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Sebanyak 200 mL larutan SrCl 2 ​ 0,05 M ditambahkan ke dalam 200 mL larutan K 2 ​ CrO 4 ​ 0,05 M. Berapakah massa zat yang mengendap? ( K sp ​ SrCrO 4 ​ = 3 , 6 × 1 0 − 5 ; A r ​ : Sr = 88 g mol − 1 , Cl = 35 , 5 g mol − 1 , K = 39 g mol − 1 , Cr = 52 g mol − 1 , dan O = 16 g mol − 1 ) .

Sebanyak 200 mL larutan  0,05 M ditambahkan ke dalam 200 mL larutan  0,05 M. Berapakah massa zat yang mengendap? , dan . 

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massa zatyang mengendap adalah 2,04 gram.

massa zat yang mengendap adalah 2,04spacegram.

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Reaksi antara dengan akan terjadi sesuai reaksi berikut ini. akan mengendap jika Tahap 1. Menentukan terbentuk atau tidaknya endapan Diketahui : setelah pencampuran setelah pencampuran Terbentuk endapan , karena . Tahap 2. Menentukan massa Reaksi antara 200 mL larutan 0,05 M dengan 200 mL larutan 0,05 M memiliki stoikiometri sebagai berikut. Jadi, massa zatyang mengendap adalah 2,04 gram.

Reaksi antara begin mathsize 14px style Sr Cl subscript 2 end style dengan begin mathsize 14px style K subscript 2 Cr O subscript 4 end style akan terjadi sesuai reaksi berikut ini.

begin mathsize 14px style Sr Cl subscript 2 and K subscript 2 Cr O subscript 4 yields Sr Cr O subscript 4 and 2 K Cl end style 

begin mathsize 14px style Sr Cr O subscript 4 end style akan mengendap jika begin mathsize 14px style Q subscript sp space Sr Cr O subscript 4 greater than K subscript sp space Sr Cr O subscript 4 end style

Tahap 1. Menentukan terbentuk atau tidaknya endapan undefined 

Diketahui :

begin mathsize 14px style space space space space space open square brackets Sr Cl subscript 2 close square brackets equals 0 comma 05 space M space space space space V space Sr Cl subscript 2 equals 200 space mL space open square brackets K subscript 2 Cr O subscript 4 close square brackets equals 0 comma 05 space M V space K subscript 2 Cr O subscript 4 equals 200 space mL end style 

begin mathsize 14px style straight V subscript 2 equals straight V space total space space space space equals straight V space SrCl subscript 2 plus straight V space straight K subscript 2 CrO subscript 4 space space space space equals 200 space mL plus 200 space mL space space space space equals 400 space mL end style 

begin mathsize 14px style open square brackets Sr Cl subscript 2 close square brackets end style setelah pencampuran

begin mathsize 14px style space space space space space space space space space space space space space space space space space space straight V subscript 1 straight M subscript 1 equals straight V subscript 2 straight M subscript 2 200 space mL cross times 0 comma 05 space straight M equals 400 space mL cross times straight M subscript 2 space space space space space space space space space space space space space space space space space space space space space space straight M subscript 2 equals 0 comma 025 thin space straight M end style 

begin mathsize 14px style space space space Sr Cl subscript 2 space space rightwards arrow space Sr to the power of 2 plus sign space space plus space space 2 Cl to the power of minus sign 0 comma 025 space M space space space 0 comma 025 space M space space space 0 comma 05 space M end style 

begin mathsize 14px style open square brackets straight K subscript 2 CrO subscript 4 close square brackets end style setelah pencampuran

begin mathsize 14px style space space space space space space space space space space space space space space space space space space straight V subscript 1 straight M subscript 1 equals straight V subscript 2 straight M subscript 2 200 space mL cross times 0 comma 05 space straight M equals 400 space mL cross times straight M subscript 2 space space space space space space space space space space space space space space space space space space space space space space straight M subscript 2 equals 0 comma 025 thin space straight M end style 

begin mathsize 14px style space K subscript 2 Cr O subscript 4 yields space space 2 K to the power of plus sign space space plus space Cr O subscript 4 to the power of 2 minus sign end exponent 0 comma 025 space M space space space space 0 comma 05 space M space space space space 0 comma 025 space M end style 

