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Pertanyaan

Larutan HCl 0,8 M sebanyak 100 mL ditambahkan pada 100 mL larutan N H subscript 3 0,8 M (Kb N H subscript 3 = 1 space x space 10 to the power of negative 5 end exponent, Kw  = 10 to the power of negative 14 end exponent), menurut reaksi :

space NH subscript 3 left parenthesis aq right parenthesis plus HCl left parenthesis aq right parenthesis rightwards arrow space NH subscript 4 Cl left parenthesis aq right parenthesis

Harga pH larutan yang terjadi adalah .....

 

 

  1. 5 minus sign log space 2

  2. 5 plus log space 2

  3. 5 plus log space 4

  4. 9 minus sign log space 4

  5. 9 plus log space 2

M. Robo

Master Teacher

Jawaban terverifikasi

Pembahasan

Diketahui :

M HCl = 0,8 M (100 mL)

N H subscript 3 0,8 M (100 mL)

Kb N H subscript 3 = 1 space x space 10 to the power of negative 5 end exponent, Kw  = 10 to the power of negative 14 end exponent

Ditanya : 

pH larutan

Jawab :

left square bracket NH subscript 4 Cl right square bracket space equals fraction numerator space mol over denominator straight V subscript total end fraction space equals space fraction numerator 0 comma 08 space mol over denominator 0 comma 2 space straight L end fraction space equals space 0 comma 4 space straight M  NH subscript 4 Cl space merupakan space garam space asam comma space sehingga space colon  left square bracket straight H plus right square bracket equals square root of fraction numerator Kw cross times open square brackets gram close square brackets over denominator Kb end fraction end root equals square root of fraction numerator 10 to the power of negative 14 end exponent cross times open square brackets 0 comma 4 close square brackets over denominator 10 to the power of negative 5 end exponent end fraction end root equals square root of 4 cross times 10 to the power of negative 10 end exponent end root equals 2 cross times 10 to the power of negative 5 end exponent space straight M  pH equals space minus space log space 2 space cross times space 10 to the power of negative 5 end exponent equals space 5 space minus space log space 2

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