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Dalam satu liter larutan terdapat garam Kalsium Klorida memiliki konsentrasi 0,01 M. Jika ditambahkan 67 mg garam Natrium Oksalat sebanyak 250 mL. (Mr Natrium Oksalat =134 ; Ksp Kalsium Oksalat=2,3 x10−9). Apakah terbentuk endapan ?

Pertanyaan

Dalam satu liter larutan terdapat garam Kalsium Klorida memiliki konsentrasi 0,01 M. Jika ditambahkan 67 mg garam Natrium Oksalat sebanyak 250 mL.

(Mr space Natrium space Oksalat space equals 134 ; Ksp space Kalsium space Oksalat equals 2 comma 3 space x 10 to the power of negative sign 9 end exponent).

Apakah terbentuk endapan ?

G. Suprobo

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Pembahasan

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Natrium space oksalat end cell equals cell fraction numerator 67 x 10 to the power of negative sign 3 end exponent space mol over denominator 134 space g forward slash mol end fraction end cell row blank equals cell 5 x 10 to the power of negative sign 4 end exponent space mol end cell row cell Molaritas space Natrium space oksalat end cell equals cell fraction numerator 5 x 10 to the power of negative sign 4 end exponent space mol over denominator 0 comma 25 space L end fraction end cell row blank equals cell 2 x 10 to the power of negative sign 3 end exponent M end cell row cell Molaritas space kalsium space klorida end cell equals cell 0 comma 01 space M end cell end table

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

Oleh karena nilai Qc greater than Kc, maka kalsium oksalat akan mengendap.

21

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