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pH larutan 0,01 M suatu asam lemah HA adalah 3,5. Konstanta ionisasi asam (Ka​) dari larutan tersebut adalah ....

Pertanyaan

pH larutan 0,01 M suatu asam lemah HA adalah 3,5. Konstanta ionisasi asam open parentheses K subscript italic a close parentheses dari larutan tersebut adalah ....space

  1. 5 cross times 10 to the power of negative sign 8 end exponentspace

  2. 1 cross times 10 to the power of negative sign 7 end exponentspace

  3. 1 cross times 10 to the power of negative sign 5 end exponentspace

  4. 2 cross times 10 to the power of negative sign 3 end exponentspace

  5. 1 cross times 10 to the power of negative sign 2 end exponentspace

Pembahasan Soal:

Asam lemah menurut Arrhenius adalah suatu zat yang apabila dilarutkan dalam air akan terurai sebagian dan menghasilkan ion H to the power of plus sign. Konsentrasi ion H to the power of plus sign yang dihasilkan dapat dihitung dengan persamaan berikut:


open square brackets H to the power of plus sign close square brackets equals square root of K subscript a cross times M subscript a end root


Sedangkan hubungan pH dan konsentrasi H to the power of plus sign dituliskan sebagai berikut:


pH equals minus sign log space open square brackets H to the power of plus sign close square brackets


Maka, untuk mengetahui besarnya K subscript a pada soal ini langkahnya adalah:

Langkah 1. Tentukan konsentrasi H to the power of plus sign dari pH

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 cell open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign pH end exponent end cell row blank equals cell 10 to the power of negative sign 3 comma 5 end exponent end cell end table


Langkah 2. Substitusi nilai open square brackets H to the power of plus sign close square brackets space dan space M subscript a ke dalam persamaan open square brackets H to the power of plus sign close square brackets equals square root of K subscript a cross times M subscript a end root


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 subscript a end root end cell row cell 10 to the power of negative sign 3 comma 5 end exponent end cell equals cell square root of K subscript a cross times 10 to the power of negative sign 2 end exponent end root end cell row cell left parenthesis 10 to the power of negative sign 3 comma 5 end exponent right parenthesis squared end cell equals cell left parenthesis square root of K subscript a cross times 10 to the power of negative sign 2 end exponent end root right parenthesis squared space end cell row cell 10 to the power of negative sign 7 end exponent end cell equals cell K subscript a cross times 10 to the power of negative sign 2 end exponent end cell row cell K subscript a end cell equals cell 10 to the power of negative sign 7 end exponent over 10 to the power of negative sign 2 end exponent end cell row blank equals cell 10 to the power of negative sign 5 end exponent end cell row blank blank blank end table


Berdasarkan perhitungan tersebut, maka nilai K subscript a equals 1 cross times 10 to the power of negative sign 5 end exponentspace.

Jadi, jawaban yang benar adalah C.spacespace

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 15 Oktober 2021

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Pembahasan Soal:

Derajat ionisasi asam merupakan perbandingan jumlah yang terionisasi dengan molekul zat mula - mula.

Konsentrasi open square brackets H to the power of plus sign close square brackets:

space space space pH equals minus sign log open square brackets H to the power of plus sign close square brackets space space space space space 3 equals minus sign log open square brackets H to the power of plus sign close square brackets open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 3 end exponent space M 

Derajat ionisasi asam lemah tersebut yaitu sebagai berikut :

alpha equals fraction numerator konsentrasi space ion space H to the power of plus sign over denominator konsentrasi space asam space lemah end fraction cross times 100 percent sign space space equals fraction numerator 10 to the power of negative sign 3 end exponent over denominator 0 comma 01 end fraction cross times 100 percent sign space space equals 10 to the power of negative sign 1 end exponent cross times 100 percent sign space space equals 10 percent sign 

Jadi, derajat ionisasi asam lemah tersebut adalah 10%.

0

Roboguru

Jika pH larutan 0,01 M suatu asam lemah HA adalah 3,5. maka tetapan asam (Ka​) adalah...

