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Untuk menetapkan kadar iodin dalam zat yang mengandung Na I O subscript 3, diambil 2 gram zat, kemudian dilarutkan ke dalam 100 mL air. Sebanyak 20 mL larutan tersebut diasamkan dan direaksikan dengan larutan KI berlebih. I subscript 2 yang dibebaskan tepat habis bereaksi dengan 12 mL larutan Na subscript 2 S subscript 2 O subscript 3 0,1 M. Jika A subscript r Na=23, O=16, dan I=127, tentukan kadar iodin dalam zat tersebut.space  

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

kadar iodin dalam sampel adalah 6,35%.space 

Pembahasan

A. Reaksi-reaksi yang terjadi pada percobaan adalah sebagai berikut:

  1. Reaksi penguraian Na I O subscript bold 3 dalam air
    Na I O subscript 3 left parenthesis italic a italic q right parenthesis yields Na to the power of plus sign left parenthesis italic a italic q right parenthesis plus I O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis 
  2. Reaksi oksihalogen I O subscript bold 3 to the power of bold minus sign dengan halida I to the power of bold minus sign  
    I O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis plus 5 I to the power of minus sign left parenthesis italic a italic q right parenthesis plus 6 H to the power of plus sign left parenthesis italic a italic q right parenthesis yields 3 I subscript 2 left parenthesis italic a italic q right parenthesis plus 3 H subscript 2 O left parenthesis italic a italic q right parenthesis 
  3. Reaksi I subscript bold 2 dengan Na subscript bold 2 S subscript bold 2 O subscript bold 3 
    I subscript 2 left parenthesis italic a italic q right parenthesis plus 2 S subscript 2 O subscript 3 to the power of minus sign left parenthesis italic a italic q right parenthesis yields 2 I to the power of minus sign left parenthesis italic a italic q right parenthesis plus S subscript 4 O subscript 6 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis 

 

B. Menentukan mol Na subscript bold 2 S subscript bold 2 O subscript bold 3 yang terpakai dalam reaksi (3)

mol space Na subscript 2 S subscript 2 O subscript 3 double bond V space Na subscript 2 S subscript 2 O subscript 3 cross times M space Na subscript 2 S subscript 2 O subscript 3 mol space Na subscript 2 S subscript 2 O subscript 3 equals 12 space mL cross times 0 comma 1 space M mol space Na subscript 2 S subscript 2 O subscript 3 equals 1 comma 2 cross times 10 to the power of negative sign 3 end exponent space mol  
 

C. Menentukan mol I subscript bold 2 dalam reaksi (3)

mol space I subscript 2 equals fraction numerator koefisien space I subscript 2 over denominator koefisien space Na subscript 2 S subscript 2 O subscript 3 end fraction cross times mol space Na subscript 2 S subscript 2 O subscript 3 mol space I subscript 2 equals 1 half cross times 1 comma 2 cross times 10 to the power of negative sign 3 end exponent mol mol space I subscript 2 equals 6 cross times 10 to the power of negative sign 4 end exponent space mol  
 

D. Menentukan mol I subscript bold 2 dalam reaksi (2)

I subscript 2 pada persamaan (3) merupakan hasil dari reaksi (2). Jadi, jumlah I subscript 2 yang dihasilkan pada persamaan (2) adalah 6 cross times 10 to the power of negative sign 4 end exponent space mmol.
 

E. Menentukan mol I O subscript bold 3 to the power of bold minus sign saat dititrasi dalam reaksi (2) (volume 20 mL)

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space I O subscript 3 to the power of minus sign end cell equals cell fraction numerator koefisien space I O subscript 3 to the power of minus sign over denominator koefisien space I subscript 2 end fraction cross times mol space I subscript 2 end cell row cell mol space I O subscript 3 to the power of minus sign end cell equals cell 1 third cross times 6 cross times 10 to the power of negative sign 4 end exponent space mol end cell row cell mol space I O subscript 3 to the power of minus sign end cell equals cell 2 cross times 10 to the power of negative sign 4 end exponent space mol end cell end table 
 

F. Menentukan mol I O subscript bold 3 to the power of bold minus sign dalam reaksi (1) (volume 100 mL)

Volume larutan yang dititrasi adalah 20 mL. Jadi jumlah mol I O subscript 3 to the power of minus sign yang ada pada 100 mL larutan adalah:

mol space I O subscript 3 to the power of minus sign equals fraction numerator V space larutan space awal over denominator V space titrasi end fraction cross times mol space I O subscript 3 to the power of minus sign space saat space titrasi mol space I O subscript 3 to the power of minus sign equals fraction numerator 100 space mL over denominator 20 space mL end fraction cross times 2 cross times 10 to the power of negative sign 4 end exponent space M mol space I O subscript 3 to the power of minus sign equals 1 cross times 10 to the power of negative sign 3 end exponent space M  
 

G. Menentukan mol Na I O subscript bold 3 dalam reaksi (1) 

mol space Na I O subscript 3 equals fraction numerator koefisien space Na I O subscript 3 space over denominator koefisien space I O subscript 3 to the power of minus sign end fraction cross times mol space I O subscript 3 to the power of minus sign mol space Na I O subscript 3 equals 1 over 1 cross times 1 cross times 10 to the power of negative sign 3 end exponent space mol mol space Na I O subscript 3 equals 1 cross times 10 to the power of negative sign 3 end exponent space mol  
 

H. Menentukan massa  Na I O subscript bold 3 pada reaksi (1)

mol space Na I O subscript 3 equals fraction numerator massa space Na I O subscript 3 over denominator M subscript r space Na I O subscript 3 end fraction massa space Na I O subscript 3 double bond mol space Na I O subscript 3 cross times M subscript r space Na I O subscript 3 massa space Na I O subscript 3 equals 1 cross times 10 to the power of negative sign 3 end exponent space mol cross times 198 space gram space mol to the power of negative sign 1 end exponent massa space Na I O subscript 3 equals 0 comma 198 space gram  
 

I. Menentukan Massa I dalam Na I O subscript bold 3

massa space I space dalam space Na I O subscript 3 equals fraction numerator n cross times Ar space I space over denominator Mr space Na I O subscript 3 space end fraction cross times massa space Na I O subscript 3 massa space I space dalam space Na I O subscript 3 equals fraction numerator 1 cross times 127 over denominator 198 end fraction cross times 0 comma 198 massa space I space dalam space Na I O subscript 3 equals 0 comma 127 space gram   

I. Menentukan kadar unsur I dalam zat
Kadar space I space dalam space zat equals fraction numerator Massa space I space over denominator Massa space zat space end fraction cross times 100 percent sign Kadar space I space dalam space zat equals fraction numerator 0 comma 127 over denominator 2 end fraction cross times 100 percent sign Kadar space I space dalam space zat equals 6 comma 35 percent sign 

Jadi kadar iodin dalam sampel adalah 6,35%.space 

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