begin mathsize 14px style straight Q subscript sp space SrCrO subscript 4 equals open square brackets Sr to the power of 2 plus end exponent close square brackets open square brackets CrO subscript 4 to the power of 2 minus end exponent close square brackets space space space space space space space space space space space space space space space space space space equals left parenthesis 0 comma 025 right parenthesis cross times left parenthesis 0 comma 025 right parenthesis space space space space space space space space space space space space space space space space space space equals 6 comma 25 cross times 10 to the power of negative 4 end exponent straight K subscript sp space SrCrO subscript 4 equals 3 comma 6 cross times 10 to the power of negative 5 end exponent straight Q subscript sp space SrCrO subscript 4 equals 62 comma 5 cross times 10 to the power of negative 5 end exponent end style 

Terbentuk endapan undefined, karena begin mathsize 14px style straight Q subscript sp space SrCrO subscript 4 greater than straight K subscript sp space SrCrO subscript 4 end style.

Tahap 2. Menentukan massa undefined 

Reaksi antara 200 mL larutan begin mathsize 14px style Sr Cl subscript 2 end style 0,05 M dengan 200 mL larutan begin mathsize 14px style K subscript 2 Cr O subscript 4 end style 0,05 M memiliki stoikiometri sebagai berikut.

begin mathsize 14px style open square brackets Sr Cl subscript 2 close square brackets equals fraction numerator n space Sr Cl subscript 2 over denominator V space Sr Cl subscript 2 end fraction n space Sr Cl subscript 2 equals open square brackets Sr Cl subscript 2 close square brackets cross times V space Sr Cl subscript 2 n space Sr Cl subscript 2 equals 0 comma 05 space mol space L to the power of negative sign 1 end exponent cross times 200 space mL n space Sr Cl subscript 2 equals 0 comma 05 space mol space L to the power of negative sign 1 end exponent cross times 0 comma 2 space L n space Sr Cl subscript 2 equals 0 comma 01 space mol end style 

begin mathsize 14px style space space left square bracket straight K subscript 2 CrO subscript 4 right square bracket equals fraction numerator straight n space straight K subscript 2 CrO subscript 4 over denominator straight V space straight K subscript 2 CrO subscript 4 end fraction straight n space straight K subscript 2 CrO subscript 4 equals left square bracket straight K subscript 2 CrO subscript 4 right square bracket cross times straight V space straight K subscript 2 CrO subscript 4 straight n space straight K subscript 2 CrO subscript 4 equals 0 comma 05 space mol space straight L to the power of negative 1 end exponent cross times 200 space mL straight n space straight K subscript 2 CrO subscript 4 equals 0 comma 05 space mol space straight L to the power of negative 1 end exponent cross times 0 comma 2 space straight L straight n space straight K subscript 2 CrO subscript 4 equals 0 comma 01 space mol end style 

 begin mathsize 14px style Mr space SrCrO subscript 4 equals Ar space Sr plus Ar thin space Cr plus space left parenthesis 4 cross times Ar space straight O right parenthesis Mr space SrCrO subscript 4 equals 88 space straight g space mol to the power of negative 1 end exponent plus 52 space straight g space mol to the power of negative 1 end exponent plus left parenthesis 4 cross times 16 space straight g space mol to the power of negative 1 end exponent right parenthesis Mr space SrCrO subscript 4 equals 140 space straight g space mol to the power of negative 1 end exponent plus space 64 space straight g space mol to the power of negative 1 end exponent Mr space SrCrO subscript 4 equals 204 space straight g space mol to the power of negative 1 end exponent  massa space SrCrO subscript 4 space yang space mengendap equals 0 comma 01 space mol cross times 204 space straight g space mol to the power of negative 1 end exponent massa space SrCrO subscript 4 space yang space mengendap equals 2 comma 04 space straight g 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 space space space space space space space end style 

Jadi, massa zat yang mengendap adalah 2,04spacegram.

Konsep Kilat

Kelarutan

Hasil Kali Kelarutan (Kₛₚ)

Faktor-Faktor Kelarutan dan Pengendapan

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