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Berapa gram (Mr = 53,5) yang harus ditambahkan ke dalam 200 ml larutan 0,2 M agar diperoleh larutan dengan pH = 9 ?

Berapa gram begin mathsize 14px style N H subscript 4 C I space end style(Mr = 53,5) yang harus ditambahkan ke dalam 200 ml larutan begin mathsize 14px style N H subscript 4 O H end style  0,2 M agar diperoleh larutan dengan pH = 9  begin mathsize 14px style left parenthesis Kb equals 10 to the power of negative sign 5 end exponent right parenthesis end style ?

Jawaban:

Rumus larutan penyangga basa:

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 Kb space x space fraction numerator mol space basa space lemah over denominator mol space asam space konjugasi space atau space mol space garam end fraction end cell row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row pH equals cell 14 minus sign pOH end cell end table end style 

Berikut adalah penentuan massa garam begin mathsize 14px style N H subscript 4 C I space end styleyang harus ditambahkan ke dalam larutan basa lemah begin mathsize 14px style N H subscript 4 O H end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space N H subscript 4 O H end cell equals cell 0 comma 2 space M space x space 0 comma 2 space L equals 0 comma 04 space mol end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 9 equals cell 14 minus sign pOH end cell row pOH equals 6 end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 6 equals cell negative sign log 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 10 to the power of negative sign 6 end exponent end cell end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space N H subscript 4 Cl end cell equals cell fraction numerator Kb space x space mol space N N H subscript 4 O H over denominator open square brackets O H to the power of minus sign close square brackets end fraction end cell row blank equals cell fraction numerator 10 to the power of negative sign 5 end exponent space x space left parenthesis 4 space x space 10 to the power of negative sign 2 end exponent space mol right parenthesis over denominator 10 to the power of negative sign 6 end exponent end fraction end cell row blank equals cell 0 comma 4 space mol end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell massa space N H subscript 4 Cl end cell equals cell 0 comma 4 space mol space x space 53 comma 5 space g forward slash mol end cell row blank equals cell 21 comma 4 space g end cell end table end style 

Jadi, massa  garam yang harus ditambahkan ke dalam larutan basa lemah adalah 21,4 g.

 

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