Latihan 8.1
1.
a. Na2CO3 mengalami hidrolisis parsial Na2CO3 → 2Na+ + CO32Na+ + H2O ( tidak ada reaksi ) ⇒ basa kuat 2CO3 + 2H2O → H2CO3 + 2OH⇒ asam lemah Maka sifat larutannya basa b. (NH4)2SO4 mengalami hidrolisis parsial (NH4)2SO4 → 2NH4+ + SO42NH4+ + H2O ⇔ NH3 + H3O+ ⇒ basa lemah SO42- + H2O ( tidak ada reaksi ) ⇒ asam kuat Maka sifat larutannya asam c. Na2S mengalami hidrolisis parsial Na2S → 2Na+ + S2Na+ + 2H2O ( tidak ada reaksi ) ⇒ basa kuat S2 + 2H2O → H2S + 2OH- ⇒ asam lemah Maka sifat larutannya basa d. (NH4)2CO3 mengalami hidrolisis total (NH4)2CO3 → 2NH4+ + CO32NH4+ + H2O ⇔ NH3 + H3O+ ⇒ basa lemah CO32- + 2H2O ⇔ H2CO3 + 2OH- ⇒ asam lemah Sifat larutannya tergantung dari kekuatan relatif asam dan basanya ( tergantung pada nilai K a dan K b ) e. K 2SO4 tidak mengalami hidrolisis K 2SO4 → 2K + + SO42K + + H2O ( tidak ada reaksi ) ⇒ basa kuat 2SO4 + H2O ( tidak ada reaksi ) ⇒ asam kuat Sifat larutan netral f. Ba(CH3COO)2 mengalami hidrolisis parsial Ba(CH3COO)2 → Ba2+ + (CH3COO)Ba2+ + H2O ( tidak ada reaksi ) ⇒ basa kuat CH3COO + H2O ⇔ CH3COOH + OH- ⇒ asam lemah Sifat larutan basa -
2.a. [OH ] =
K w K a
×
M
=
10
14
−
6,3 ×10
5
−
×
0.1 = 3,98 × 10-6
pOH = - log ( 3,98 × 10-6) = 6 – log 3,98 pH = 14 – (6 – log 3,98) = 8 + log 3,98 = 8,59 -
b. [OH ] =
K w K a
× M =
10
14
−
4,9 ×10
10
−
×
0.1 = 1,43 × 10-3
pOH = - log ( 1,43 × 10-3 ) = 3 – log 1,43 pH = 14 – ( 3 – log 1,43 ) = 11,15 1
K w
+
c. [H ] =
10
× M =
K b
14
−
1,8 ×10
5
−
×
0.1 = 7,45 × 10-6
pH = - log 7,45 × 10-6 ) = 6 – log 7,45 = 5,13 3.
NaOH + CH3COOH → CH3COONa + 50mL 0,1M 50mL 0,1M awal : 5 mmol 5 mmol reaksi : 5 mmol 5 mmol 5 mmol sisa : 5 mmol + CH3COONa → CH3COO + Na CH3COO- + H2O ⇔ CH3COOH + OH5 mmol
[CH3COO-] =
100 mL
K w
-
[OH ] =
K a
×
[CH
3
=
H2O
0.05 M
COO
-
]
10
=
14
−
1,8 ×10
5
−
×
0,05 = 5,27 × 10-6
pOH = - log ( 5,27 × 10-6 ) = 6 – log 5,27 pH = 14 – ( 6 – log 5,27 ) = 8 + log 5,27 = 8,72 4.
