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Larutan H 2 ​ SO 4 ​ 0,02 M akan memberikan warna yang sama dengan larutan HA 0,2 M jika ditetesi indikator yang sama. Tetapan ionisasi ( K a ​ ) HA adalah ....

Larutan 0,02 M akan memberikan warna yang sama dengan larutan HA 0,2 M jika ditetesi indikator yang sama. Tetapan ionisasi HA adalah ....

  1. begin mathsize 14px style 1 cross times 10 to the power of negative sign 3 end exponent end style

  2. begin mathsize 14px style 2 cross times 10 to the power of negative sign 3 end exponent end style

  3. begin mathsize 14px style 4 cross times 10 to the power of negative sign 3 end exponent end style

  4. begin mathsize 14px style 8 cross times 10 to the power of negative sign 3 end exponent end style

  5. begin mathsize 14px style 9 cross times 10 to the power of negative sign 3 end exponent end style

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Y. Rochmawatie

Master Teacher

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Pembahasan

Pembahasan
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Tetapan ionisasi asam ( ) menyatakan ukuran kekuatan dari suatu asam secara kuantitatif. Diketahui Ditanya Tetapan ionisasi HA Jawab Nilai a = Jumlah ion = 2 pH larutan pH larutan HA Larutan memberikan warna yang sama dengan larutan HA, maka pH kedua larutan ini sama. Konsentrasi ion HA Tetapan ionisasi HA Jadi, jawaban yang benar adalah D.

Tetapan ionisasi asam (begin mathsize 14px style K subscript a end style) menyatakan ukuran kekuatan dari suatu asam secara kuantitatif.

Diketahui

begin mathsize 14px style M subscript H subscript 2 S O subscript 4 end subscript equals space 0 comma 02 space M M subscript HA equals 0 comma 2 space M end style

Ditanya

Tetapan ionisasi begin mathsize 14px style open parentheses K subscript a close parentheses end style HA

Jawab

  • Nilai begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space H subscript 2 S O subscript 4 end style

            begin mathsize 14px style H subscript 2 S O subscript 4 yields 2 H to the power of plus sign and S O subscript 4 to the power of 2 minus sign end exponent end style

            a = Jumlah ion begin mathsize 14px style open square brackets H to the power of plus sign close square brackets end style = 2

            begin mathsize 14px style open square brackets H to the power of plus sign close square brackets double bond a middle dot M open square brackets H to the power of plus sign close square brackets equals 2 middle dot 0 comma 02 open square brackets H to the power of plus sign close square brackets equals 0 comma 04 open square brackets H to the power of plus sign close square brackets equals 4 middle dot 10 to the power of negative sign 2 end exponent end style 

  • pH larutan begin mathsize 14px style H subscript 2 S O subscript 4 end style

            begin mathsize 14px style 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 4 middle dot 10 to the power of negative sign 2 end exponent pH equals 2 minus sign log space 4 end style 

  • pH larutan HA

            Larutan undefined memberikan warna yang sama dengan larutan HA, maka pH kedua larutan ini sama.

            begin mathsize 14px style pH space H subscript 2 S O subscript 4 space equals space pH space HA space equals space 2 minus sign log space 4 end style

  • Konsentrasi ion undefined HA

            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 cell 2 minus sign log space 4 end cell 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 anti space log space 2 minus sign log space 4 end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 4 middle dot 10 to the power of negative sign 2 end exponent space M end cell end table end style 

  • Tetapan ionisasi HA

          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 middle dot M end root end cell row cell 4 middle dot 10 to the power of negative sign 2 end exponent end cell equals cell square root of K subscript a middle dot 0 comma 2 end root end cell row cell open parentheses 4 middle dot 10 to the power of negative sign 2 end exponent close parentheses squared end cell equals cell left parenthesis square root of K subscript a middle dot 0 comma 2 end root right parenthesis squared space end cell row cell 16 middle dot 10 to the power of negative sign 4 end exponent end cell equals cell K subscript a middle dot 0 comma 2 end cell row cell K subscript a end cell equals cell fraction numerator 16 middle dot 10 to the power of negative sign 4 end exponent over denominator 2 middle dot 10 to the power of negative sign 1 end exponent end fraction end cell row cell K subscript a end cell equals cell 8 middle dot 10 to the power of negative sign 3 end exponent end cell end table end style 

Jadi, jawaban yang benar adalah D.

 

 

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