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pH dari 100 mL larutan  0,1 M adalah 5, maka pH da...

pH dari 100 mL larutan begin mathsize 14px style H C N end style 0,1 M adalah 5, maka pH dari 50 mL larutan undefined 0,04 M adalah...undefined 

Jawaban:

Senyaw begin mathsize 14px style H C N end style merupakan asam lemah sehingga perlu ditentukan nilai begin mathsize 14px style K subscript a end style dari nilai pH


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals 5 row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign 5 end exponent end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of K subscript a cross times space open square brackets Asam close square brackets end root end cell row cell 10 to the power of negative sign 5 end exponent end cell equals cell square root of K subscript a cross times 0 comma 1 end root end cell row cell 10 to the power of negative sign 10 end exponent end cell equals cell K subscript a cross times 0 comma 1 end cell row cell K subscript a end cell equals cell 10 to the power of negative sign 9 end exponent end cell end table end style


Setelah ditentukan harga begin mathsize 14px style K subscript a end style maka dapat ditentukan pH dari undefined 0,04 M


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of 10 to the power of negative sign 9 end exponent cross times 0 comma 04 end root end cell row blank equals cell square root of 4 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell 2 cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row pH equals cell negative sign log space 2 cross times 10 to the power of negative sign 5 comma 5 end exponent end cell row blank equals cell 5 comma 5 minus sign log space 2 end cell end table end style


Jadi, pH larutan HCN 0,04 M adalah 5,5-log 2.undefined 

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