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Berapa gram amonia (NH3​) 0,25 M, pH-nya 11 + log 2 direaksikan dengan larutan asam formiat (HCOOH) 0,09 M yang pH-nya 3 - log 6. Hitunglah derajat hidrolisis garam HCOONH4​ dalam larutan 0,15 M tersebut!

Pertanyaan

Berapa gram amonia begin mathsize 14px style open parentheses N H subscript 3 close parentheses end style 0,25 M, pH-nya 11 + log 2 direaksikan dengan larutan asam formiat begin mathsize 14px style open parentheses H C O O H close parentheses end style 0,09 M yang pH-nya 3 - log 6. Hitunglah derajat hidrolisis garam begin mathsize 14px style H C O O N H subscript 4 end style dalam larutan 0,15 M tersebut!

Pembahasan Video:

Pembahasan:

Derajat hidrolisis (h) adalah akar hasil bagi antara Kh dengan konsentrasi garam.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row h equals cell fraction numerator square root of K subscript h end root over denominator 1 plus square root of K subscript h end root end fraction space end cell row cell K subscript h end cell equals cell square root of fraction numerator K subscript w over denominator K subscript a cross times K subscript b end fraction end root space end cell end table end style  

Maka kita cari terlebih dahulu undefined dan begin mathsize 14px style K subscript a end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH space N H subscript 3 end cell equals cell 11 plus log space 2 end cell row pOH equals cell 14 minus sign left parenthesis 11 plus log space 2 right parenthesis end cell row pOH equals cell 3 minus sign log space 2 end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 3 end exponent end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of K subscript b cross times M subscript b end root end cell row cell 2 cross times 10 to the power of negative sign 3 end exponent end cell equals cell square root of K subscript b cross times 0 comma 25 end root space left parenthesis kuadratkan space kedua space ruas right parenthesis end cell row cell 4 cross times 10 to the power of negative sign 6 end exponent end cell equals cell K subscript b cross times 0 comma 25 end cell row cell K subscript b end cell equals cell 16 cross times 10 to the power of negative sign 6 end exponent end cell row cell K subscript b end cell equals cell 1 comma 6 cross times 10 to the power of negative sign 5 end exponent end cell end table end style   

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH space H C O O H end cell equals cell 3 minus sign log space 6 end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 6 cross times 10 to the power of negative sign 3 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 M subscript a end root end cell row cell 6 cross times 10 blank to the power of negative sign 3 end exponent end cell equals cell square root of K subscript a cross times 0 comma 09 end root space left parenthesis kuadratkan space kedua space ruas right parenthesis end cell row cell 36 cross times 10 to the power of negative sign 6 end exponent end cell equals cell K subscript a cross times 0 comma 09 end cell row cell K subscript a end cell equals cell 4 cross times 10 to the power of negative sign 4 end exponent end cell end table end style 

Sebelum mencari derajat hidrolisis garam (h), kita cari terlebih dahulu nilai begin mathsize 14px style K subscript h end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript h end cell equals cell square root of fraction numerator K subscript w over denominator K subscript a cross times K subscript b end fraction end root end cell row cell K subscript h end cell equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 4 cross times 10 to the power of negative sign 4 end exponent cross times 1 comma 6 cross times 10 to the power of negative sign 5 end exponent end fraction end root end cell row cell K subscript h end cell equals cell square root of 10 to the power of negative sign 4 end exponent over 64 end root equals 1 over 8 cross times 10 to the power of negative sign 2 end exponent end cell row cell K subscript h end cell equals cell 0 comma 125 cross times 10 to the power of negative sign 2 end exponent end cell end table end style 

 

table attributes columnalign right center left columnspacing 0px end attributes row h equals cell fraction numerator square root of K subscript h end root over denominator 1 plus square root of K subscript h end root end fraction space end cell row h equals cell fraction numerator square root of 0 comma 125 cross times 10 to the power of negative sign 2 end exponent end root over denominator 1 plus square root of 0 comma 125 cross times 10 to the power of negative sign 2 end exponent end root end fraction space end cell row h equals cell fraction numerator 0 comma 0354 over denominator 1 plus 0 comma 0354 end fraction end cell row h equals cell fraction numerator 0 comma 0354 over denominator 1 comma 0354 end fraction end cell row h equals cell 0 comma 034 end cell end table    

Jadi, derajat hidrolisis garam begin mathsize 14px style H C O O N H subscript 4 end style  adalah 0,034.

Jawaban terverifikasi

Dijawab oleh:

N. Puspita

Terakhir diupdate 07 Oktober 2021

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Beberapa gram amonia (NH3​) 0,25 M, pH-nya 11+ log 2 direaksikan dengan larutan asam formiat (HCOOH) 0,09 M yang pH-nya 3 - log 6. Hitunglah : Derajat hidrolisis garam HCOONH4​ dalam larutan 0,15 M t...

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