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Garam kalium asetat C H 3 ​ COO K dapat dibuat dengan mereaksikan 20 mL larutan asam asetat 0,5 M dan 20 mL larutan kalium hidroksida 0,5 M dengan reaksi: C H 3 ​ COO H ( a q ) + K O H ( a q ) → C H 3 ​ COO K ( a q ) + H 2 ​ O ( l ) . Besarnya pH larutan setelah dicampur adalah ... ( K w = 1.1 0 − 14 , K a C H 3 ​ COO H = 1.1 0 − 5 )

Garam kalium asetat dapat dibuat dengan mereaksikan 20 mL larutan asam asetat 0,5 M dan 20 mL larutan kalium hidroksida 0,5 M dengan reaksi: . Besarnya pH larutan setelah dicampur adalah ...

  1. 5,5 - log 5

  2. 5,5 + log 5

  3. 8,5 - log 5

  4. 8,5 + log 5

  5. 11 + log 5

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S. Susanti

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Mahasiswa/Alumni Universitas Jayabaya

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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 C H subscript 3 C O O H space plus space K O H space rightwards arrow space C H subscript 3 C O O K space plus space H subscript 2 O  M u l a space space space space space space space space space space space space space space space space space space space 10 space m m o l space space space space 10 space m m o l  R e a k s i space space space space space space space space space space space space minus 10 space m m o l space minus 10 space m m o l space space space plus 10 space m m o l space space plus 10 space m m o l  S i s a 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 0 space space space space space space space space space space space space space space space 10 space m m o l space space space space space space 10 space m m o l
 

left square bracket C H subscript 3 C O O K right square bracket equals fraction numerator m o l over denominator V subscript t o t a l end subscript end fraction equals fraction numerator 0 comma 01 space m o l over denominator 0 comma 04 space L end fraction equals space 0 comma 25 space M    C H subscript 3 C O O K space m e r u p a k a n space g a r a m space b a s a comma space s e h i n g g a colon  left square bracket O H minus right square bracket equals square root of fraction numerator K w cross times 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 cross times left square bracket 0 comma 25 right square bracket over denominator 10 to the power of negative 5 end exponent end fraction end root equals square root of 25 cross times 20 to the power of negative 11 end exponent end root equals 5 cross times 10 space M  p O H equals negative log space 5 cross times 10 to the power of negative 5 comma 5 end exponent equals 5 comma 5 minus log space 5  p H space space space equals 14 minus p O H  space space space space space space space space equals 14 minus left parenthesis 5 comma 5 minus log space 5 right parenthesis  space space space space space space space space equals 8 comma 5 plus log space 5

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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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