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Untuk reaksi hidrolisis:  Cu(H2​O4​)2+(aq)+H2​O(l)⇌Cu(H2​O)3​(OH)+(aq)+H3​O+(aq),  nilai tetapan kesetimbangan hidrolisisnya adalah 1×10−8. Jika CuCl2​.2H2​O (Mr​=170,5) dilarutkan dalam air sampai volume larutan mencapai 0,250 L, temyata larutan yang terbentuk memiliki pH= 5,00. Berapakah massa (gram) hidrat tersebut yang larut dalam air?

Pertanyaan

Untuk reaksi hidrolisis: 

Cu open parentheses H subscript 2 O subscript 4 close parentheses to the power of 2 plus end exponent left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses equilibrium Cu open parentheses H subscript 2 O close parentheses subscript 3 left parenthesis O H right parenthesis to the power of plus end exponent left parenthesis italic a italic q right parenthesis plus H subscript 3 O to the power of plus sign left parenthesis italic a italic q right parenthesis

nilai tetapan kesetimbangan hidrolisisnya adalah 1 cross times 10 to the power of negative sign 8 end exponent. Jika Cu Cl subscript 2 point 2 H subscript 2 O space left parenthesis italic M subscript r equals 170 comma 5 right parenthesis dilarutkan dalam air sampai volume larutan mencapai 0,250 L, temyata larutan yang terbentuk memiliki pH= 5,00. Berapakah massa (gram) hidrat tersebut yang larut dalam air?

Q. 'Ainillana

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Jawaban terverifikasi

Jawaban

massa (gram) hidrat tersebut yang larut dalam air dalah 0,43 gram.

Pembahasan

Mengubah pH ke dalam bentuk open square brackets H to the power of plus sign close square brackets:

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 end table 

Misal open square brackets Cu open parentheses H subscript 2 O close parentheses subscript 4 to the power of 2 plus end exponent close square brackets subscript awal equals open square brackets Cu Cl subscript 2 point 2 H subscript 2 O close square brackets equals x space M, maka nilai x dapat dihitung melalui:

 

table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript h end cell equals cell K subscript a equals fraction numerator open square brackets Cu open parentheses H subscript 2 O close parentheses subscript 3 left parenthesis O H right parenthesis to the power of plus close square brackets open square brackets H subscript 3 O to the power of plus sign close square brackets over denominator open square brackets Cu open parentheses H subscript 2 O close parentheses subscript 4 to the power of 2 plus end exponent close square brackets end fraction end cell row cell 1 comma 0 cross times 10 to the power of negative sign 8 end exponent end cell equals cell fraction numerator open parentheses 1 cross times 10 to the power of negative sign 5 end exponent close parentheses squared over denominator open square brackets x minus sign 1 cross times 10 to the power of negative sign 5 end exponent close square brackets end fraction end cell row cell x minus sign 1 cross times 10 to the power of negative sign 5 end exponent end cell equals cell fraction numerator open parentheses 1 cross times 10 to the power of negative sign 5 end exponent close parentheses squared over denominator open square brackets 1 comma 0 cross times 10 to the power of negative sign 8 end exponent close square brackets end fraction end cell row x equals cell 0 comma 01 space M end cell end table 

open square brackets Cu open parentheses H subscript 2 O close parentheses subscript 4 to the power of 2 plus end exponent close square brackets subscript awal equals open square brackets Cu Cl subscript 2 point 2 H subscript 2 O close square brackets equals 0 comma 01 space M 

Maka, adapat diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell massa space Cu Cl subscript 2 point 2 H subscript 2 O end cell equals cell M cross times V cross times Mr end cell row blank equals cell 0 comma 01 cross times 0 comma 5 cross times 170 comma 5 end cell row blank equals cell 0 comma 43 space gram end cell end table  

Jadi massa (gram) hidrat tersebut yang larut dalam air dalah 0,43 gram.

53

5.0 (1 rating)

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