NH3 + HCl NH4Cl → 50mL 0,1M 50mL 0,1M awal : 5 mmol 5 mmol reaksi : 5 mmol 5 mmol 5 mmol sisa : 5 mmol + NH4Cl + H2O ⇔ NH3 + H3O 5 mmol
[NH4+] =
K w
+
[H ] =
=
100 mL ×
K b
[
NH 4
0.05 M
+
]
10
=
14
−
1,8 ×10
5
−
×
0,05 = 5,27 × 10-6
pH = - log ( 5,27 × 10-6 ) = 6 – log 5,27 = 5,27 5. pH = 5 → [H+] = 10-5 NH4Cl → NH4+ + ClNH4+ + H2O ⇔ NH3 + H3O+ ; Mr NH4 = 53,5 a +
[NH4 ] = 53 ,5
=
0,5 +
[H ] = 10-5 = 10-10 =
K w K b
2a 53,5 2a 53,5
×
2a 53,5
[
NH 4 10
×
×10
+
]
M
=
10
14
−
1.10
5
−
×
2a 53,5
−9
9
−
2
10-1 =
2a 53,5
2a = 5,35→ a = 2,675 jadi, berat NH4Cl yang diperlukan = 2,675 gr 6. a. CH3COOH + NaOH 0,1M 50mL 0,1M 0mL = 5 mmol = 0 mmol → pH hanya ditentukan larutan CH3COOH [H+] = K a × M = 10 −5 ×10 −1 = 10 −6 =10 −3 pH = - log 10-3 = 3 b. CH3COOH + NaOH → CH3COONa + H2O 0,1M 50mL 0,1M 25ml awal : 5 mmol 2,5 mmol reaksi : 2,5 mmol 2,5 mmol 2,5 mmol sisa: 2,5 mmol 2,5 mmol *) terbentuk buffer CH3COOH/CH3COO[H+] = K a ×
mmol CH3 COOH mmol CH3 COO
-
= 10
−
5
×
2,5 2,5
= 10
pH = - log 10-5 ) = 5 c. CH3COO + NaOH → CH3COONa 0,1M 50mL 0,1M 50mL awal : 5 mmol 5 mmol reaksi : 5 mmol 5 mmol 5 mmol sisa : 5 mmol *) terbentuk garam terhidrolisis CH3COONa → CH3COO- + Na+ CH3COO- + H2O ⇔ CH3COOH + OH-
[OH ] =
K w K a
×
[CH
3 COO
-
]
=
10
5 mmol 150 mL 1
=
1 30
+
5
H2O
14
−
10
5
−
×
0,05 = 7,07 × 10-6
pOH = - log (7,07 × 10-6 ) = 6 – log 7,07 pH = 14 – ( 6 – log 7,07 ) = 8,849 d. CH3COOH + NaOH → CH3COONa 0,1M 50ml 0,1M 100mL awal : 5 mmol 10mmol reaksi : 5 mmol 5 mmol 5 mmol sisa : 5 mmol 5 mmol *) pH ditentukan oleh basa tersisa NaOH → Na+ + OH5 mmol 5 mmol 5 mmol [OH-] =
−
+ H2O
M
30
pOH = - log
3
1 1 ) = 14 + log = 12,52 30 30
pH = 14 – (- log 7. a.
NH3 + HCl 0,1M 50mL 0,1M 0mL = 5 mmol = 0 mmol *) pH hanya ditentukan oleh larutan basa [OH-] = K b × M = 10 −5 × 0,1 = 10 −6 = 10-3 pOH = - log ( 10-3 ) = 3 pH = 14 – 3 =11 b. NH3 + HCl → NH4Cl 0,1M 50mL 0,1M 25mL awal : 5 mmol 2,5 mmol reaksi : 2,5 mmol 2,5 mmol 2,5 mmol sisa : 2,5 mmol 2,5 mmol *) terbentuk larutan Buffer NH3/NH4+ mmol NH 3 2,5 5 5 = 10 × = 10 [OH-] = K b × 2,5 mmol NH 4 −
−
+
pOH = - log 10-5 = 5 pH = 14 – 5 = 9 c. NH3 + HCl → NH4Cl 0,1M 50mL 0,1M 50mL awal : 5 mmol 5 mmol reaksi : 5 mmol 5 mmol 5 mmol sisa : 5 mmol *) terbentuk garam terhidrolisis NH4Cl → NH4+ + Cl NH4+ + H2O ⇔ NH3 + H3O+ [NH4+] = +
[H ] =
5 mmol 100 mL K w K b
×
= 0,05 M
[ NH ] +
4
=
10
−14
10
−5
× 0,05 = 7,07 × 10-6
pH = - log ( 7,07 × 10-6 ) = 6 – log 7,07 = 8,85 d. NH3 + HCl NH4Cl → 0,1M 50mL 0,1M 100mL awal : 5 mmol 10 mmol reaksi : 5 mmol 5 mmol 5 mmol sisa : 5 mmol 5 mmol *) pH dikatakan oleh asam yang tersisa : HCl H+ + Cl→ 5 mmol 5 mmol 5 mmol [H+] =
5 mmol 150 mL
=
1 30
M
4
1 = 1,48 30
pH = - log
5
