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Ke dalam 100 mL larutan C H 3 ​ COO H 0 , 2 M ( K a = 1 × 1 0 − 5 ) ditambahkan 100 mL larutan N a O H 0 , 2 M sehingga terjadi reaksi: C H 3 ​ COO H ( a q ) + N a O H ( a q ) → C H 3 ​ COO H N a ( a q ) + H 2 ​ O ( l ) Harga pH larutan yang terjadi sebesar .... ( K w = 1 0 − 14 )

Ke dalam 100 mL larutan ditambahkan 100 mL larutan sehingga terjadi reaksi:

Harga pH larutan yang terjadi sebesar ....

 

  1. 5

  2. 6

  3. 9

  4. 10

  5. 11

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A. Ratna

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

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space space space space space space space space space space space space space space space C H subscript 3 C O O H space space plus space space space N a O H rightwards arrow C H subscript 3 C O O N a space space plus space space space H subscript 2 O space  M u l a space space space space space space space 20 space m m o l space space space space space space space space space 20 space m m o l space  R e a k s i minus 20 space m m o l space space space space space space space minus space 20 space m m o l space plus space 20 space m m o l space space space space space space plus 20 space m m o l space  S i s a space space space space space space space space 0 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 0 space space space space space space space space space space space space space space space space space space space 20 space m m o l space space space space space space space space 20 space m m o l space space space space space space space

 

table attributes columnalign right center left columnspacing 0px end attributes row cell left square bracket C H subscript 3 C O O N a right square bracket end cell equals cell fraction numerator m o l over denominator V t o t a l end fraction equals fraction numerator 0 comma 02 m o l over denominator 0 comma 2 L end fraction equals 0 comma 1 M end cell row blank blank cell C H subscript 3 C O O N a space m e r u p a k a n space g a r a m space b a s a comma s e h i n g g a colon end cell row cell left square bracket O H to the power of – right square bracket end cell equals cell square root of fraction numerator K w x left square bracket g a r a m right square bracket over denominator K a end fraction end root equals square root of fraction numerator 10 to the power of negative 14 end exponent x left square bracket 0 comma 1 right square bracket over denominator 10 to the power of negative 5 end exponent end fraction end root equals square root of 10 to the power of negative 10 end exponent end root equals 10 to the power of negative 5 end exponent M end cell row cell p O H end cell equals cell – l o g 10 to the power of negative 5 end exponent equals 5 end cell row cell p H end cell equals cell 14 – p O H end cell row blank equals cell 14 – 5 end cell row cell p H end cell equals 9 end table

Latihan Bab

Konsep Kilat

Pendahuluan Hidrolisis Garam

Hidrolisis Parsial Garam Asam

Hidrolisis Parsial Garam Basa

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

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Suatu larutan HCN 0,1 M mempunyai pH sebesar 3 – log 2, berapa gram kalium sianida yang terlarut dalam 500 cm 3 larutan agar diperoleh pH sebesar 9 + log 2? ( A r ​ K = 39; C = 12; dan N = 14).

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