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Sebanyak 100 mL larutan Pb(NO3)2 0,4 M dicampurkan dengan 100 mL larutan K2SO4 0,4 M. Jika Ksp PbSO4 = 4 x 10-8, maka massa PbSO4 yang mengendap sebanyak .... (Ar Pb = 207; S = 32; O = 16; K = 39)

Pertanyaan

Sebanyak 100 mL larutan Pb(NO3)0,4 M dicampurkan dengan 100 mL larutan K2SO0,4 M. Jika Ksp PbSO4 = 4 x 10-8, maka massa PbSO4 yang mengendap sebanyak ....

(Ar Pb = 207; S = 32; O = 16; K = 39)

  1. 12,12 gram

  2. 24,24 gram

  3. 30,30 gram

  4. 60,60 gram

  5. 303 gram

Pembahasan Soal:

Kesetimbangan reaksi :

begin mathsize 14px style text PbSO end text subscript text 4 end text end subscript text ↔Pb end text to the power of text 2+ end text end exponent text +SO end text subscript text 4 end text end subscript to the power of text 2- end text end exponent  text Pb end text open parentheses text NO end text subscript text 3 end text end subscript close parentheses subscript text 2 end text end subscript text  terionisasi sempurna: end text  text mol Pb end text open parentheses text NO end text subscript text 3 end text end subscript close parentheses subscript text 2 end text end subscript text =0,4 M× end text 100 over 1000 text mL=0,04 mL end text  text Pb end text open parentheses text NO end text subscript text 3 end text end subscript close parentheses subscript text 2 end text end subscript text  (aq)→ end text Pb to the power of text 2+ end text end exponent text (aq)+2NO end text subscript text 3 end text end subscript text (aq) end text  text 0,04 mol             0,04 mol    0,08 mol end text    text K end text subscript text 2 end text end subscript text SO end text subscript text 4 end text end subscript text  terionisasi sempurna: end text  text mol  end text text K end text subscript text 2 end text end subscript text SO end text subscript text 4 end text end subscript text =0,4 M end text cross times 100 over 1000 text mL=0,04 mL end text  text K end text subscript text 2 end text end subscript text SO end text subscript text 4 end text end subscript text  (aq)→2K end text to the power of text + end text end exponent text (aq)+SO end text subscript text 4 end text end subscript to the power of text 2- end text end exponent text (aq) end text  text 0,04 mol        0,08 mol   0,04 mol end text end style

Untuk total volume larutan 200 mL atau 0,2 L, diperoleh:

begin mathsize 14px style open square brackets text Pb end text to the power of text 2+ end text end exponent close square brackets subscript text awal end text end subscript equals fraction numerator text 0,04 mol end text over denominator text 0,2 L end text end fraction equals text 0,2 mol/L end text  open square brackets text SO end text subscript 4 to the power of text 2- end text end exponent close square brackets subscript text awal end text end subscript equals fraction numerator text 0,04 mol end text over denominator text 0,2 L end text end fraction equals text 0,2 mol/L end text  text Hasil kali ion-ion= end text open square brackets text Pb end text to the power of text 2+ end text end exponent close square brackets subscript text awal end text end subscript text × end text open square brackets text SO end text subscript 4 to the power of text 2- end text end exponent close square brackets subscript text awal end text end subscript text =0,2×0,2=0,04 mol end text to the power of text 2 end text end exponent text /L end text to the power of text 2 end text end exponent end style

(hasil kali ion-ion PbSO4 > Ksp PbSO4, maka terbukti terbentuk endapan)

Kelarutan PbSO4begin mathsize 14px style square root of text 0,04 mol end text to the power of text 2 end text end exponent text /L end text to the power of text 2 end text end exponent end root text =0,2 mol/L end text end style
dalam 1 L terdapat 0,2 mol PbSO4

Jadi, dalam 0,2 L larutan, mol zat yang dapat mengendap adalah :
begin mathsize 14px style fraction numerator text 0,2 L end text over denominator 1 space text L end text end fraction text ×0,2 mol=0,04 mol end text end style

Massa PbSO4 yang mengendap = 0,04 x 303 = 12,12 gram.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puji

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 16 Desember 2020

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Pembahasan Soal:

 

Ca(NO3)2 + Na2CO3

rightwards arrow

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Mula

0,5 mmol   0,5 mmol

 

 

