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Tentukan pH larutan berikut  Asam asetat 0,1 mol yang terionisasi 60 % dalam 500 ml larutan

Pertanyaan

Tentukan pH larutan berikut 

Asam asetat 0,1 mol yang terionisasi 60 % dalam 500 ml larutanundefined 

  1. ...undefined 

  2. ...undefined 

Pembahasan Video:

Pembahasan Soal:

Asam asetat begin mathsize 14px style open parentheses C H subscript 3 C O O H close parentheses end style merupakan asam lemah, untuk menghitung nilai pH larutan asam asetat, terlebih dahulu kita mencari nilai begin mathsize 14px style K subscript a end style dari derajat ionisasi.


begin mathsize 14px style alpha equals square root of fraction numerator K subscript a over denominator open square brackets C H subscript 3 C O O H close square brackets end fraction end root 0 comma 6 equals square root of fraction numerator K subscript a over denominator open parentheses begin display style fraction numerator 0 comma 1 space mol over denominator 0 comma 5 space L end fraction end style close parentheses end fraction end root 0 comma 6 squared equals fraction numerator K subscript a over denominator 0 comma 2 space M end fraction K subscript a equals 7 comma 2 cross times 10 to the power of negative sign 2 end exponent   open square brackets H to the power of plus sign close square brackets equals square root of K subscript a cross times M subscript C H subscript 3 C O O H end subscript end root open square brackets H to the power of plus sign close square brackets equals square root of 7 comma 2 cross times 10 to the power of negative sign 2 end exponent cross times 0 comma 2 space M end root open square brackets H to the power of plus sign close square brackets equals square root of 144 cross times 10 to the power of negative sign 4 end exponent end root open square brackets H to the power of plus sign close square brackets equals 12 cross times 10 to the power of negative sign 2 end exponent thin space M open square brackets H to the power of plus sign close square brackets equals 1 comma 2 cross times 10 to the power of negative sign 1 end exponent thin space M   pH equals minus sign log space open square brackets H to the power of plus sign close square brackets pH equals minus sign log space left parenthesis 1 comma 2 cross times 10 to the power of negative sign 1 end exponent right parenthesis pH equals 1 minus sign log space 1 comma 2 end style 


Jadi, pH larutan asam asetat 0,1 mol yang terionisasi 60% dalam 500 ml larutan adalah undefined.space space space

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 20 Juli 2021

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