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Hasil Kali Kelarutan ( K sp ​ ) PbSO 4 ​ adalah 2 × 1 0 − 8 . Jika 100 mL larutan Pb ( NO 3 ​ ) 2 ​ 0,03 M ditambahkan kedalam 400 ml larutan Na 2 ​ SO 4 ​ 0,04 M. Perkirakan apakah hasil reaksi akan membentuk endapan?

Hasil Kali Kelarutan  adalah . Jika 100 mL larutan  0,03 M ditambahkan kedalam 400 ml larutan  0,04 M. Perkirakan apakah hasil reaksi akan membentuk endapan?
 

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A. Budi

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

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Pembahasan

Dengan demikian, maka jawaban yang tepat adalah terbentuk endapan.

begin mathsize 14px style Pb open parentheses N O subscript 3 close parentheses subscript 2 open parentheses aq close parentheses and Na subscript 2 S O subscript 4 open parentheses aq close parentheses yields space Pb S O subscript 4 open parentheses s close parentheses and 2 Na N O subscript 3 open parentheses aq close parentheses  open square brackets Pb to the power of 2 plus sign close square brackets subscript 1 cross times V subscript 1 equals open square brackets Pb to the power of 2 plus sign close square brackets subscript 2 cross times V subscript 2 0 comma 03 cross times 100 equals open square brackets Pb to the power of 2 plus sign close square brackets subscript 2 cross times 500 open square brackets Pb to the power of 2 plus sign close square brackets subscript 2 equals 0 comma 006 space M  open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets subscript 1 cross times V subscript 1 equals open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets subscript 2 cross times V subscript 2 0 comma 04 cross times 400 equals open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets subscript 2 cross times 500 open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets subscript 2 equals 0 comma 032 space M  Pb S O subscript 4 open parentheses s close parentheses equilibrium Pb to the power of 2 plus sign open parentheses aq close parentheses and S O subscript 4 to the power of 2 minus sign end exponent open parentheses aq close parentheses Q subscript sp equals open square brackets Pb to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets Q subscript sp equals open parentheses 0 comma 006 close parentheses cross times open parentheses 0 comma 032 close parentheses Q subscript sp equals 1 comma 92 cross times 10 to the power of negative sign 4 end exponent  Q subscript sp greater than K subscript sp 1 comma 92 cross times 10 to the power of negative sign 4 end exponent greater than 2 cross times 10 to the power of negative sign 8 end exponent space left parenthesis terbentuk space endapan right parenthesis space end style

Dengan demikian, maka jawaban yang tepat adalah terbentuk endapan.

 

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