JAWABAN DAN PEMBAHASAN id. docx

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1. Ksp AgI = s2 10-16_{= s}2 10-8_{mol L}-1_{= s}

Jadi kelarutan AgI dalam air adalah 10-8_{mol L}-1 2. Kelarutan dalam MgI2 0.005 M dan [I-] = 0.01 M

Ksp AgI = [Ag+_{] [I}-_{] = s x (0.01 + s)} S diabaikan terhadap 0.01

10-16_{= s x 0.01} s = 10-14_{mol L}-1

Jadi kelarutan AgI dalam larutan MgI2 0.005M adalah 10–14 mol L-1 3. MgC2O4 → Mg2+ + C2O4

2-Ksp = [Mg2+_][C 2O42-]

Ksp = s2_{= (43 x 10}-4₎2_{= 1849 x 10}-8

4. s (AgCl) = 1,5 x 10-3 _{/ 150 = 10}-5 _mol/L

Tentukan [Ag+] dan [Cl-]:

AgCl (s) <==> Ag+ (aq) + Cl- (aq) 10-5₁₀-5₁₀-5

Jadi:

[Ag+] = Cl- = 10-5 _mol/L

Menghitung Ksp Ksp = [Ag+] . [Cl-] Ksp = 10-5_{. 10}-5_{= 10}-10

5. Ca(OH)2 (s) <==> Ca2+ (aq) + 2OH- (aq) Pada reaksi ini m = 1 dan n = 2, Jadi

s = 2 x 10-4

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s = 2 x 10-4

Massa Ca(OH)2 yang dapat larut dalam 500 mL air: Massa = s x (500 / 1000)

Massa = 2 x 10-4 x (500 / 1000) = 10-4 _gram

7. AgCl (s) <==> Ag+ (aq) + Cl- (aq) n n n

NaCl (aq) <==> Na+ (aq) + Cl- (aq) 0,1 0,1 0,1

Jadi didalam sistem terdapat: [Ag+] = n mol/L

[Cl-] = (n + 0,1) mol/L = 0,1 mol/L ----> Ini karena Cl- yang berasal dari AgCL sangat lebih sedikit dibanding Cl- yang berasal dari NaCl sehingga Cl-dari AgCl diabaikan.

Ksp AgCl = [Ag+] [Cl-] 4 x 10-10_{= n . 0,1}

n = 4 x 10-9_mol/L

8. Ksp AgI = s2

10-16_{= s}2

10-8_{mol L}-1_{= s}

Jadi kelarutan AgI dalam air adalah 10-8_{mol L}-1

9. Kelarutan dalam KI 0.01M Ksp AgI = [Ag+_{] [I}-_]

= s x (0.01 + s)

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10-16_{= s x 0.01}

s = 10-14_{mol L}-1

Jadi kelarutan AgI dalam larutan KI 0.01M adalah 10–14_{mol L}-1

10. Kelarutan dalam AgNO3 0.1 M

Ksp AgI = [Ag+_{] [I}-_]

= (0.1 +s ) x s

S diabaikan terhadap 0.01 10-16_{= 0.1 x s}

s = 10-15_{mol L}-1

11. Kelarutan dalam MgI2 0.005 M dan [I-] = 0.01 M

Ksp AgI = [Ag+_{] [I}-_]

= s x (0.01 + s)

S diabaikan terhadap 0.01 10-16_{= s x 0.01}

s = 10-14_{mol L}-1

Jadi kelarutan AgI dalam larutan MgI2 0.0o5M adalah 10–14 mol L-1

Penyelesaian :

AgIO3  Ag+ + IO3–

S s s

Konsentrasi ion Ag+_{= konsentrasi ion IO}

3– = s = kelarutan AgIO3 = 2 × 10–6 mol/L

Ksp = [Ag+][IO3–]

Ksp = (s)(s)

Ksp = (2 × 10–6)(2 × 10–6) = 4 × 10–12

12. Jawaban :

Ksp Ag2CrO4 = 4 s3 = 4(10–4)3 = 4 × 10–12

Ksp Ag2CrO4 = [Ag+]2 [CrO42–]

4 × 10–12_{= [Ag}+_]2_{× 10}–2

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Ag2CrO4 → 2Ag+ + CrO42–

Kelarutan Ag2CrO4 = ½ x 2 × 10–5 = 10–5 M

Jadi, kelarutan Ag2CrO4 dalam larutan K2CrO4 adalah 10–5 M. 13.Pembahasan :

Ksp Mg(OH)2 = [Mg2+] [OH–]2

3 × 10–11_{= 3 × 10}–11_[OH–_]2

[OH–_]2 _{= 10}–10

[OH–_{] = 10}–5_M

pOH = 5

pH = 14 – pOH

pH = 14 – 5 = 9

14.Ksp = (Ag+₎2_(CrO 4-2)

4.10-12_{= (2s)}2_(s)

4.10-12_{= 4s}3

10-12_{= s}3

S = 10-4_(CrO

4-2) = s=10-4

15.Ag3PO4 → 3Ag+_{+ PO4} 3-a 33-a 3-a

Ksp = [Ag+_]3_[PO43-_{] = (3a)}3_{(a) = 27a}4

16.Ag2CrO4 → 2Ag+_{+ CrO4} 2-1×10-4 _2×10-4 _1×10-4

Ksp = [Ag+_]2_[CrO42-_]

