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Pertanyaan

pH larutan asam etanoat 0,05 M yang terionisasi sebanyak 6% adalah ....

  1. 0,52

  2. 1,00

  3. 1,52

  4. 2,00

  5. 2,52

Y. Rochmawatie

Master Teacher

Jawaban terverifikasi

Pembahasan

Pada soal diketahui beberapa data sebagai berikut:

  • begin mathsize 14px style left square bracket asam space etanoat right square bracket equals open square brackets C H subscript 3 C O O H close square brackets equals 0 comma 05 space M equals 5 cross times 10 to the power of negative sign 2 end exponent space M end style
  • begin mathsize 14px style alpha equals 6 percent sign equals 6 over 100 equals 6 cross times 10 to the power of negative sign 2 end exponent end style 

Data yang ingin diketahui adalah pH larutan dari asam etanoat yang termasuk asam lemah. Sebelum menentukan pH, ada beberapa data yang harus ditentukan terlebih dahulu. Data yang pertama kali ditentukan adalah konstanta asam lemah (begin mathsize 14px style K subscript a end style) melalui derajat ionisasi (begin mathsize 14px style alpha end style).

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row alpha equals cell square root of fraction numerator K subscript a over denominator open square brackets C H subscript 3 C O O H close square brackets end fraction end root end cell row cell alpha squared end cell equals cell fraction numerator K subscript a over denominator open square brackets C H subscript 3 C O O H close square brackets end fraction end cell row cell K subscript a end cell equals cell alpha squared middle dot open square brackets C H subscript 3 C O O H close square brackets end cell row cell K subscript a end cell equals cell left parenthesis 6 cross times 10 to the power of negative sign 2 end exponent right parenthesis squared middle dot 5 cross times 10 to the power of negative sign 2 end exponent end cell row cell K subscript a end cell equals cell 36 cross times 10 to the power of negative sign 4 end exponent middle dot 5 cross times 10 to the power of negative sign 2 end exponent end cell row cell K subscript a end cell equals cell 180 cross times 10 to the power of negative sign 6 end exponent end cell row cell K subscript a end cell equals cell 1 comma 8 cross times 10 to the power of negative sign 4 end exponent end cell row blank blank blank end table end style 

Langkah selanjutnya adalah menentukan nilai begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style.

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 point open square brackets C H subscript 3 C O O H close square brackets end root end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of 1 comma 8 cross times 10 to the power of negative sign 4 end exponent.5 cross times 10 to the power of negative sign 2 end exponent end root end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of 9 cross times 10 to the power of negative sign 6 end exponent end root end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 3 cross times 10 to the power of negative sign 3 end exponent end cell end table end style 

Dari nilai begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style yang telah diketahui, maka pH larutan dapat dihitung.

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 pH equals cell negative sign log space 3 cross times 10 to the power of negative sign 3 end exponent end cell row pH equals cell 3 minus sign log space 3 end cell row pH equals cell 3 minus sign 0 comma 48 end cell row pH equals cell 2 comma 52 end cell end table end style 

Berdasarkan perhitungan yang telah dilakukan, pH larutan asam etanoat adalah 2,52.

Jadi, jawaban yang benar adalah E.space 

 

 

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