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Sebanyak 5 gram cuplikan yang mengandung NaOH dilarutkan dalam 200 mL akuades. Jika pH larutan tersebut 13 + lo g 5 , kadar NaOH dalam cuplikan adalah .... ( A r ​ : Na = 23 g mol − 1 , H = 1 g mol − 1 , dan O = 16 g mol − 1 )

Sebanyak 5 gram cuplikan yang mengandung NaOH dilarutkan dalam 200 mL akuades. Jika pH larutan tersebut , kadar NaOH dalam cuplikan adalah .... (: Na = , H = , dan O = ) 

  1. 80%undefined 

  2. 70%undefined 

  3. 60%undefined 

  4. 50%undefined 

  5. 40%undefined 

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I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

jawaban yang benar adalah A.

Pembahasan

Pembahasan
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Diketahui: Berikut langkah penyelesaiannya: Mencari Mencari [NaOH] Mencari mol NaOH Mencari massa NaOH Mencari kadar NaOH Jadi, jawaban yang benar adalah A.

Diketahui:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell m space cuplikan end cell equals cell space 5 space gram end cell row cell V space akuades end cell equals cell 200 space mL equals 0 comma 2 space L end cell row cell pH space larutan end cell equals cell 13 plus log space 5 end cell row cell Mr space Na O H end cell equals cell Ar space Na and Ar space O and Ar space H end cell row blank equals cell left parenthesis 23 plus 16 plus 1 right parenthesis g forward slash mol end cell row blank equals cell 40 space g forward slash mol end cell row blank blank blank end table end style  

Berikut langkah penyelesaiannya:

  1. Mencari begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets end style

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell 14 minus sign pH end cell row blank equals cell 14 minus sign left parenthesis 13 plus log space 5 right parenthesis end cell row blank equals cell 1 minus sign log space 5 end cell row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row cell 1 minus sign log space 5 end cell equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 5 cross times 10 to the power of negative sign 1 space end exponent M end cell end table end style 
     
  2. Mencari [NaOH]

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell M cross times valensi space basa end cell row M equals cell fraction numerator 5 cross times 10 to the power of negative sign 1 end exponent M over denominator 1 end fraction end cell row blank equals cell 5 cross times 10 to the power of negative sign 1 end exponent space M end cell end table end style
     
  3. Mencari mol NaOH

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell M cross times V end cell row blank equals cell 5 cross times 10 to the power of negative sign 1 end exponent space M cross times 0 comma 2 space L end cell row blank equals cell 10 to the power of negative sign 1 end exponent space mol end cell end table end style
     
  4. Mencari massa NaOH

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row m equals cell n cross times Mr end cell row blank equals cell 0 comma 1 space mol cross times 40 space g forward slash mol end cell row blank equals cell 4 space g end cell end table end style 
     
  5. Mencari kadar NaOH

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Kadar space Na O H end cell equals cell fraction numerator m thin space Na O H over denominator m space cuplikan end fraction cross times 100 percent sign end cell row blank equals cell fraction numerator 4 space g over denominator 5 space g end fraction cross times 100 percent sign end cell row blank equals cell 80 percent sign end cell end table end style
     

Jadi, jawaban yang benar adalah A.

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