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Sebanyak 1,92 gram  dicampurkan ke dalam 100 mL la...

Sebanyak 1,92 gram begin mathsize 14px style C subscript 2 H subscript 5 COOX end style dicampurkan ke dalam 100 mL larutan begin mathsize 14px style C subscript 2 H subscript 5 C O O H end style 2 M (Ka= begin mathsize 14px style 1 cross times 10 to the power of negative sign 5 end exponent end style) sehingga membentuk larutan penyangga dengan pH= 4. Unsur begin mathsize 14px style X end style yang terkandung dalam senyawa tersebut adalah...

 begin mathsize 14px style Ar space C equals 12 space g space mol to the power of negative sign 1 end exponent semicolon space H equals 1 space g space mol to the power of negative sign 1 end exponent semicolon space O equals 16 space g space mol to the power of negative sign 1 end exponent semicolon K equals space 39 space g space mol to the power of negative sign 1 end exponent semicolon space Li equals 3 space g space mol to the power of negative sign 1 end exponent semicolon space Ba equals space 137 space g space mol to the power of negative sign 1 end exponent semicolon Ca equals space 40 space g space mol to the power of negative sign 1 end exponent semicolon space Na equals space 23 space g space mol to the power of negative sign 1 end exponent end style

  1. begin mathsize 14px style K end style

  2. begin mathsize 14px style Li end style

  3. begin mathsize 14px style Ba end style

  4. begin mathsize 14px style Ca end style

  5. begin mathsize 14px style Na end style

Jawaban:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals 4 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 4 end exponent space M end cell end table 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 Ka cross times begin inline style a over g end style end cell row cell 10 to the power of negative sign 4 end exponent end cell equals cell 10 to the power of negative sign 5 end exponent cross times begin inline style fraction numerator M cross times V open parentheses L close parentheses over denominator g end fraction end style end cell row cell 10 to the power of negative sign 4 end exponent end cell equals cell 10 to the power of negative sign 5 end exponent cross times begin inline style fraction numerator 2 cross times 0 comma 1 over denominator g end fraction end style end cell row g equals cell begin inline style fraction numerator 2 cross times 10 to the power of negative sign 6 end exponent over denominator 10 to the power of negative sign 4 end exponent end fraction end style end cell row blank equals cell 2 cross times 10 to the power of negative sign 2 end exponent space mol end cell end table end style


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell begin inline style m over Mr end style end cell row cell 2 cross times 10 to the power of negative sign 2 end exponent end cell equals cell begin inline style fraction numerator 1 comma 92 over denominator Mr end fraction end style end cell row Mr equals 96 end table end style


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Mr space C subscript 2 H subscript 5 COOX end cell equals cell 3 left parenthesis 12 right parenthesis plus 5 left parenthesis 1 right parenthesis plus 2 left parenthesis 16 right parenthesis plus X end cell row 96 equals cell 36 plus 5 plus 32 plus X end cell row X equals 23 end table end style

Unsur yang memiliki Ar= 23 adalah unsur begin mathsize 14px style Na end style.

Jadi, jawaban yang benar adalah E.

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