= (2×10-4₎2_(1×10-4_{) = 4×10}-12

17. Kelarutan (s) PbI2 = 0,922 g/liter

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PbI2 → Pb2+ + 2I2

0,002 0,002 0,004 Ksp PbI2 = [Pb2+][I–]2

= (0,002)(0,004)2

= (2×10-3_)(4×10-3₎2

= 32 x 10-9_{= 3,2 x 10}-8

18.AgCl → Ag+_{+ Cl}–_{, misal kelarutan = x} x x x

Ksp AgCl = [Ag+_{][ Ag}+_]

= x x

= x2

x2 _{= 1,96 x 10}-10

=√(1,96 x 10-10₎

= 1,4 x 10-5_{mol/liter = 1,4 x 10}-5_{mol/liter x 143,5 gram/mol}

= 200,9 x 10-5_gram/liter

= 20,09 x 10-4_gram/liter

19.pH = 10 → pOH = 14 – 10 = 4 Þ [OH–_{] = 10}-4 Mg(OH)2 → Mg2+_{+ 2OH}–

0,5×10-4 _0,5×10-4 ₁₀-4

Ksp = [Mg+_{][ OH}–_]2

= (0,5×10-4_{)( 10}-4₎

= 0,5 x 10-12_{= 5 x 10}-13

20.MgCl2 → Mg+_{+ 2Cl}– 2×10-3_2×10-3

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Ksp = [Mg+_][OH–_]2

2×10-11_{= 2×10}-3_[OH–_]2

[OH–_]2_{= (2×10}-11_{)/ (2×10}-3_{) = 1×10}-8

[OH–_{] =}_√_(1×10-8_{) = 1×10}-4

pOH = -log[OH–_{] = -log(1×10}-4_{) = 4}

pH = 14 – 4 = 10

21.NaOH → Na+_{+ OH}– 0,1

Mg(OH)2 → Mg2+_{+ 2OH}– 0,1

Ksp = [Mg2+_][OH–_]2

1,8 x 10-1 _{= (x)(0,1)}2

x = (1,8×10-11_)/0,01

= 1,8 x 10-9

22.CaF2 → Ca2+_{+ 2F}– s 2s

Ksp CaF2 = (Ca2+_)(F–₎2

= (s)(2s)2_{= 4s}3

23.Mg(OH)2 → Mg2+_{+ 2OH}– 10-2_s

Ksp Mg(OH)2 = [Mg2+_][OH–_]2

1×10-12_{= (10}-2_)(s)2

s2 _{= 10}-12_/10-2_{= 10}-10

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[OH–_{] = 10}-5

pOH = -log[OH–_]

= -log(10-5_{) = 5}

pH = 14 – 5 = 9

24.PbSO4 → Pb2+_{+ SO4} 2-1,4×10-4 _1,4×10-4 _1,4×10-4

Ksp PbSO4 = [Pb2+_][SO42-_{] = (1,4×10}-4_{)( 1,4×10}-4_{) = 1,96×10}-8

K2SO4 → 2K+_{+ SO4} 2-0,05 2-0,05

Ksp PbSO4 = [Pb2+_][SO42+_]

1,96×10-8_{= s (0,05)}

s = (1,96×10-8_)/(5×10-2_{) = 0,392 x 10}-6

= 0,4 x 10-6_M

25.KbrO3 → K+_{+ BrO3}– 0,1 0,1

[BrO3–_{] = (0,1M x 0,1 L)/1 L = 0,01 M = 10}-2_M

Ksp AgBrO = [Ag+_][BrO3–_]

6×10-5_{= [Ag}+_](10-2₎

[Ag+_{] = (6×10}-5_)/10-2_{= 6×10}-3_M

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Ksp = [Pb2+][I-]2 Ksp = s.(2s)2 = 4s3

27.CdS ⇌ Cd2+ + S s s s Ksp = [Cd2+][S2-] Ksp = s. s = s2

28.Ca(OH)2⇌ Ca2+ + 2OH

s s 2s Ksp = [Ca2+][OH-]2

Ksp = s.(2s)2 = 4s3

29.Ag2CrO4⇌ 2Ag+ + CrO S 2s s Ksp = [Ag+]2[CrO42-] Ksp = (2s)2.s = 4s3

30.Cu2S ⇌ 2Cu+ + S s 2s s Ksp = [Cu+]2[S2-] Ksp = (2s)2.s = 4s3

31.PEMBAHASAN

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33.PEMBAHASAN

34.PEMBAHASAN

Menentukan [I-] dari KI KI → K+ + I

-0,2 -0,2 -0,2 M

Menentukan kelarutan PbI2 dalam KI 0,2 M PbI2 ⇌ Pb2+ + 2I

-s -s 2-s

Ksp = [Pb2+][I-]2 ([I-] dari dua sumber tapi dari PbI2 diabaikan karena terlalu kecil) Ksp = s.(2s + 0,2)2

7 x 10-9 = s . 4 x 10-2

S= 1,75 x 10-2

35.PEMBAHASAN

Menentukan [OH-] dari larutan dengan pH = 12 pH = 12

pOH = 14 - 12 = 2 [OH-] = 10-2 M

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Mg(OH) ⇌ Mg2+ + s s 2s

Ksp = [Mg2+][OH-]2 ([OH-] dari dua sumber tapi dari Mg(OH)diabaikan karena kecil) Ksp = s.(2s + 10-2)2

3,2 x 10-11 = s . 10-4 S= 3,2 x 10-7

36.Pembahasan :

37.Pembahasan :

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39.Pembahasan :