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Ke dalam 200 mL larutan  0,02 M ditambahkan 200 mL larutan  0,02 M. Jika , massa zat yang mengendap adalah .... (Ar Ca = 40 g/mol, S = 32 g/mol dan O = 16 g/mol)

Pertanyaan

Ke dalam 200 mL larutan begin mathsize 14px style Ca Cl subscript 2 end style 0,02 M ditambahkan 200 mL larutan begin mathsize 14px style Na subscript 2 S O subscript 4 end style 0,02 M. Jika begin mathsize 14px style K subscript sp space Ca S O subscript 4 equals 9 cross times 10 to the power of negative sign 6 end exponent end style, massa zat yang mengendap adalah .... (Ar Ca = 40 g/mol, S = 32 g/mol dan O = 16 g/mol)
 

  1. 0,544 gramspace 

  2. 0,741 gramspace 

  3. 0,804 gramspace 

  4. 0,849 gramspace 

  5. 0,920 gramspace 

Pembahasan Soal:

Padatan begin mathsize 14px style Ca S O subscript 4 end style yang terbentuk:

begin mathsize 14px style n subscript Ca Cl subscript 2 end subscript equals 0 comma 02 cross times 0 comma 2 equals 0 comma 004 space mol n subscript Na subscript 2 S O subscript 4 end subscript equals 0 comma 02 cross times 0 comma 2 equals 0 comma 004 space mol Ca Cl subscript 2 open parentheses aq close parentheses and Na subscript 2 S O subscript 4 open parentheses aq close parentheses yields Ca S O subscript 4 open parentheses s close parentheses and 2 Na Cl open parentheses aq close parentheses 0 comma 004 space space space space space space space space space space space space 0 comma 004 space space bottom enclose negative sign 0 comma 004 space space space space space space space space minus sign 0 comma 004 space space space space space space space space space space space space space plus 0 comma 004 space space space space space space space space plus 0 comma 008 end enclose 0 space space space space space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space space space space space space space space space 0 comma 004 space space space space space space space space space space space space 0 comma 008 m subscript Ca S O subscript 4 end subscript equals 0 comma 004 cross times 136 equals 0 comma 544 space g end style

Padatan begin mathsize 14px style Ca S O subscript 4 end style yang larut:

begin mathsize 14px style Ca S O subscript 4 open parentheses s close parentheses equilibrium Ca 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 bottom enclose negative sign s space space space space space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space s end enclose  K subscript sp equals open square brackets Ca 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 K subscript sp equals s squared 9 cross times 10 to the power of negative sign 6 end exponent equals s squared s equals 3 cross times 10 to the power of negative sign 3 end exponent space M n subscript Ca S O subscript 4 end subscript equals 3 cross times 10 to the power of negative sign 3 end exponent cross times 0 comma 4 n subscript Ca S O subscript 4 end subscript equals 1 comma 2 cross times 10 to the power of negative sign 3 end exponent space mol m subscript Ca S O subscript 4 end subscript equals 1 comma 2 cross times 10 to the power of negative sign 3 end exponent cross times 136 m subscript Ca S O subscript 4 end subscript equals 0 comma 1632 space g end style

Padatan begin mathsize 14px style Ca S O subscript 4 end style yang tersisa:

begin mathsize 14px style m subscript Ca S O subscript 4 end subscript equals 0 comma 544 minus sign 0 comma 1632 m subscript Ca S O subscript 4 end subscript equals 0 comma 3808 space g end style

Dengan demikian, maka jawaban yang tepat adalah 0,3808 g.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 April 2021

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