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Campuran  dan  padat yang massanya 4,8 gram dapat menetralkan 100 mL larutan  1 M. Berapa gram massa  dan  dalam campuran tersebut?

Pertanyaan

Campuran Na O H dan K O H padat yang massanya 4,8 gram dapat menetralkan 100 mL larutan H Cl 1 M. Berapa gram massa Na O H dan K O H dalam campuran tersebut?

Pembahasan Soal:

Na O H yields Na to the power of plus sign and O H to the power of minus sign K O H yields K to the power of plus sign and O H to the power of minus sign H Cl yields H to the power of plus sign and Cl to the power of minus sign 


Pada proses penetralan (netralisasi) berlaku :


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H to the power of plus sign end cell equals cell mol space O H to the power of minus sign end cell row cell a cross times M space H Cl cross times V space H Cl end cell equals cell b cross times mol space O H to the power of minus sign subscript total end cell row cell 1 cross times 1 space M cross times 0 comma 1 space L end cell equals cell 1 cross times mol space O H to the power of minus sign subscript total end cell row cell mol space O H to the power of minus sign subscript total end cell equals cell 0 comma 1 space mol end cell end table 


Maka, nilai mol total untuk O H to the power of minus sign (dari campuran K O H dan Na O H) adalah 0,1 mol.
 

mol space O H to the power of minus sign subscript total equals n space O H to the power of minus sign space K O H and n space O H to the power of minus sign space Na O H 


Jika kita anggap massa K O H double bond x , maka massa Na O H equals 4 comma 8 minus sign x 


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na O H end cell equals cell fraction numerator massa space Na O H over denominator Mr end fraction end cell row blank equals cell fraction numerator 4 comma 8 minus sign x over denominator 40 end fraction end cell end table 


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space K O H end cell equals cell fraction numerator massa space K O H over denominator Mr end fraction end cell row blank equals cell x over 56 end cell end table 


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space O H to the power of minus sign subscript total end cell equals cell mol space K O H and mol space Na O H end cell row cell 0 comma 1 end cell equals cell x over 56 plus fraction numerator 4 comma 8 minus sign x over denominator 40 end fraction end cell row cell 0 comma 1 end cell equals cell fraction numerator 40 x plus 268.8 minus sign 56 x over denominator 2240 end fraction end cell row 224 equals cell 268 comma 8 minus sign 16 x end cell row cell 16 x end cell equals cell 44 comma 8 end cell row x equals cell 2 comma 8 end cell end table 


Massa KOH = 2,8 gram

Massa NaOH = 4,8 - 2,8 = 2,0 gram

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 05 Mei 2021

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