Pembahasan Soal:

Salah satu ciri-ciri asam lemah yaitu memiliki tetapan ionisasi asam begin mathsize 14px style open parentheses K subscript a close parentheses end style. Untuk mengetahui pH senyawa asam digunakan rumus:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets H to the power of plus sign close square brackets end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of K subscript a cross times M subscript a end root end cell end table 

1. Menghitung nilai [H+]

pH3,5[H+]===log[H+]log[H+]1×103,5M  

2. Menghitung nilai Ka

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 subscript a end root end cell row cell 1 cross times 10 to the power of negative sign 3 comma 5 end exponent end cell equals cell square root of K subscript a cross times 0 comma 01 space end root end cell row cell left parenthesis 1 cross times 10 to the power of negative sign 3 comma 5 end exponent right parenthesis squared end cell equals cell left parenthesis square root of K subscript a cross times left parenthesis 1 cross times 10 to the power of negative sign 2 end exponent right parenthesis space end root right parenthesis squared end cell row cell 1 cross times 10 to the power of negative sign 7 end exponent end cell equals cell K subscript a cross times left parenthesis 1 cross times 10 to the power of negative sign 2 end exponent space right parenthesis end cell row cell K subscript a end cell equals cell fraction numerator 1 cross times 10 to the power of negative sign 7 end exponent over denominator 1 cross times 10 to the power of negative sign 2 end exponent end fraction space end cell row blank equals cell 1 cross times 10 to the power of negative sign 5 end exponent space M end cell end table  

Ka asam lemah HA adalah 1 cross times 10 to the power of negative sign 5 end exponent 

Jadi, jawaban yang benar adalah C.

0

Roboguru

pH suatu larutan asam lemah, MOH 0,001 M sebesar 5-log 2, maka harga Ka​ asam tersebut adalah ...

Pembahasan Soal:

Tahap 1. Menentukan nilai open square brackets H to the power of plus sign close square brackets

table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 5 minus sign log space 2 end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent end cell end table

 

Tahap 2. Menentukan nilai K subscript a

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 cell 2 cross times 10 to the power of negative sign 5 end exponent end cell equals cell square root of K subscript a cross times 10 to the power of negative sign 3 end exponent end root end cell row cell left parenthesis 2 cross times 10 to the power of negative sign 5 end exponent right parenthesis squared end cell equals cell open parentheses square root of K subscript a cross times 10 to the power of negative sign 3 end exponent end root close parentheses squared end cell row cell 4 cross times 10 to the power of negative sign 10 end exponent end cell equals cell K subscript a cross times 10 to the power of negative sign 3 end exponent end cell row cell K subscript a end cell equals cell fraction numerator 4 cross times 10 to the power of negative sign 10 end exponent over denominator 10 to the power of negative sign 3 end exponent end fraction end cell row cell K subscript a end cell equals cell 4 cross times 10 to the power of negative sign 7 end exponent end cell row blank blank blank end table

Jadi, jawaban tidak ada di dalam opsi jawaban.

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Roboguru

Asam HX 0,1 M mengalami ionisasi 1%. Nilai tetapan ionisasi HX adalah ....

Pembahasan Soal:

Diketahui:
M HX = 0,1 M
begin mathsize 14px style alpha equals 1 percent sign equals 0 comma 01 equals 1 cross times 10 to the power of negative sign 2 end exponent end style

Ditanya: Ka HX=?

begin mathsize 14px style alpha equals square root of Ka over Ma end root open parentheses alpha close parentheses squared equals left parenthesis square root of Ka over Ma end root right parenthesis squared space alpha squared equals Ka over Ma Ka equals alpha squared cross times Ma Ka equals left parenthesis 1 cross times 10 to the power of negative sign 2 end exponent right parenthesis squared cross times 1 cross times 10 to the power of negative sign 1 end exponent Ka equals 1 cross times 10 to the power of negative sign 5 end exponent end style 

Jadi, jawaban yang benar adalah B.

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Roboguru

Larutan CH3​COOH0,05M yang terionisasi dengan derajat ionisasi mempunyai pH sebesar ....

Pembahasan Soal:

Diketahui :

  • M  Error converting from MathML to accessible text.
  • begin mathsize 14px style open parentheses italic alpha close parentheses equals 0 , 02 end style

Ditanya : pH?

Jawab :

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 space end cell equals cell space alpha space cross times space M subscript C H subscript 3 C O O H end subscript end cell row blank equals cell space 0 comma 02 space cross times space 0 comma 05 space M end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell space 1 cross times 10 to the power of negative sign 3 end exponent end cell row blank blank blank row cell pH space end cell equals cell space minus sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell space minus sign log space 1 cross times 10 to the power of negative sign 3 end exponent end cell row blank equals cell space 3 space minus sign space log space 1 end cell row blank equals cell space 3 space minus sign space 0 end cell row pH equals 3 end table end style 

Jadi pH larutan Error converting from MathML to accessible text. adalah 3.

Jadi, jawaban yang benar adalah C.

0

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