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Harga pH suatu larutan basa lemah LOH 0,1 M adalah 9, maka harga  dari basa tersebut adalah ....

Pertanyaan

Harga pH suatu larutan basa lemah LOH 0,1 M adalah 9, maka harga K subscript b dari basa tersebut adalah ....space 

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell 14 minus sign pH end cell row blank equals cell 14 minus sign 9 end cell row blank equals 5 end table 


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 5 equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets 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 5 end exponent end cell end table 
 

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 b cross times M end root end cell row cell 10 to the power of negative sign 5 end exponent space end cell equals cell square root of K subscript b cross times 10 to the power of negative sign 1 end exponent end root end cell row cell left parenthesis 10 to the power of negative sign 5 end exponent right parenthesis squared end cell equals cell K subscript b end subscript cross times 10 to the power of negative sign 1 end exponent end cell row cell K subscript b end cell equals cell 10 to the power of negative sign 10 end exponent over 10 to the power of negative sign 1 end exponent end cell row cell K subscript b end cell equals cell 10 to the power of negative sign 9 end exponent end cell end table 


Jadi, harga Kb dari basa tersebut adalah bottom enclose bold 10 to the power of bold minus sign bold 9 end exponent bold. end enclose 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Agnia

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 23 September 2021

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Gas  dialirkan ke dalam 800 mL air pada STP. Jika konsentrasi ion  dalam larutan yang terjadi sebesar  M, tentukan volume gas  yang telah dialirkan!

Pembahasan Soal:

  • Menentukan mol undefined yang dialirkan

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 b cross times M subscript b end root end cell row cell 3 cross times 10 to the power of negative sign 3 end exponent end cell equals cell square root of 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times M subscript b end root end cell row cell M subscript b end cell equals cell space fraction numerator 9 cross times 10 to the power of negative sign 6 end exponent over denominator 1 comma 8 cross times 10 to the power of negative sign 5 end exponent end fraction end cell row cell M subscript b end cell equals cell 0 comma 5 space M end cell row M equals cell n over V end cell row n equals cell M cross times V open parentheses L close parentheses end cell row blank equals cell 0 comma 5 space M cross times 0 comma 8 space L end cell row blank equals cell 0 comma 4 space mol end cell end table end style

  • Menentukan volume begin mathsize 14px style N H subscript 3 end style yang dialirkan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell fraction numerator V over denominator 22 comma 4 space L end fraction end cell row V equals cell 0 comma 4 cross times 22 comma 4 end cell row blank equals cell 8 comma 96 space L end cell end table end style 


Jadi, volume basa yang dialirkan sebesar 8,96 L.undefined 

0

Roboguru

Massa serbuk  yang harus dilarutkan untuk membuat 200 mL larutan basa dengan pH = 8 + log 9 sebanyak ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 8 plus log space 9 end cell row pOH equals cell 14 minus sign pH end cell row blank equals cell 14 minus sign left parenthesis 8 plus log space 9 right parenthesis end cell row blank equals cell 6 minus sign log space 9 end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell negative sign antilog space pOH end cell row blank equals cell negative sign antilog space 6 minus sign log space 9 end cell row blank equals cell 9 cross times 10 to the power of negative sign 6 end exponent end cell end table end style 

Tentukan M nya,

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 Kb middle dot M end root end cell row cell 9 cross times 10 to the power of negative sign 6 end exponent end cell equals cell square root of 9 cross times 10 to the power of negative sign 10 end exponent middle dot M end root end cell row cell open parentheses 9 cross times 10 to the power of negative sign 6 end exponent close parentheses squared end cell equals cell open parentheses square root of 9 cross times 10 to the power of negative sign 10 end exponent middle dot M end root close parentheses squared end cell row cell 81 cross times 10 to the power of negative sign 12 end exponent end cell equals cell 9 cross times 10 to the power of negative sign 10 end exponent middle dot M end cell row M equals cell fraction numerator 81 cross times 10 to the power of negative sign 12 end exponent over denominator 9 cross times 10 to the power of negative sign 10 end exponent end fraction end cell row blank equals cell 9 cross times 10 to the power of negative sign 2 end exponent end cell end table end style

