Roboguru

pH dari larutan   0,1 M dengan nilai   adalah ....

Pertanyaan

pH dari larutan open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript  0,1 M dengan nilai Kb space N H subscript 3 equals space 1 space x space 10 to the power of negative sign 5 end exponent  adalah ....

  1. 8,85space

  2. 6,23space

  3. 3,88space

  4. 4,73

  5. 4,85space

Pembahasan Soal:

open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript adalah larutan garam yang bersifat asam.

Rumus garam asam:

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 square root of Kw over Kb end root x space n space x space open square brackets G close square brackets 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 end table

[G]= konsentrasi kation (ion positif) garam

n=valensi  

 

open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript yields 2 N H subscript 4 to the power of plus sign and S O subscript 4 to the power of 2 minus sign

jumlah ion positif pada N H subscript 4 to the power of plus sign adalah 2 sehingga n = 2.

 

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 square root of fraction numerator 1 x 10 to the power of negative sign 14 end exponent over denominator 1 x 10 to the power of negative sign 5 end exponent end fraction end root x space 2 space x space space 10 to the power of negative sign 1 end exponent end cell row blank equals cell 1 comma 4 space x space 10 to the power of negative sign 5 end exponent 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 log space left parenthesis 1 comma 4 space x space 10 to the power of negative sign 5 end exponent right parenthesis end cell row blank equals cell 5 minus sign log space 1 comma 4 end cell row blank blank blank end table

 

Jadi, pH dari larutan open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript adalah 5-log 1,4.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 04 Mei 2021

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