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Pertanyaan

Larutan 100 mL NH3 0,8 M dicampur dengan 100 mL HCl 0,8 M, menurut reaksi:

N H subscript 3 left parenthesis a q right parenthesis plus H C l left parenthesis a q right parenthesis space rightwards arrow space N H subscript 4 C l left parenthesis a q right parenthesis

left parenthesis K w space equals space 10 to the power of negative 14 end exponent comma space K b space N H subscript 3 space equals space 10 to the power of negative 5 end exponent right parenthesis

Harga pH larutan tersebut adalah...

  1. 5 - log 2

  2. 5 + log 2

  3. 9 - log 2

  4. 9 + log 2

  5. 12 + log 2

D. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Pembahasan

N H subscript 3 left parenthesis a q right parenthesis plus H C l left parenthesis a q right parenthesis space rightwards arrow space N H subscript 4 C l left parenthesis a q right parenthesis

0,8 M             0,8 M

100 mL          100 mL

harga pH larutan yang terjadi ?

space space space space space space space space space N H subscript 3 left parenthesis a q right parenthesis space space space space space space space space space space space space space space space plus space space space space space H C l left parenthesis a q right parenthesis space space space space space space space space space space space space space space space rightwards arrow space space N H subscript 4 C l left parenthesis a q right parenthesis  space space space space space space space space space 0 comma 8 space M cross times 100 m L space space space space space space space space space space space space 0 comma 8 space M cross times 100 space m L  M space equals space 80 space m m o l space space space space space space space space space space space space space space space space space space space space space space equals space 80 space m m o l space space space space space space space space space space space space space space space space space space space space space space minus  fraction numerator R space equals space 80 space m m o l space space space space space space space space space space space space space space space space space space space space space space space space space space space 80 space m m o l space space space space space space space space space space space space space space space space 80 space m m o l over denominator S space equals space space space space space minus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus space space space space space space space space space space space space space space space space space space space space space space space space space space space 80 space m m o l space end fraction

M o l a r i t a s space g a r a m space equals space fraction numerator m m o l space g a r a m over denominator v o l u m e space t o t a l end fraction space equals space fraction numerator 80 space m m o l over denominator 200 space m L end fraction space equals space 0 comma 4 space M

Karena semua reaktan habis, berarti merupakan campuran hidrolisis. Hidrolisis dilihat dari sifat yang kuat, disini yang kuat adalah HCl (asam).

Rumus Hidrolisis untuk Garam asam:

open square brackets H to the power of plus close square brackets space equals space square root of fraction numerator K w cross times M o l a r i t a s space G a r a m cross times v a l e n s i over denominator K b end fraction end root  open square brackets H to the power of plus close square brackets space equals square root of fraction numerator 10 to the power of negative 14 end exponent cross times 0 comma 4 space M cross times 1 over denominator 10 to the power of negative 5 end exponent end fraction end root  open square brackets H to the power of plus close square brackets space equals square root of 4 cross times 10 to the power of negative 10 end exponent end root  open square brackets H to the power of plus close square brackets space equals space 2 cross times 10 to the power of negative 5 end exponent

pH = -log[H+] = -log [2 x 10-5] = 5-log 2

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