Roboguru

Jika Ksp​Hg2​SO4​=6×10−7, konsentrasi ion Hg+ dan ion SO42−​ dalam larutan jenuh Hg2​SO4​ berturut-turut adalah ...

Pertanyaan

Jika begin mathsize 14px style K subscript sp space Hg subscript 2 S O subscript 4 equals 6 cross times 10 to the power of negative sign 7 end exponent end style, konsentrasi ion begin mathsize 14px style Hg to the power of plus sign end style dan ion begin mathsize 14px style S O subscript 4 to the power of 2 minus sign end style dalam larutan jenuh begin mathsize 14px style Hg subscript 2 S O subscript 4 end style berturut-turut adalah ...space

  1. begin mathsize 14px style 21 comma 2 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style dan begin mathsize 14px style 5 comma 3 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style 

  2. begin mathsize 14px style 15 comma 9 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style dan begin mathsize 14px style 21 comma 2 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style 

  3. begin mathsize 14px style 10 comma 6 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style dan begin mathsize 14px style 21 comma 2 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style   

  4. begin mathsize 14px style 10 comma 6 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style dan begin mathsize 14px style 5 comma 3 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style 

  5. begin mathsize 14px style 5 comma 3 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style dan begin mathsize 14px style 10 comma 6 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent end style 

Pembahasan:

Larutan jenuh adalah larutan yang sudah mencapai batas maksimum kelarutannya dan sama dengan nilai hasil kali kelarutannya (begin mathsize 14px style K subscript sp end style). Persamaan reaksi ionisasi senyawa begin mathsize 14px style Hg subscript 2 S O subscript 4 end style adalah sebagai berikut:space


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

Nilai konsentrasi ion begin mathsize 14px style Hg to the power of plus sign end styledan begin mathsize 14px style S O subscript 4 to the power of 2 minus sign end style dapat diketahui dengan cara mencari nilai kelarutan (s) berdasarkan data begin mathsize 14px style K subscript sp end style begin mathsize 14px style Hg subscript 2 S O subscript 4 end style yaitu sebagai berikut.space


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp end cell equals cell open square brackets Ag to the power of plus sign close square brackets squared middle dot open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell row cell K subscript sp end cell equals cell left parenthesis 2 s right parenthesis squared middle dot open parentheses s close parentheses end cell row cell K subscript sp end cell equals cell 4 s cubed end cell row cell 6 cross times 10 to the power of negative sign 7 end exponent end cell equals cell 4 s cubed end cell row cell s cubed end cell equals cell fraction numerator 6 cross times 10 to the power of negative sign 7 end exponent over denominator 4 end fraction end cell row cell s cubed end cell equals cell 1 comma 5 cross times 10 to the power of negative sign 7 end exponent end cell row cell s cubed end cell equals cell 0 comma 15 cross times 10 to the power of negative sign 6 end exponent end cell row s equals cell cube root of 0 comma 15 cross times 10 to the power of negative sign 6 end exponent end root end cell row s equals cell 0 comma 53 cross times 10 to the power of negative sign 2 end exponent end cell row s equals cell 5 comma 3 cross times 10 to the power of negative sign 3 end exponent end cell end table end style    

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell 2 s end cell row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell 2 cross times 5 comma 3 middle dot 10 to the power of negative sign 3 end exponent end cell row cell open square brackets Ag to the power of plus sign close square brackets end cell equals cell 10 comma 6 cross times 10 to the power of negative sign 3 end exponent end cell row blank blank blank row cell open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell equals s row cell open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end cell equals cell 5 comma 3 cross times 10 to the power of negative sign 3 end exponent end cell end table end style  


Jadi, jawaban yang tepat adalah D.space

Jawaban terverifikasi

Dijawab oleh:

A. Kusbardini

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 06 Oktober 2021

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