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 $begin mathsize 14px style I to the power of minus sign end style$ dicampurkan 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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