of negative sign 5 end exponent right parenthesis end cell row blank equals cell 5 minus sign log space 2 end cell end table end style
 

Maka, Hitunglah pH larutan begin mathsize 14px style open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript end style 0,4 M adalah 5 - log 2.

Roboguru

Sebanyak 100,7 gram dilarutkan dalam air hingga volume larutan 500 mL. Hitunglah tetapan hidrolisis garam dan pH larutan tersebut!

Pembahasan Soal:

Digunakan harga begin mathsize 14px style K subscript b space N H subscript 4 O H equals 10 to the power of negative sign 5 end exponent end style

Menghitung tetapan hidrolisis begin mathsize 14px style open parentheses K subscript h close parentheses end style


begin mathsize 14px style K subscript h equals K subscript w over K subscript b K subscript h equals 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent K subscript h equals 10 to the power of negative sign 9 end exponent end style


Menghitung mol begin mathsize 14px style N H subscript 4 Cl end style


begin mathsize 14px style mol space N H subscript 4 Cl equals massa over M subscript r mol space N H subscript 4 Cl equals fraction numerator 100 comma 7 over denominator 53 comma 3 end fraction mol space N H subscript 4 Cl equals 1 comma 88 space mol end style


Menghitung begin mathsize 14px style open square brackets N H subscript 4 Cl close square brackets end style


begin mathsize 14px style open square brackets N H subscript 4 Cl close square brackets equals mol over volum open square brackets N H subscript 4 Cl close square brackets equals fraction numerator 1 comma 88 space mol over denominator 0 comma 5 space L end fraction open square brackets N H subscript 4 Cl close square brackets equals 0 comma 94 space M end style


Menghitung pH


begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K subscript 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 9 end exponent cross times 0 comma 94 end root open square brackets H to the power of plus sign close square brackets equals 3 comma 07 cross times 10 to the power of negative sign 5 end exponent 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 open parentheses 3 comma 07 cross times 10 to the power of negative sign 5 end exponent close parentheses pH equals 4 comma 5 end style


Jadi, pH larutan tersebut adalah 4,5.space

Roboguru

Hitunglah pH larutan  0,01 M !

Pembahasan Soal:

Larutan N H subscript 4 Cl termasuk larutan garam yang mengalami hidrolisis parsial dan bersifat asam dengan reaksi sebagai berikut.


N H subscript 4 Cl space rightwards arrow space N H subscript 4 to the power of plus space plus space Cl to the power of minus sign N H subscript 4 to the power of plus space plus space H subscript 2 O space rightwards harpoon over leftwards harpoon space N H subscript 4 O H space plus space H to the power of plus sign Cl to the power of minus sign space plus space H subscript 2 O space rightwards arrow space tidak space bereaksi


Dengan demikian, pH larutan N H subscript 4 Cl 0,01 M left parenthesis Kb space N H subscript 4 O H equals 2 cross times 10 to the power of negative sign 5 end exponent right parenthesis yaitu :


table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript N H subscript 4 to the power of plus end subscript end cell equals cell M subscript N H subscript 4 Cl end subscript equals 0 comma 01 space M end cell row blank equals cell 10 to the power of negative sign 2 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 Kw over Kb cross times M subscript N H subscript 4 to the power of plus end subscript 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 2 cross times 10 to the power of negative sign 5 end exponent end fraction cross times 10 to the power of negative sign 2 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 7 end exponent end root end cell row blank equals cell square root of 5 cross times 10 to the power of negative sign 8 end exponent end root end cell row blank equals cell 2 comma 24 cross times 10 to the power of negative sign 4 end exponent 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 blank equals cell negative sign log left parenthesis 2 comma 24 cross times 10 to the power of negative sign 4 end exponent right parenthesis end cell row blank equals cell 4 minus sign log space 2 comma 24 end cell end table


Jadi, pH larutan N H subscript 4 Cl tersebut adalah 4 - log 2,24.space 

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