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Larutan 100 mL  0,2 M  dicampurkan dengan 50 mL la...

Larutan 100 mL begin mathsize 14px style space N H subscript 4 O H space end style 0,2 M begin mathsize 14px style left parenthesis Kb equals 10 to the power of negative sign 5 end exponent space right parenthesis end style dicampurkan dengan 50 mL larutan begin mathsize 14px style H subscript 2 S O subscript 4 end style  0,1 M, pH larutan yang terjadi adalah ....

Jawaban:

Persamaan reaksi setara:

begin mathsize 14px style 2 N H subscript 4 O H left parenthesis italic a italic q right parenthesis plus space H subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis rightwards arrow open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 left parenthesis italic a italic q right parenthesis plus 2 H subscript 2 O space left parenthesis italic l right parenthesis end style 

  • mmol begin mathsize 14px style space N H subscript 4 O H space end style = 0,2 M x 100 mL = 20 mmol = 0,02 mol
  • mmol begin mathsize 14px style H subscript 2 S O subscript 4 end style= 0,1 M x 50 mL= 5 mmol=0,005 mol

Stoikiometri:

Rumus pH larutan penyangga basa:

begin mathsize 14px style open square brackets O H to the power of minus sign close square brackets double bond Kb space x space fraction numerator mol space basa space lemah over denominator mol space garam end fraction end style 

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 10 to the power of negative sign 5 end exponent space x space fraction numerator 0 comma 01 space mol over denominator 0 comma 005 space mol end fraction end cell row blank equals cell fraction numerator 10 to the power of negative sign 7 end exponent over denominator 5 x 10 to the power of negative sign 3 end exponent end fraction end cell row blank equals cell 2 x 10 to the power of negative sign 5 end exponent end cell row pOH equals cell negative sign log left parenthesis 2 x 10 to the power of negative sign 5 end exponent right parenthesis end cell row blank equals cell 5 minus sign log space 2 end cell row pH equals cell 14 minus sign left parenthesis 5 minus sign log space 2 right parenthesis end cell row blank equals cell 9 plus log space 2 end cell end table end style 

Jadi, pH larutan yang terjadi adalah 9+log 2.

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