Roboguru

Suatu larutan mengandung ion  dan masing-masing 0,...

Suatu larutan mengandung ion undefined dan undefined masing-masing 0,1 M. Jika larutan tersebut ditetesi dengan larutan begin mathsize 14px style Na subscript 2 S O subscript 4 space 0 comma 1 space M point end style

Berapa persen begin mathsize 14px style Ba S O subscript 4 end style yang mengendap sebelum begin mathsize 14px style Ca S O subscript 4 end style mulai mengendap? Perubahan volume larutan karena penambahan larutan undefined dapat diabaikan.

(Ksp begin mathsize 14px style Ca S O subscript 4 space equals space 2 comma 4 x 10 to the power of negative sign 5 end exponent comma space dan space Ba S O subscript 4 space end subscript equals space 1 x 10 to the power of negative sign 10 end exponent end style )

Jawaban:

Reaksi peruraian kedua zat:
begin mathsize 14px style Ca S O subscript 4 yields and is yielded by Ca to the power of 2 plus sign and S O subscript 4 to the power of 2 minus sign end exponent Ba S O subscript 4 yields and is yielded by Ba to the power of 2 plus sign and S O subscript 4 to the power of 2 minus sign end exponent  end style

Karena keduanya terionisasi dengan jumlah ion yang sama, kita dapat membandingkan nilai Ksp secara langsung untuk melihat mana yang lebih mudah larut. Karena begin mathsize 14px style Ksp space Ca S O subscript 4 greater than Ksp space Ba S O subscript 4 end style maka begin mathsize 14px style Ca S O subscript 4 end style akan mudah larut dengan kata lain ia akan membentuk endapan belakangan.

Berapakah begin mathsize 14px style open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets end style ketika Ca S O subscript size 14px 4  mulai mengendap dan begin mathsize 14px style open square brackets Ca to the power of 2 plus sign close square brackets equals 0 comma 1 M end style?

Suatu larutan akan mulai membentuk endapan jika nilai Ksp = Hkki (Hasil kali kelarutan ion–ion)

begin mathsize 14px style Ksp space Ca S O subscript 4 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 space space space 2 comma 4.10 to the power of negative sign 5 end exponent equals open square brackets 10 to the power of negative sign 1 end exponent close square brackets open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets space space space space space space open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets equals 2 comma 4.10 to the power of negative sign 4 end exponent M end style

Dengan begin mathsize 14px style S O subscript 4 to the power of 2 minus sign end exponent end style mulai membentuk endapan, ketika itu pula Ba S O subscript size 14px 4sudah mengendap maka begin mathsize 14px style Ba to the power of 2 plus sign end style dapat dihitung:
Error converting from MathML to accessible text.

Persen begin mathsize 14px style Ba S O subscript 4 end style yang mengendap sebelum begin mathsize 14px style Ca S O subscript 4 end style mulai mengendap :

begin mathsize 14px style equals fraction numerator 4 comma 17.10 to the power of negative sign 7 end exponent over denominator 1.10 to the power of negative sign 5 end exponent end fraction x 100 percent sign equals 4 comma 17 percent sign end style

 


 

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