Roboguru

Sebanyak 100 mL larutan  0,2 M direaksikan dengan 100 mL  0,2 M. Jika garam yang dihasilkan diencerkan dengan 600 mL air, pH larutan garam setelah pengenceran adalah ()

Pertanyaan

Sebanyak 100 mL larutan begin mathsize 14px style H N O subscript 3 end style 0,2 M direaksikan dengan 100 mL begin mathsize 14px style N H subscript 4 O H end style 0,2 M. Jika garam yang dihasilkan diencerkan dengan 600 mL air, pH larutan garam setelah pengenceran adalah undefined(begin mathsize 14px style K subscript b N H subscript 3 equals 10 to the power of negative sign 5 end exponent end style)

  1. 2 - log 5space 

  2. 5 - log 6space 

  3. 5 - log 1undefined 

  4. 6 - log 5undefined 

  5. 6 - log 1undefined 

Pembahasan Soal:

begin mathsize 14px style space space mmol space H N O subscript 3 space equals space M space x space V space equals space 0 comma 2 space x space 100 space equals space 20 space mmol mmol space N H subscript 4 O H space equals M space x space V space equals space 0 comma 2 space x space 100 space equals space 20 space mmol end style

Karena mmol begin mathsize 14px style H N O subscript 3 end style = mmol begin mathsize 14px style N H subscript 4 O H end style maka terbentuk garam undefined.

begin mathsize 14px style open square brackets N H subscript 4 N O subscript 3 close square brackets equals mmol over Vtotal equals 20 over 200 equals 0 comma 1 space M end style

Pengenceran :

begin mathsize 14px style space space space V subscript 1 space x space M subscript 1 double bond V subscript 2 space x space M subscript 2 200 space x space 0 comma 1 equals open parentheses 200 plus 600 close parentheses space x space M subscript 2 space space space space space space space space space 20 space space equals 800 space x space M subscript 2 space space space space space space space space space M subscript 2 space equals space 2 comma 5.10 to the power of negative sign 2 end exponent  end style

 Rumus pH garam asam  : 

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space equals square root of Kw over Kb open square brackets N H subscript 4 N O subscript 3 close square brackets val end root  open square brackets H to the power of plus sign close square brackets space equals square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent open square brackets 2 comma 5.10 to the power of negative sign 2 end exponent close square brackets space x space 1 end root open square brackets H to the power of plus sign close square brackets space equals 5.10 to the power of negative sign 6 end exponent space M  pH space equals minus sign log space open parentheses H to the power of plus sign close parentheses space space space space space space equals minus sign log space open parentheses 5.10 to the power of negative sign 6 end exponent close parentheses space space space space space space equals 6 space minus sign space log space 5  end style

Jadi, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Februari 2021

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