Reaksi

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+0,5 mmol  +1,0 mmol

Sisa

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table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets Mg to the power of 2 plus sign close square brackets cross times open square brackets O H to the power of minus sign close square brackets squared end cell equals cell K subscript sp end cell row cell fraction numerator 0 comma 04 over denominator 2 end fraction cross times open square brackets O H to the power of minus sign close square brackets squared end cell equals cell K subscript sp end cell row cell 0 comma 02 cross times open square brackets O H to the power of minus sign close square brackets squared end cell equals cell 3 comma 2 cross times 10 to the power of negative sign 11 end exponent end cell row cell open square brackets O H to the power of minus sign close square brackets squared end cell equals cell fraction numerator 3 comma 2 cross times 10 to the power of negative sign 11 end exponent over denominator 0 comma 02 end fraction end cell row cell open square brackets O H to the power of minus sign close square brackets squared end cell equals cell 1 comma 6 cross times 10 to the power of negative sign 9 end exponent end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of 1 comma 6 cross times 10 to the power of negative sign 9 end exponent end root end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 4 cross times 10 to the power of negative sign 5 end exponent end cell row pOH equals cell negative sign log space open square brackets O H to the power of minus sign close square brackets end cell row pOH equals cell negative sign log space open parentheses 4 cross times 10 to the power of negative sign 5 end exponent close parentheses end cell row pOH equals cell 4 comma 4 end cell row pH equals cell 14 minus sign pOH end cell row pH equals cell 14 minus sign 4 comma 4 end cell row bold pH bold equals cell bold 9 bold comma bold 6 end cell end table


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Pembahasan Soal:

begin mathsize 14px style Ag N O subscript bold 3 space 0 comma 4 space M comma space V space equals space 0 comma 1 space L M space equals space n over V n space equals space M point V n space equals space 0 comma 4 space space mol forward slash L space x space 0 comma 1 space L n space equals space 0 comma 04 space mol space  Ca Cr O subscript bold 4 space 0 comma 4 space M comma space V space equals space 0 comma 1 space L M space equals space n over V n space equals space M point V n space equals space 0 comma 4 space space mol forward slash L space x space 0 comma 1 space L n space equals space 0 comma 04 space mol space end style 

Persamaan reaksinya adalah 

 

Error converting from MathML to accessible text.
 

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Pembahasan Soal:

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Ca Cl subscript 2 space space yields Ca to the power of 2 plus sign space space space plus space space 2 Cl to the power of minus sign 0 comma 01 space M space space space space 0 comma 01 space M space space space 0 comma 01 space M  maka space left square bracket Ca to the power of 2 plus sign space right square bracket equals 0 comma 01 space M   Na subscript 2 C subscript 2 O subscript 4 yields 2 Na to the power of plus sign and C subscript 2 O subscript 4 to the power of 2 minus sign 0 comma 002 space M space space space 0 comma 002 space M space space 0 comma 002 space M space space space space space  maka space open square brackets C subscript 2 O subscript 4 to the power of 2 minus sign close square brackets equals space 0 comma 002 space M space

 

table attributes columnalign right center left columnspacing 0px end attributes row cell Qc space Ca C subscript 2 O subscript 4 end cell equals cell open square brackets Ca to the power of 2 plus sign close square brackets open square brackets C subscript 2 O subscript 4 to the power of 2 minus sign close square brackets end cell row blank equals cell left parenthesis 0 comma 01 right parenthesis left parenthesis 0 comma 002 right parenthesis end cell row blank equals cell open parentheses 1 x 10 to the power of negative sign 2 end exponent close parentheses left parenthesis 2 x 10 to the power of negative sign 3 end exponent right parenthesis end cell row blank equals cell 2 x 10 to the power of negative sign 5 end exponent end cell row blank blank blank row blank blank blank end table

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Pembahasan Soal:

 

Ba left parenthesis NO subscript 3 right parenthesis subscript 2 space left parenthesis aq right parenthesis

  +

Na subscript 2 CrO subscript 4 space left parenthesis aq right parenthesis

rightwards arrow

BaCrO subscript 4 space left parenthesis straight s right parenthesis

 +

2 space NaNO subscript 3 space left parenthesis aq right parenthesis

Mula

1 mmol

 

1 mmol

 

 

 

 

Reaksi

- 1 mmol

 

- 1 mmol

 

+1 mmol

 

+2 mmol

Sisa

  1.  

 

  1.  

 

1 mmol

 

2 mmol

M a s s a space B a C r O subscript 4 equals m o l cross times M r equals 0 comma 001 space m o l cross times 253 space g divided by m o l equals 0 comma 253 space g r a m 

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