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Sebanyak 2,6 gram garam ditambahkan ke dalam 100 mL larutan  0,04 M (Ka = ) sehingga pH larutan menjadi 5 - log 2. Unsur X yang terkandung dalam garam tersebut adalah... (: H = 1 g , C= 1 2 g , O = 1 6 g , Mg = 24 g , Ca = 40 g , Fe = 56 g , dan Ba = 137 g )

Pertanyaan

Sebanyak 2,6 gram garam begin mathsize 14px style open parentheses H C O O close parentheses subscript 2 X space end styleditambahkan ke dalam 100 mL larutan begin mathsize 14px style H C O O H end style 0,04 M (Ka begin mathsize 14px style H C O O H end style = begin mathsize 14px style 2 cross times 10 to the power of negative sign 4 end exponent end style) sehingga pH larutan menjadi 5 - log 2. Unsur X yang terkandung dalam garam tersebut adalah...

(begin mathsize 14px style italic A subscript italic r end style: H = 1 g begin mathsize 14px style mol to the power of negative sign 1 end exponent end style, C= 1 2 g begin mathsize 14px style mol to the power of negative sign 1 end exponent end style, O = 1 6 g begin mathsize 14px style mol to the power of negative sign 1 end exponent end style, Mg = 24 g begin mathsize 14px style mol to the power of negative sign 1 end exponent end style, Ca = 40 g begin mathsize 14px style mol to the power of negative sign 1 end exponent end style, Fe = 56 g begin mathsize 14px style mol to the power of negative sign 1 end exponent end style , dan Ba = 137 g begin mathsize 14px style mol to the power of negative sign 1 end exponent end style)

Pembahasan Video:

Pembahasan Soal:

Diketahui pH larutan = begin mathsize 14px style 5 minus sign log space 2 space end style

Sehingga konsentrasi begin mathsize 14px style open square brackets H to the power of plus sign close square brackets space end style 

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

Berdasarkan pH larutan penyangga yang diketahui, maka mol begin mathsize 14px style H C O O to the power of minus sign end style dapat dicari dengan perhitungan sebagai berikut:
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell H C O O H end cell rightwards harpoon over leftwards harpoon cell H to the power of plus sign and H C O O to the power of minus sign end cell row Ka equals cell fraction numerator open square brackets H to the power of plus sign close square brackets open square brackets H C O O to the power of minus sign close square brackets over denominator open square brackets H C O O H close square brackets end fraction end cell row cell open square brackets H C O O to the power of minus sign close square brackets end cell equals cell K subscript a cross times fraction numerator open square brackets H C O O H close square brackets over denominator open square brackets H to the power of plus sign close square brackets end fraction end cell row blank equals cell 2 cross times 10 to the power of negative sign 4 end exponent cross times fraction numerator 4 cross times 10 to the power of negative sign 2 end exponent space M over denominator 2 cross times 10 to the power of negative sign 5 end exponent space M end fraction end cell row blank equals cell 0 comma 4 space M space end cell end table end style 

Menentukan mol begin mathsize 14px style H C O O to the power of minus sign space end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H C O O to the power of minus sign end cell equals cell M space H C O O to the power of minus sign cross times space V space H C O O to the power of minus sign end cell row blank equals cell space 0 comma 4 space M space cross times 0 comma 1 space L end cell row blank equals cell space 4 cross times 10 squared space mol space end cell end table end style 


Menentukan X dari senyawa begin mathsize 14px style open parentheses H C O O close parentheses subscript 2 X space end style 

 Error converting from MathML to accessible text. 


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space open parentheses H C O O close parentheses subscript 2 X space end cell equals cell space fraction numerator koefesien space open parentheses H C O O close parentheses subscript 2 X space over denominator koefesien space H C O O to the power of minus sign end fraction cross times mol space H C O O to the power of minus sign end cell row blank equals cell 1 half cross times 0 comma 04 space mol end cell row blank equals cell 0 comma 02 space mol space end cell end table end style 

 

Error converting from MathML to accessible text. 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Mr space open parentheses H C O O close parentheses subscript 2 X end cell equals cell 2 left parenthesis Ar space H right parenthesis plus 2 left parenthesis Ar space C right parenthesis plus 4 left parenthesis Ar space O right parenthesis plus space Ar space X end cell row 130 equals cell 2 plus 24 plus 64 plus space Ar space X end cell row 130 equals cell 90 plus Ar space X end cell row cell Ar space X end cell equals cell 40 space end cell end table end style 

Jadi, unsur X yang terkandung dalam garam tersebut adalah Ca.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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