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Sebanyak p gram amonium sulfat ( ( NH 4 ​ ) 2 ​ SO 4 ​ ) dilarutkan ke dalam 400 mL air sehingga terhidrolisis menurut reaksi berikut. NH 4 + ​ ( a q ) + H 2 ​ O ( l ) ⇌ NH 4 ​ OH ( a q ) + H + ( a q ) Jika pH yang terbentuk sebesar 5,5 dan K b ​ NH 4 ​ OH = 1 , 8 × 1 0 − 5 ,banyaknya p garam yang harus dilarutkan adalah ....

Sebanyak p gram amonium sulfat   dilarutkan ke dalam 400 mL air sehingga terhidrolisis menurut reaksi berikut.
 


 

Jika pH yang terbentuk sebesar 5,5 dan , banyaknya p garam yang harus dilarutkan adalah .... begin mathsize 14px style M subscript r space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript equals 132 space g space mol to the power of minus sign to the power of 1 end style   

  1. 0,24 gramundefined

  2. 0,48 gram

  3. 0,96 gram

  4. 2,40 gram

  5. 2,50 gram

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Y. Rochmawatie

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B

jawaban yang benar adalah B

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Menentukan Molaritas garam karena koefisien adalah 2 maka valensinya =2 Menentukan massa Jadi, jawaban yang benar adalah B

Menentukan Molaritas garam begin mathsize 14px style open parentheses open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 close parentheses end style 

begin mathsize 14px style left parenthesis open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript right parenthesis yields 2 N H subscript 4 to the power of plus sign and S O subscript 4 to the power of 2 minus sign end style

karena koefisien begin mathsize 14px style N H subscript 4 to the power of plus sign end style adalah 2 maka valensinya =2

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 5 comma 5 end cell 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 comma 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 w over K subscript b space cross times open square brackets N H subscript 4 to the power of plus sign close square brackets cross times valensi space end root end cell row cell 10 to the power of negative sign 5 comma 5 end exponent end cell equals cell square root of fraction numerator 10 to the power of negative sign 14 end exponent over denominator 1 comma 8 space cross times space 10 to the power of negative sign 5 end exponent end fraction space cross times space open square brackets N H subscript 4 to the power of plus sign close square brackets cross times 2 end root end cell row cell left parenthesis 10 to the power of negative sign 5 comma 5 end exponent right parenthesis squared end cell equals cell fraction numerator 2 cross times 10 to the power of negative sign 14 end exponent over denominator 1 comma 8 space cross times space 10 to the power of negative sign 5 end exponent end fraction space cross times space open square brackets N H subscript 4 to the power of plus sign close square brackets end cell row cell 10 to the power of negative sign 11 end exponent end cell equals cell fraction numerator 2 cross times 10 to the power of negative sign 14 end exponent over denominator 1 comma 8 space cross times space 10 to the power of negative sign 5 end exponent end fraction space cross times space open square brackets N H subscript 4 to the power of plus sign close square brackets end cell row cell 1 comma 8 cross times 10 to the power of negative sign 16 end exponent end cell equals cell 2 cross times 10 to the power of negative sign 14 end exponent cross times open square brackets N H subscript 4 to the power of plus sign close square brackets end cell row cell open square brackets N H subscript 4 to the power of plus sign close square brackets end cell equals cell fraction numerator 1 comma 8 cross times 10 to the power of negative sign 16 end exponent over denominator 2 cross times 10 to the power of negative sign 14 end exponent end fraction end cell row blank equals cell 0 comma 9 cross times 10 to the power of negative sign 2 end exponent end cell end table end style      

Menentukan massa begin mathsize 14px style left parenthesis open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end subscript right parenthesis end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell M space cross times space V end cell row n equals cell 0 comma 9 space cross times space 10 to the power of negative sign 2 end exponent space cross times space 0 comma 4 end cell row n equals cell 3 comma 6 space cross times space 10 to the power of negative sign 3 end exponent end cell row massa equals cell n space cross times space Mr end cell row massa equals cell 3 comma 6 space cross times space 10 to the power of negative sign 3 end exponent space cross times space 132 end cell row massa equals cell 0 comma 48 space gram end cell end table end style

Jadi, jawaban yang benar adalah B

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