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Suatu campuran penyangga yang terbentuk dari 500 mL larutan HCOOH 0,1 M dan 500 mL larutan HCOONa 0,1 M, ditambah 10 mL larutan yang pH-nya 13. Hitunglah pH sesudah dilakukan penambahan! ( K a ​ HCOOH = 2 × 1 0 − 4 )

Suatu campuran penyangga yang terbentuk dari 500 mL larutan HCOOH 0,1 M dan 500 mL larutan HCOONa 0,1 M, ditambah 10 mL larutan yang pH-nya 13. Hitunglah pH sesudah dilakukan penambahan! (

 

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Y. Rochmawatie

Master Teacher

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Jawaban

pH sesudah dilakukan penambahan adalah 4 - log 1,92.

pH sesudah dilakukan penambahan adalah 4 - log 1,92.

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Diketahui: Berikut penyelesaiannya: Menentukan mol komponen penyangga Menentukan mol larutan dengan pH=13 Mencari pH akhir Penambahan basa akan menambah jumlah mol basa konjugasi dan mengurangi mol asam. Jadi, pH sesudah dilakukan penambahan adalah 4 - log 1,92.

Diketahui:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H C O O H close square brackets end cell equals cell 0 comma 1 space M end cell row cell V space H C O O H end cell equals cell 500 space mL equals 0 comma 5 space L end cell row cell open square brackets H C O O Na close square brackets end cell equals cell 0 comma 1 space M end cell row cell V space H C O O Na end cell equals cell space 500 space mL equals 0 comma 5 space L end cell row cell V space larutan end cell equals cell 10 space mL end cell row cell pH space larutan end cell equals 13 row cell K subscript a space H C O O H end cell equals cell 2 cross times 10 to the power of negative sign 4 end exponent end cell end table end style

Berikut penyelesaiannya:

  1. Menentukan mol komponen penyangga

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n thin space H C O O H end cell equals cell M cross times V end cell row blank equals cell 0 comma 1 cross times 0 comma 5 space L end cell row blank equals cell 5 cross times 10 to the power of negative sign 2 end exponent space mol end cell row blank equals cell 50 space mmol end cell end table end style

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n space H C O O Na end cell equals cell M cross times V end cell row blank equals cell 0 comma 1 space M cross times 0 comma 5 space L end cell row blank equals cell 5 cross times 10 to the power of negative sign 2 end exponent space mol end cell row blank equals cell 50 space mmol end cell end table end style

    begin mathsize 14px style nHCOOH double bond nHCOO to the power of minus sign equals 50 space mmol end style 
     
  2. Menentukan mol larutan dengan pH=13

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row 13 equals cell 14 minus sign pOH end cell row pOH equals 1 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 1 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 10 to the power of negative sign 1 end exponent M end cell row n equals cell M cross times V end cell row blank equals cell 0 comma 1 space M cross times 0 comma 01 space L end cell row blank equals cell 10 to the power of negative sign 3 end exponent space mol equals 1 space mmol end cell end table end style 
     
  3. Mencari pH akhir

    Penambahan basa akan menambah jumlah mol basa konjugasi dan mengurangi mol asam.

    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 K subscript a fraction numerator open parentheses nHCOOH bond nOH to the power of minus sign close parentheses over denominator left parenthesis nHCOO to the power of minus sign and nOH to the power of minus sign right parenthesis end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 4 end exponent fraction numerator left parenthesis 50 space mmol minus sign 1 space mmol right parenthesis over denominator left parenthesis 50 space mmol plus 1 space mmol right parenthesis end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 4 end exponent cross times 0 comma 96 end cell row blank equals cell 1 comma 92 space cross times 10 to the power of negative sign 4 end exponent space M end cell row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row blank equals cell negative sign space log space 1 comma 92 space cross times 10 to the power of negative sign 4 end exponent end cell row blank equals cell 4 minus sign log space 1 comma 92 space end cell end table end style

Jadi, pH sesudah dilakukan penambahan adalah 4 - log 1,92.

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