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Larutan  merupakan asam lemah bivalen yang terionisasi sesuai reaksi berikut.   Harga pH larutan 0,002 M adalah ....

Pertanyaan

Larutan H subscript 2 C O subscript 3 merupakan asam lemah bivalen yang terionisasi sesuai reaksi berikut.
H subscript 2 C O subscript 3 open parentheses aq close parentheses equilibrium H C O subscript 3 to the power of minus sign open parentheses aq close parentheses and H to the power of plus sign open parentheses aq close parentheses space K subscript a 1 end subscript equals 4 comma 5 cross times 10 to the power of negative sign 7 end exponent H C O subscript 3 to the power of minus sign open parentheses aq close parentheses equilibrium C O subscript 3 to the power of 2 minus sign end exponent open parentheses aq close parentheses and H to the power of plus sign open parentheses aq close parentheses space K subscript a 2 end subscript equals 4 comma 7 cross times 10 to the power of negative sign 11 end exponent 
Harga pH larutan H subscript 2 C O subscript 3 0,002 M adalah ....

  1. 3 - log 6space 

  2. 3 - log 5space 

  3. 4 - log 3space 

  4. 5 - log 3space 

  5. 6 - log 3space 

Pembahasan Soal:

H subscript 2 C O subscript 3 equilibrium H C O subscript 3 to the power of minus sign plus H to the power of plus sign space K subscript a 1 end subscript equals 4 comma 5 cross times 10 to the power of negative sign 7 end exponent bottom enclose H C O subscript 3 to the power of minus sign rightwards harpoon over leftwards harpoon C O subscript 3 to the power of 2 minus sign end exponent plus H to the power of plus sign space K subscript a 2 end subscript equals 4 comma 7 cross times 10 to the power of negative sign 11 end exponent end enclose H subscript 2 C O subscript 3 equilibrium C O subscript 3 to the power of 2 minus sign end exponent plus 2 H to the power of plus sign space K subscript a equals 2 comma 115 cross times 10 to the power of negative sign 17 end exponent K subscript a equals fraction numerator open square brackets H to the power of plus sign close square brackets squared open square brackets C O subscript 3 to the power of 2 minus sign end exponent close square brackets over denominator open square brackets H subscript 2 C O subscript 3 close square brackets end fraction 2 comma 115 cross times 10 to the power of negative sign 17 end exponent cross times open square brackets H subscript 2 C O subscript 3 close square brackets equals open square brackets H to the power of plus sign close square brackets cubed 2 comma 115 cross times 10 to the power of negative sign 17 end exponent cross times 0 comma 002 equals open square brackets H to the power of plus sign close square brackets cubed open square brackets H to the power of plus sign close square brackets cubed equals 4 comma 23 cross times 10 to the power of negative sign 20 end exponent open square brackets H to the power of plus sign close square brackets equals 3 comma 5 cross times 10 to the power of negative sign 7 end exponent pH equals 7 minus sign log space 3 comma 5 pH equals 6 comma 46

Namun penyelesaiannya dapat disederhanakan dengan asumsi bahwa kesetimbangan yang paling berkontribusi menghasilkan H to the power of plus sign adalah kesetimbangan ke-1. Sehingga

open square brackets H to the power of plus sign close square brackets equals square root of K subscript italic a cross times open square brackets H subscript 2 C O subscript 3 close square brackets end root open square brackets H to the power of plus sign close square brackets equals square root of 4 comma 5 cross times 10 to the power of negative sign 7 end exponent cross times 0 comma 002 end root open square brackets H to the power of plus sign close square brackets equals 3 cross times 10 to the power of negative sign 5 end exponent pH equals 5 minus sign log space 3

Dengan demikian maka jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Budi

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 30 April 2021

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Suatu larutan diketahui memiliki nilai pH sebesar 3. Tentukan besar konsentrasi ion  dalam larutan tersebut!

Pembahasan Soal:

Nilai pH menunjukkan konsentrasi ion begin mathsize 14px style H to the power of plus sign end style dalam larutan. Secara matematika hubungan pH dan konsentrasi ion begin mathsize 14px style H to the power of plus sign end style dapat dituliskan sebagai berikut.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jika space open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign n end exponent comma space maka space pH double bond n end cell row cell Jika space open square brackets H to the power of plus sign close square brackets end cell equals cell x point 10 to the power of negative sign n end exponent comma space maka space pH double bond n bond log space x end cell row blank blank cell Sebaliknya comma end cell row cell Jika space pH end cell equals cell n comma space maka space open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign n end exponent end cell end table end style


Dengan demikian, jika suatu larutan memiliki pH = 3, maka konsentrasi ion begin mathsize 14px style H to the power of plus sign end style adalah begin mathsize 14px style 10 to the power of negative sign 3 end exponent end style.space

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begin mathsize 14px style alpha equals square root of Ka over M end root 0 comma 02 equals square root of fraction numerator Ka over denominator 0 comma 05 end fraction end root Ka 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 Ka cross times M end root open square brackets H to the power of plus sign close square brackets equals square root of 2 cross times 10 to the power of negative sign 5 end exponent cross times 5 cross times 10 to the power of negative sign 2 end exponent end root open square brackets H to the power of plus sign close square brackets equals square root of 1 cross times 10 to the power of negative sign 6 end exponent end root open square brackets H to the power of plus sign close square brackets equals 1 cross times 10 to the power of negative sign 3 end exponent pH equals 3 end style

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Pembahasan Soal:

