Pertanyaan

Ke dalam 100 mL larutan netral yang merupakan campuran dari larutan garam KCl 0,001 M, Na 2 CrO 4 0,001 M, dan K 2 SO 4 0,001 M ditambah 100 mL larutan Pb(NO 3 ) 2 0,002 M. Campuran diaduk sampai tercampur rata. Ksp PbCl 2 = 1,7.10 -5 Ksp PbCrO 4 = 1,8.10 -14 Ksp PbSO 4 = 1,8.10 -8 Endapan yang terjadi adalah garam-garam ….

Ke dalam 100 mL larutan netral yang merupakan campuran dari larutan garam KCl 0,001 M, Na₂CrO₄ 0,001 M, dan K₂SO₄ 0,001 M ditambah 100 mL larutan Pb(NO₃)₂ 0,002 M. Campuran diaduk sampai tercampur rata.

Ksp PbCl₂ = 1,7.10^-5

Ksp PbCrO₄ = 1,8.10^-14

Ksp PbSO₄ = 1,8.10^-8

Endapan yang terjadi adalah garam-garam ….

PbCl₂ saja
PbSO₄ saja
PbCrO₄ saja
PbCrO₄ dan PbSO₄
PbCl₂ dan PbSO₄

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S. Utari

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

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Pembahasan

Diketahui: [KCl] = 0,001 M ; [Na 2 CrO 4 ] = 0,001 M ; [K 2 SO 4 ] = 0,001 M [Pb(NO 3 ) 2 ] =0,002 M ; VolumePb(NO 3 ) 2 = 100 mL Ksp PbCl 2 = 1,7.10 -5 Ksp PbCrO 4 = 1,8.10 -14 Ksp PbSO 4 = 1,8.10 -8 Dijawab: V KCl = v Na 2 CrO 4 = v K 2 SO 4 = 100 mL M KCl = M Na 2 CrO 4 = M K 2 SO 4 = 0,001 M V Pb(NO 3 ) 2 = 100 mL M Pb(NO 3 ) 2 = 0,002 M V total = 200 mL Sehingga, garam yang mengendap adalahPbCrO 4 dan PbSO 4.

Diketahui: [KCl] = 0,001 M ; [Na₂CrO₄] = 0,001 M ; [K₂SO₄] = 0,001 M

[Pb(NO₃)₂] = 0,002 M ; Volume Pb(NO₃)₂ = 100 mL

Ksp PbCl₂ = 1,7.10^-5

Ksp PbCrO₄ = 1,8.10^-14

Ksp PbSO₄= 1,8.10^-8

Dijawab:

V KCl = v Na₂CrO₄ = v K₂SO₄ = 100 mL

M KCl = M Na₂CrO₄ = M K₂SO₄ = 0,001 M

V Pb(NO₃)₂= 100 mL

M Pb(NO₃)₂= 0,002 M

V total = 200 mL

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$begin mathsize 12px style P b S O subscript 4 space space space space space space space rightwards harpoon over leftwards harpoon space space space space space space space P b to the power of 2 plus space end exponent space space space plus space space space space space space space S O subscript 4 to the power of 2 minus end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space 0 comma 2 space m m o l space space space space space space space space space 0 comma 1 space m m o l space Q s p space P b S O subscript 4 space equals space left square bracket P b to the power of 2 plus end exponent right square bracket left square bracket S O subscript 4 to the power of 2 minus end exponent right square bracket Q s p space P b S O subscript 4 space equals left parenthesis fraction numerator n space P b to the power of 2 plus end exponent over denominator V o l u m e space t o t a l end fraction right parenthesis left parenthesis fraction numerator n space S O subscript 4 to the power of 2 minus end exponent over denominator V o l u m e space t o t a l end fraction right parenthesis Q s p space P b S O subscript 4 space equals left parenthesis fraction numerator 0 comma 2 space m m o l over denominator 200 space m L end fraction right parenthesis left parenthesis fraction numerator 0 comma 1 space m m o l over denominator 200 space m L end fraction right parenthesis Q s p space P b S O subscript 4 space equals left parenthesis 0 comma 001 right parenthesis left parenthesis 0 comma 0005 right parenthesis space Q s p space P b S O subscript 4 space equals 5 space x space 10 to the power of negative 7 end exponent space rightwards arrow space Q s p space greater than space K s p space equals space m e n g e n d a p space space space space space space end style$

Sehingga, garam yang mengendap adalah PbCrO₄dan PbSO_4.

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