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Jika244 mg AlPO 4 ​ dapat larut dalam 200 mL larutan H 3 ​ PO 4 ​ 0 , 05 M .Tetapan hasil kelarutan ( A r ​ Al = 27, P = 31, O = 16)adalah ….

Jika 244 mg  dapat larut dalam 200 mL larutan . Tetapan hasil kelarutan begin mathsize 14px style Al P O subscript 4 end style ( Al = 27, P = 31, O = 16) adalah ….space 

  1. begin mathsize 14px style 1 cross times 10 to the power of negative sign 2 end exponent end style 

  2. begin mathsize 14px style 1 cross times 10 to the power of negative sign 3 end exponent end style 

  3. begin mathsize 14px style 5 cross times 10 to the power of negative sign 4 end exponent end style 

  4. begin mathsize 14px style 5 cross times 10 to the power of negative sign 5 end exponent end style 

  5. begin mathsize 14px style 5 cross times 10 to the power of negative sign 6 end exponent end style  

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Menentukan mol . Menentukan konsentrasi . Menentukan konsentrasi berdasarkan persamaan reaksi berikut. Menentukan konsentrasi dari . Menentukan dalam . Jadi, jawaban yang benar adalah C.

Menentukan mol begin mathsize 14px style Al P O subscript 4 end style
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell m over italic M subscript r end cell row n equals cell fraction numerator m over denominator italic A subscript r space Al plus italic A subscript r space P plus left parenthesis italic A subscript r space O cross times 4 right parenthesis end fraction end cell row n equals cell fraction numerator 244 cross times 10 to the power of negative sign 3 end exponent space g over denominator 122 space g forward slash mol end fraction end cell row n equals cell 2 cross times 10 to the power of negative sign 3 end exponent space mol end cell end table end style 
 

Menentukan konsentrasi begin mathsize 14px style Al P O subscript 4 end style
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row M equals cell n over V end cell row M equals cell fraction numerator 2 cross times 10 to the power of negative sign 3 end exponent space mol over denominator 0 comma 2 end fraction end cell row M equals cell 10 to the power of negative sign 2 end exponent space mol end cell end table end style 
 

Menentukan konsentrasi begin mathsize 14px style Al to the power of 3 plus sign end style berdasarkan persamaan reaksi berikut.
 

begin mathsize 14px style Al P O subscript 4 yields Al to the power of 3 plus sign and P O subscript 4 to the power of 3 minus sign end exponent end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space Al to the power of 3 plus sign end cell equals cell fraction numerator koefisien space Al to the power of 3 plus sign over denominator koefisien space Al P O subscript 4 end fraction cross times M space Al P O subscript 4 end cell row cell M space Al to the power of 3 plus sign end cell equals cell 1 over 1 cross times 10 to the power of negative sign 2 end exponent space M end cell row cell M space Al to the power of 3 plus sign end cell equals cell 10 to the power of negative sign 2 end exponent space M end cell end table end style 
 

Menentukan konsentrasi begin mathsize 14px style P O subscript 4 to the power of 3 minus sign end exponent end style dari begin mathsize 14px style H subscript 3 P O subscript 4 space end subscript 0 comma 05 space M end style
 

begin mathsize 14px style H subscript 3 P O subscript 4 equilibrium 3 H to the power of plus sign and P O subscript 4 to the power of 3 minus sign end exponent end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space P O subscript 4 to the power of 3 minus sign end exponent end cell equals cell fraction numerator koefisien space P O subscript 4 to the power of 3 minus sign end exponent over denominator koefisien space H subscript 3 P O subscript 4 end fraction cross times M space H subscript 3 P O subscript 4 end cell row cell M space P O subscript 4 to the power of 3 minus sign end exponent end cell equals cell 1 over 1 cross times 0 comma 05 space M end cell row cell M space P O subscript 4 to the power of 3 minus sign end exponent end cell equals cell 0 comma 05 space M end cell end table end style 
 

Menentukan begin mathsize 14px style K subscript sp end style begin mathsize 14px style Al P O subscript 4 end style dalam begin mathsize 14px style H subscript 3 P O subscript 4 space end subscript 0 comma 05 space M end style
 

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 Al to the power of 3 plus sign close square brackets open square brackets P O subscript 4 to the power of 3 minus sign end exponent close square brackets end cell row cell K subscript sp end cell equals cell open square brackets 10 to the power of negative sign 2 end exponent close square brackets open square brackets 0 comma 05 close square brackets end cell row cell K subscript sp end cell equals cell 5 cross times 10 to the power of negative sign 4 end exponent end cell end table end style 
 

Jadi, jawaban yang benar adalah C.undefined 

 

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