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Jika dalam gelas kimia terdapat 5 L larutan kalsium asetat 0,004 M, maka pH larutan tersebut adalah....

Jika dalam gelas kimia terdapat 5 L larutan kalsium asetat  begin mathsize 14px style open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca end style 0,004 M, maka pH larutan tersebut adalah....

begin mathsize 14px style open parentheses K subscript a space C H subscript 3 C O O H equals 2 cross times 10 to the power of negative sign 5 end exponent close parentheses end style 

  1. begin mathsize 14px style 5 minus sign log space 2 end style 

  2. begin mathsize 14px style 6 minus sign log space 2 end style 

  3. begin mathsize 14px style 9 plus log space 2 end style 

  4. begin mathsize 14px style 8 plus log space 2 end style 

  5. begin mathsize 14px style 9 end style 

Jawaban:

Diketahui:
Larutan kalsium asetat  begin mathsize 14px style open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca end style 0,004 M, 5 L
begin mathsize 14px style open parentheses K subscript a space C H subscript 3 C O O H equals 2 cross times 10 to the power of negative sign 5 end exponent close parentheses end style

Ditanya:
pH larutan

Penyelesaian

1. Menuliskan persamaan reaksi ionisasi senyawa garam

open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca yields 2 C H subscript 3 C O O to the power of minus sign and Ca to the power of 2 plus sign 

begin mathsize 14px style open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca end style merupakan garam yang bersifat basa.

2. Menghitung konsentrasi O H to the power of minus sign

Konsentrasi anion garam:

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets C H subscript 3 C O O to the power of minus sign close square brackets end cell equals cell 2 over 1 cross times left square bracket open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca right square bracket end cell row blank equals cell 2 cross times left parenthesis 0 comma 004 space M right parenthesis end cell row blank equals cell 0 comma 008 space M end cell row blank equals cell 8 cross times 10 to the power of negative sign 3 end exponent space M end cell end table 

open square brackets O H to the power of minus sign close square brackets equals square root of K subscript w over K subscript a cross times left square bracket anion space garam right square bracket end root space space space space space space space space space equals square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 2 middle dot 10 to the power of negative sign 5 end exponent end fraction cross times 8 middle dot 10 to the power of negative sign 3 end exponent end root open square brackets O H to the power of minus sign close square brackets equals 2 middle dot space 10 to the power of negative sign 6 end exponent space M 


3. Menghitung pOH larutan garam

 table attributes columnalign right center left columnspacing 0px end attributes row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row cell space space space space space end cell equals cell negative sign log space left parenthesis 2 middle dot 10 to the power of negative sign 6 end exponent right parenthesis end cell row blank equals cell 6 minus sign log space 2 end cell row blank blank blank end table

4. Menghitung pH larutan garam

 table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 14 minus sign pOH end cell row blank equals cell 14 minus sign left parenthesis 6 minus sign log space 2 right parenthesis end cell row blank equals cell 8 plus log space 2 end cell end table 

Jadi, jawaban yang tepat adalah D.

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