Tentukan massanya,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row M equals cell massa over Mr cross times 1000 over pelarut end cell row cell 9 cross times 10 to the power of negative sign 2 end exponent end cell equals cell massa over 107 cross times 1000 over 200 end cell row cell 1000 space massa end cell equals cell 9 cross times 10 to the power of negative sign 2 end exponent space cross times space 107 space cross times space 200 end cell row massa equals cell 1 comma 926 space gram end cell end table end style

Jadi, jawaban yang benar adalah A

0

Roboguru

Larutan  dengan konsentrasi 0,2 M memilki pH sebesar 11+log 2. Jika  dan maka tentukanlah: a. Nilai  larutan tersebut b. Persentase banyaknya  yang terionisasi (derajad ionisasinya)

Pembahasan Soal:

Larutan begin mathsize 14px style N H subscript 4 O H end style merupakan larutan basa lemah. Konsentrasi begin mathsize 14px style O H to the power of minus sign end style pada basa lemah, dipengaruhi oleh konstanta disosiasi basa (Kb). Untuk menentukan Kb, dapat menggunakan rumus hubungan antara Kb dan  begin mathsize 14px style O H to the power of minus sign end style. Namun sebelumnya, perlu dihitung konsentrasi begin mathsize 14px style O H to the power of minus sign end style dalam begin mathsize 14px style N H subscript 4 O H end style.

Penentuan konsentrasi begin mathsize 14px style O H to the power of minus sign end style dalam begin mathsize 14px style N H subscript 4 O H 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 pOH equals cell 14 minus sign pH end cell row blank equals cell 14 minus sign left parenthesis 11 plus log 2 right parenthesis end cell row blank equals cell 3 minus sign log 2 end cell row blank blank blank row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row cell 3 minus sign log 2 end cell equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 3 end exponent end cell end table end style 
 

a. Penentuan nilai Kb


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 b cross times M subscript N H subscript 4 O H end subscript end root end cell row cell K subscript b end cell equals cell open square brackets O H to the power of minus sign close square brackets squared over M subscript blank subscript N H subscript 4 O H end subscript end subscript end cell row blank equals cell fraction numerator left parenthesis 2 cross times 10 to the power of negative sign 3 end exponent right parenthesis squared over denominator 0 comma 2 space mol forward slash L end fraction end cell row blank equals cell fraction numerator 4 cross times 10 to the power of negative sign 6 end exponent over denominator 2 cross times 10 to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell end table end style 
 

Jadi, nilai Kb begin mathsize 14px style N H subscript 4 O H end style sebesar begin mathsize 14px style 2 cross times 10 to the power of negative sign 5 end exponent end style

b. Derajad ionisasi begin mathsize 14px style N H subscript 4 O H end style (alpha


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row alpha equals cell square root of K subscript b over M subscript N H subscript 4 O H end subscript end root end cell row blank equals cell square root of fraction numerator 2 cross times 10 to the power of negative sign 5 end exponent over denominator 0 comma 2 end fraction end root end cell row blank equals cell square root of fraction numerator 2 cross times 10 to the power of negative sign 5 end exponent over denominator 2 cross times 10 to the power of negative sign 1 end exponent end fraction end root end cell row blank equals cell square root of 1 cross times 10 to the power of negative sign 4 end exponent end root end cell row blank equals cell 1 cross times 10 to the power of negative sign 2 end exponent end cell row blank equals cell 0 comma 01 end cell end table end style 


Dari perhitungan di atas, diperoleh nilai derajad ioniasi (alpha) dari begin mathsize 14px style N H subscript 4 O H end style sebesar 0,01 atau 1%.space 


 

0

Roboguru

Reaksi kesetimbangan piridin () dalam air adalah .  Jika larutan   0,4 M dalam air memiliki pH = 11,5, maka nilai  piridin adalah ....

