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Sebanyak 100 mL larutan begin mathsize 14px style Pb open parentheses N O subscript 3 close parentheses subscript 2 end style begin mathsize 14px style 2 cross times 10 to the power of negative sign 3 end exponent end style mol ditambahkan kedalam 100 mL larutan netral merupakan campuran dari larutan-larutan garam meliputi begin mathsize 14px style K Cl end style (begin mathsize 14px style 10 to the power of negative sign 3 end exponent end style mol), begin mathsize 14px style K subscript 2 Cr O subscript 4 end style (begin mathsize 14px style 10 to the power of negative sign 3 end exponent end style mol), dan begin mathsize 14px style K subscript 2 S O subscript 4 end style (begin mathsize 14px style 10 to the power of negative sign 3 end exponent end style mol). Campuran tersebut diaduk secara merata. Jika Ksp : begin mathsize 14px style Pb Cl subscript 2 end style = begin mathsize 14px style 1 comma 7 cross times 10 to the power of negative sign 5 end exponent end style; begin mathsize 14px style Pb Cr O subscript 4 end style = begin mathsize 14px style 2 cross times 10 to the power of negative sign 14 end exponent end style; dan begin mathsize 14px style Pb S O subscript 4 end style = begin mathsize 14px style 2 cross times 10 to the power of negative sign 8 end exponent end style, endapan yang terjadi yaitu ...

  1. begin mathsize 14px style Pb S O subscript 4 end style saja    space 

  2. begin mathsize 14px style Pb Cl subscript 2 end style saja    space 

  3. begin mathsize 14px style Pb Cr O subscript 4 end style saja    space 

  4. begin mathsize 14px style Pb Cl subscript 2 end style dan begin mathsize 14px style Pb Cr O subscript 4 end style     space 

  5. begin mathsize 14px style Pb Cr O subscript 4 end style dan begin mathsize 14px style Pb S O subscript 4 end style    space 

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Jawaban terverifikasi

Jawaban

jawaban yang sesuai adalah C.space  

Pembahasan

begin mathsize 14px style open square brackets Pb to the power of 2 plus sign close square brackets end style setelah direaksikan
 

begin mathsize 14px style open square brackets Pb to the power of 2 plus sign close square brackets equals fraction numerator 2 cross times 10 to the power of negative sign 3 end exponent over denominator 200 end fraction equals 10 to the power of negative sign 5 end exponent end style 
 

begin mathsize 14px style open square brackets Cl to the power of minus sign close square brackets comma space open square brackets Cr O subscript 4 to the power of 2 minus sign close square brackets comma space open square brackets S O subscript 4 to the power of 2 minus sign close square brackets end style setelah direaksikan
 

begin mathsize 14px style open square brackets Cl to the power of minus sign close square brackets equals 10 to the power of negative sign 3 end exponent over 200 equals 5 cross times 10 to the power of negative sign 6 end exponent open square brackets Cr O subscript 4 to the power of minus sign close square brackets equals 10 to the power of negative sign 3 end exponent over 200 equals 5 cross times 10 to the power of negative sign 6 end exponent open square brackets S O subscript 4 to the power of 2 minus sign close square brackets equals 10 to the power of negative sign 3 end exponent over 200 equals 5 cross times 10 to the power of negative sign 6 end exponent end style 
 

Menentukan senyawa yang terbentuk mengendap atau larut
 

begin mathsize 14px style Q subscript c space Pb Cl subscript 2 double bond open square brackets Pb to the power of 2 plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets squared Q subscript c space Pb Cl subscript 2 equals left square bracket 10 to the power of negative sign 5 end exponent right square bracket left square bracket 5 cross times 10 to the power of negative sign 6 end exponent right square bracket squared Q subscript c space Pb Cl subscript 2 equals 2 comma 5 cross times 10 to the power of negative sign 16 end exponent  Q subscript c space Pb Cl subscript 2 space less than space K subscript sp space Pb Cl subscript 2 end style 

Tidak terbentuk endapan (larut)
 

begin mathsize 14px style Q subscript c space Pb Cr O subscript 4 double bond open square brackets Pb to the power of 2 plus sign close square brackets open square brackets Cr O subscript 4 close square brackets Q subscript c space Pb Cr O subscript 4 equals left square bracket 10 to the power of negative sign 5 end exponent right square bracket left square bracket 5 cross times 10 to the power of negative sign 6 end exponent right square bracket Q subscript c space Pb Cr O subscript 4 equals 5 cross times 10 to the power of negative sign 11 end exponent  Q subscript c space Pb Cr O subscript 4 space greater than space K subscript sp space Pb Cr O subscript 4 end style 

Terbentuk endapan
 

begin mathsize 14px style Q subscript c space Pb S O subscript 4 double bond open square brackets Pb to the power of 2 plus sign close square brackets open square brackets S O subscript 4 close square brackets Q subscript c space Pb S O subscript 4 equals left square bracket 10 to the power of negative sign 5 end exponent right square bracket left square bracket 5 cross times 10 to the power of negative sign 6 end exponent right square bracket Q subscript c space Pb S O subscript 4 equals 5 cross times 10 to the power of negative sign 11 end exponent  Q subscript c space Pb S O subscript 4 space less than space K subscript sp space Pb S O subscript 4 end style 

Tidak terbentuk endapan (larut)
 

Jadi, jawaban yang sesuai adalah C.space  

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