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Sampel yang terdiri atas campuran natrium  dan kal...

Sampel yang terdiri atas campuran natrium begin mathsize 14px style left parenthesis italic A subscript italic r space equals space 23 space g space mol to the power of negative sign 1 end exponent right parenthesis end style dan kalsium begin mathsize 14px style left parenthesis italic A subscript r space equals space 40 space g space mol to the power of negative sign 1 end exponent right parenthesis end style diuji di laboratorium untuk mengetahui massa tiap-tiap unsur. Sebanyak 74,5 gram sampel dilarutkan dalam air sampai volumenya 500 mL. Sebanyak 100 mL larutan tersebut diambil, lalu dititrasi dengan larutan asam klorida 5 M. Titik akhir titrasi tercapai saat penambahan larutan asam klorida 140 mL, tentukan:


volume gas begin mathsize 14px style H subscript 2 end style yang dihasilkan (begin mathsize 14px style 0 space degree C end style, 1 atm).undefined 

  1. ...undefined 

  2. ...undefined 

Jawaban:

Sebanyak 74,5 gram sampel campuran yang terdiri dari logam natrium dan kalsium dilarutkan ke dalam air. Massa logam Na dimisalkan sebagai begin mathsize 14px style x end style.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na space end cell equals cell space fraction numerator massa space Na over denominator italic A subscript r space Na end fraction end cell row blank equals cell space fraction numerator x space g over denominator 23 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell space x over 23 mol end cell row cell mol space Ca space end cell equals cell space fraction numerator massa space Ca over denominator italic A subscript r space Ca end fraction end cell row blank equals cell space fraction numerator left parenthesis 74 comma 5 minus sign x right parenthesis space g over denominator 40 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell space fraction numerator left parenthesis 74 comma 5 minus sign x right parenthesis over denominator 40 end fraction mol end cell end table end style

begin mathsize 14px style 2 Na and 2 H subscript 2 O yields 2 Na O H and H subscript 2 x over 23 mol space space space space space space space space space space space space space space x over 23 mol end style  

begin mathsize 14px style space space space space space Ca space space space plus 2 H subscript 2 O yields Ca open parentheses O H close parentheses subscript 2 plus H subscript 2 fraction numerator left parenthesis 74 comma 5 minus sign x right parenthesis over denominator 40 end fraction mol space space space space space space space space space fraction numerator left parenthesis 74 comma 5 minus sign x right parenthesis over denominator 40 end fraction mol end style

 
Setelah nilai mol masing-masing larutan diperoleh, dapat dihitung nilai konsentrasi 100 mL basa total yang akan dititrasi dengan asam klorida 5 M sebanyak 140 mL.

begin mathsize 14px style Na O H yields Na to the power of plus sign and O H to the power of minus sign x over 23 mol space space space space space space space space space space space space space x over 23 mol end style 
 

begin mathsize 14px style Ca open parentheses O H close parentheses subscript 2 yields Ca to the power of 2 plus sign and 2 O H to the power of minus sign left parenthesis fraction numerator 74 comma 5 minus sign x over denominator 40 end fraction right parenthesis mol space space space space space space space space 2 left parenthesis fraction numerator 74 comma 5 minus sign x over denominator 40 end fraction right parenthesis mol end style 


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space basa space total space end cell equals cell space fraction numerator mol space basa space total over denominator V space total end fraction end cell row blank equals cell space fraction numerator begin display style x over 23 space mol plus fraction numerator left parenthesis 74 comma 5 minus sign x right parenthesis over denominator 20 end fraction space mol end style over denominator 0 comma 5 space L end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 1713 comma 5 minus sign 3 x over denominator 460 end fraction space mol end style over denominator 0 comma 5 space L end fraction end cell row blank equals cell space fraction numerator 1713 comma 5 minus sign 3 x over denominator 230 end fraction space M end cell end table end style


Setelah konsentrasi basa total diketahui, dapat dihitung nilai undefined melalui persamaan titrasi.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript a cross times V subscript a space end cell equals cell space M subscript b cross times V subscript b end cell row cell 5 space M cross times 140 space mL space end cell equals cell space fraction numerator 1713 comma 5 minus sign 3 x over denominator 230 end fraction space M cross times 100 space mL end cell row cell 1713 comma 5 minus sign 3 x end cell equals 1610 row cell 3 x space end cell equals cell space 103 comma 5 end cell row cell x space end cell equals cell space massa space Na space equals space 34 comma 5 space g end cell end table end style


Selanjutnya, volume total begin mathsize 14px style H subscript 2 end style yang dihasilkan dapat dihitung.

begin mathsize 14px style 2 Na and 2 H subscript 2 O yields 2 Na O H and H subscript 2 x over 23 mol space space space space space space space space space space space space x over 23 mol space space space space x over 46 mol end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript STP space end cell equals cell space mol space H subscript 2 cross times 22 comma 4 end cell row blank equals cell space fraction numerator 34 comma 5 over denominator 46 end fraction cross times 22 comma 4 end cell row blank equals cell space 16 comma 8 space L 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 V subscript STP space end cell equals cell space fraction numerator left parenthesis 74 comma 5 minus sign 34 comma 5 right parenthesis over denominator 40 end fraction mol cross times 22 comma 4 end cell row blank equals cell space 22 comma 4 space L end cell end table end style 

begin mathsize 14px style V subscript STPtotal equals 16 comma 8 plus 22 comma 4 equals 39 comma 2 space L end style   

Jadi, volume gas hidrogen total yang dihasilkan adalah 39,2 L.space 

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