Pembahasan Soal:

Berdasarkan reaksi kesetimbangan piridin dalam air, dapat disimpulkan bahwa senyawa tersebut tergolong sebagai basa lemah. Basa lemah memiliki nilai tetapan kesetimbangan ionisasi basa yang dilambangkan dengan begin mathsize 14px style K subscript b end style. Harga pH larutan basa lemah dapat dihitung jika nilai undefined-nya diketahui menggunakan rumus-rumus 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 space end cell equals cell space square root of K subscript b cross times italic M subscript b end root end cell row cell pOH space end cell equals cell space minus sign log open square brackets O H to the power of minus sign close square brackets end cell row cell pH space end cell equals cell space 14 minus sign pOH end cell end table end style


Pada soal tersebut, ingin diketahui tetapan kesetimbangan basa dari piridin. Hal tersebut dapat diketahui dengan menggunakan informasi nilai pH dan konsentrasi larutan piridin.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH space end cell equals cell space 11 comma 5 end cell row cell pOH space end cell equals cell space 14 minus sign 11 comma 5 end cell row blank equals cell space 2 comma 5 end cell row cell open square brackets O H to the power of minus sign close square brackets space end cell equals cell space 10 to the power of negative sign 2 comma 5 end exponent end cell row cell 10 to the power of negative sign 2 comma 5 end exponent space end cell equals cell space square root of K subscript b cross times italic M subscript b end root end cell row cell left parenthesis 10 to the power of negative sign 2 comma 5 end exponent right parenthesis squared space end cell equals cell space open parentheses square root of K subscript b cross times 4 cross times 10 to the power of negative sign 1 end exponent end root close parentheses squared end cell row cell 10 to the power of negative sign 5 end exponent space end cell equals cell space K subscript b cross times 4 cross times 10 to the power of negative sign 1 end exponent end cell row cell K subscript b italic space end cell equals cell space 1 fourth cross times 10 to the power of negative sign 4 end exponent end cell row blank equals cell space 2 comma 5 cross times 10 to the power of negative sign 5 end exponent end cell end table end style


Jadi, jawaban yang benar adalah B.undefined 

0

Roboguru

Larutan dimetilamin () akan mengalami perubahan pH dari 10 menjadi 9 jika diencerkan ....

Pembahasan Soal:

Diketahui:

Larutan dimetilamin (begin mathsize 14px style K subscript italic b space equals space 5 cross times 10 to the power of negative sign 4 end exponent end style) akan mengalami perubahan pH dari 10 menjadi 9space

Ditanya:

Berapa kali larutan dimetilamin diencerkan agar pH berubah dari 10 menjadi 9?

Jawab:

1. Menghitung konsentrasi larutan saat pH=10

Untuk pH 10:


begin mathsize 14px style pH equals 10 pOH equals 4 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 open square brackets O H to the power of minus sign close square brackets equals square root of K subscript b cross times M subscript b end root 10 to the power of negative sign 4 end exponent equals square root of 5 cross times 10 to the power of negative sign 4 end exponent cross times M subscript b end root open parentheses 10 to the power of negative sign 4 end exponent close parentheses squared equals open parentheses square root of 5 cross times 10 to the power of negative sign 4 end exponent cross times M subscript b end root close parentheses squared 10 to the power of negative sign 8 end exponent equals 5 cross times 10 to the power of negative sign 4 end exponent cross times M subscript b M subscript b equals 2 cross times 10 to the power of negative sign 5 end exponent space M end style  

 

2. Menghitung konsentrasi larutan saat pH=9
Untuk pH 9: 


Error converting from MathML to accessible text. 

 

3. Menghitung pengenceran yang harus dilakukan dari pH 10 mejadi 9
Pengenceran:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 1 cross times V subscript 1 end cell equals cell M subscript 2 cross times V subscript 2 end cell row cell 2 cross times 10 to the power of negative sign 5 end exponent cross times V subscript 1 end cell equals cell 2 cross times 10 to the power of negative sign 7 end exponent cross times V subscript 2 end cell row cell fraction numerator 2 cross times 10 to the power of negative sign 5 end exponent over denominator 2 cross times 10 to the power of negative sign 7 end exponent end fraction end cell equals cell V subscript 2 end cell row cell space space space 100 V subscript 1 end cell equals cell V subscript 2 end cell end table end style  

Jadi untuk mengubah pH larutan dimetilamin dari pH 10 menjadi 9 adalah dengan pengenceran 100 kali volum semula.


Jadi, jawaban yang tepat adalah D.space space

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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