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Suatu larutan A mengandung ion  dan ion . Jika lar...

Suatu larutan A mengandung ion begin mathsize 14px style Cu to the power of 2 plus sign space 2 comma 12 cross times 10 to the power of negative sign 4 end exponent space M end style dan ion begin mathsize 14px style Pb to the power of 2 plus sign space 2 comma 8 cross times 10 to the power of negative sign 3 end exponent space M end style. Jika larutan yang mengandung ion begin mathsize 14px style I to the power of minus sign end styledicampurkan ke dalam larutan A secara terus-menerus, jelaskan endapan begin mathsize 14px style Pb I subscript 2 space left parenthesis K subscript sp equals 1 comma 4 cross times 10 to the power of negative sign 8 end exponent right parenthesis end style atau endapan begin mathsize 14px style Cu I space left parenthesis K subscript sp equals 5 comma 3 cross times 10 to the power of negative sign 12 end exponent right parenthesis end style yang akan terbentuk lebih dahulu! 

Jawaban:

Syarat terbentuknya endapan adalah Qsp > Ksp.

Untuk mengetahui endapan mana yang lebih dahulu terbentuk, maka harus mencari konsentrasi undefined pada masing-masing garam.

Konsentrasi undefined pada garam begin mathsize 14px style Pb I subscript 2 end style dihitung menggunakan persamaan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript sp space Pb I subscript 2 end cell equals cell open square brackets Pb to the power of 2 plus sign close square brackets open square brackets I to the power of minus sign close square brackets squared end cell row cell 1 comma 4 cross times 10 to the power of negative sign 8 end exponent end cell equals cell open parentheses 2 comma 8 cross times 10 to the power of negative sign 3 end exponent close parentheses cross times open square brackets I to the power of minus sign close square brackets squared end cell row cell open square brackets I to the power of minus sign close square brackets end cell equals cell square root of fraction numerator 1 comma 4 cross times 10 to the power of negative sign 8 end exponent over denominator 2 comma 8 cross times 10 to the power of negative sign 3 end exponent end fraction end root end cell row blank equals cell 2 comma 23 cross times 10 to the power of negative sign 3 end exponent space M end cell end table end style  

Konsentrasi begin mathsize 14px style I to the power of minus sign end style pada garam CuI dihitung menggunakan persamaan berikut. 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Q subscript sp space Cu I end cell equals cell open square brackets Cu to the power of plus sign close square brackets open square brackets I to the power of minus sign close square brackets end cell row cell 5 comma 3 cross times 10 to the power of negative sign 12 end exponent end cell equals cell open parentheses 2 comma 12 cross times 10 to the power of negative sign 4 end exponent close parentheses cross times open square brackets I to the power of minus sign close square brackets end cell row cell open square brackets I to the power of minus sign close square brackets end cell equals cell fraction numerator 5 comma 3 cross times 10 to the power of negative sign 12 end exponent over denominator 2 comma 12 cross times 10 to the power of negative sign 4 end exponent end fraction end cell row blank equals cell 2 comma 37 cross times 10 to the power of negative sign 8 end exponent space M end cell end table end style 

Berdasarkan konsentrasi undefined yang dibutuhkan untuk mengendapkan garam undefined dan garam CuI, dibutuhkan konsentrasi undefined yang lebih sedikit pada garam CuI.

Jadi, dapat disimpulkan endapan garam CuI lebih dulu mengendap dibandingkan dengan garam undefined

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