Tentukan Mol (n)

begin mathsize 14px style n space equals space fraction numerator P V over denominator RT end fraction end style

begin mathsize 14px style n space equals space fraction numerator 1 space x space 0 comma 024 over denominator 0 comma 08 space x space 300 end fraction end style 

n space equals space 0 comma 001

Tentukan Molaritas (M)

begin mathsize 14px style M space equals space fraction numerator 0 comma 001 over denominator 0 comma 25 end fraction end style 

begin mathsize 14px style M space equals space 4 space x space 10 to the power of negative sign 3 space end exponent end style

Tentukan pH

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals jumlah space asam space x space M open square brackets H to the power of plus sign close square brackets equals 1 space x space 4 space x space 10 to the power of negative sign 3 end exponent open square brackets H to the power of plus sign close square brackets equals 4 space x space 10 to the power of negative sign 3 end exponent pH equals minus sign space log space open square brackets H to the power of plus sign close square brackets pH equals minus sign space log space open square brackets 4 space x space 10 to the power of negative sign 3 end exponent close square brackets pH equals 3 space minus sign space log space 4 pH equals 3 space minus sign space 0 comma 6 pH equals 2 comma 4 end style

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Pembahasan Soal:

Diketahui:
Sebuah jeruk massanya 80 gram mengandung vitamin C sebesar 56 mg.
Vitamin C yang dihasilkan dilarutkan ke dalam air sampai volume 100 mL. 
undefined asam askorbat = begin mathsize 14px style 1 comma 2 cross times 10 to the power of negative sign 13 end exponent end style.
begin mathsize 14px style C subscript 6 H subscript 8 O subscript 6 space left parenthesis A subscript r space C equals 12 comma space H equals 1 comma space O equals 16 right parenthesis end style.

Ditanya:
pH yang terjadi jiika ke dalam 100 mL larutan tersebut ditambahkan lagi air sebanyak 100 mL,

Jawab:

1. Menghitung konsentrasi asam askorbat dalam 100 mL air

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript 1 end cell equals cell m over M subscript r cross times fraction numerator 1.000 over denominator V end fraction end cell row blank equals cell fraction numerator 0 comma 056 space g over denominator 176 space g space mol to the power of negative sign 1 end exponent end fraction cross times fraction numerator 1.000 over denominator 100 space mL end fraction end cell row blank equals cell 3 comma 18 cross times 10 to the power of negative sign 3 end exponent space M end cell end table end style  

2. Menghitung konsentrasi asam askorbat setelah diencerkan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript 1 cross times M subscript 1 end cell equals cell V subscript 2 cross times M subscript 2 end cell row cell 100 space mL cross times 3 comma 18 cross times 10 to the power of negative sign 3 end exponent space M end cell equals cell left parenthesis 100 plus 100 right parenthesis space mL cross times M subscript 2 end cell row cell M subscript 2 end cell equals cell 1 comma 59 cross times 10 to the power of negative sign 3 end exponent space M end cell end table end style  

3. Menghitung konsentrasi begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style dalam larutan asam askorbat yang telah diencerkan

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 a cross times M end root end cell row blank equals cell square root of 1 comma 2 cross times 10 to the power of negative sign 13 end exponent cross times 1 comma 59 cross times 10 to the power of negative sign 3 end exponent end root end cell row blank equals cell square root of 1 comma 91 cross times 10 to the power of negative sign 16 end exponent end root end cell row blank equals cell 1 comma 38 cross times 10 to the power of negative sign 8 end exponent space M end cell end table end style  

4. Menghitung pH larutan 

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 left parenthesis 1 comma 38 cross times 10 to the power of negative sign 8 end exponent right parenthesis end cell row blank equals cell 8 minus sign log space 1 comma 38 end cell row blank equals cell 7 comma 86 end cell end table end style 


Jadi, pH setelah diencerkan adalah 7,86.

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Tentukan  pada  0,05 M, jika diketahui  adalah asam diprotik yang reaksi peruraiannya berlangsung 2 tahap dimana nilai  , sangat besar dan  !

Pembahasan Soal:

Ka1 sangat besar sehingga dapat diartikan H2SO4 terurai secara sempurna,

Reaksi pertama,

begin mathsize 14px style H subscript 2 S O subscript 4 yields space space H to the power of plus sign space space space space plus space space space space H S O subscript 4 to the power of minus sign 0 comma 05 M space space space space space 0 comma 05 M space space space space space space space 0 comma 05 M space end style 

Reaksi kedua,

Error converting from MathML to accessible text. 

Hitung nilai x,

begin mathsize 14px style space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space K subscript italic a bold 2 end subscript bold equals fraction numerator begin bold style open square brackets H to the power of plus sign close square brackets end style begin bold style left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket end style over denominator begin bold style left square bracket H S O subscript 4 to the power of minus sign right square bracket end style end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space 0 comma 011 equals fraction numerator left parenthesis 0 comma 05 plus x right parenthesis x over denominator 0 comma 05 minus sign x end fraction space space space space space space space 0 comma 00055 minus sign 0 comma 011 x equals 0 comma 05 x plus x squared x squared plus 0 comma 061 x minus sign 0 comma 00055 equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x equals 0 comma 008 end style  

hitung nilai berapa persen nilai x, jika lebih dari 5% maka dimasukan dalam perhitungan

begin mathsize 14px style fraction numerator 0 comma 008 over denominator 0 comma 05 end fraction middle dot 100 percent sign equals 16 percent sign end style 

karena nilai x signifikan, maka

begin mathsize 14px style Total space open square brackets H to the power of plus sign close square brackets equals 0 comma 05 plus 0 comma 008 space space space space space space space space space space space space space space space space space space equals 0 comma 058 space M end style 

sehingga [H+] adalah 0,058 M

 

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