asam basa ppt
DESCRIPTION
asam basaTRANSCRIPT
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Acid-Base Equilibriaacid-base reactions
BuffersHydrolysisTitrations
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O
H
H + O
H
H O
H
H H OH-+[ ] +
Sifat-sifat Asam-Basa Air
H2O (l) H+ (aq) + OH- (aq)
H2O + H2O H3O+ + OH-
asambasa
konjugat
basaasam
konjugat
autoionisasi air
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H2O (l) H+ (aq) + OH- (aq)
Solubility Product Constant (Konstanta Hasilkali Ion Air)
Kc =[H+][OH-]
[H2O] [H2O] = konstan
Kc[H2O] = Kw = [H+][OH-]
Konstanta hasilkali ion air (Kw) adalah hasilkali antara konsentrasi molar ion H+ dan ion OH- pada suhu tertentu.
Pada suhu 250C
Kw = [H+][OH-] = 1,0 x 10-14
[H+] = [OH-]
[H+] > [OH-]
[H+] < [OH-]
Larutan bersifat
netral
asam
basa
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pH – Ukuran Keasaman
pH = -log [H+]
[H+] = [OH-]
[H+] > [OH-]
[H+] < [OH-]
Larutan bersifat
netral
asam
basa
[H+] = 1 x 10-7
[H+] > 1 x 10-7
[H+] < 1 x 10-7
pH = 7
pH < 7
pH > 7
Pada suhu 250C
pH [H+]
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pOH = -log [OH-]
[H+][OH-] = Kw = 1,0 x 10-14
-log [H+] – log [OH-] = 14,00
pH + pOH = 14,00
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pH air hujan di daerah tertentu di bagian timur-laut Amerika pada hari tertentu adalah 4,82. Berapakah konsentrasi ion H+ dalam air hujan?
pH = -log [H+]
[H+] = 10-pH = 10-4,82 = 1,5 x 10-5 M
Konsentrasi ion OH- dalam sampel darah adalah 2,5 x 10-7 M. Berapakah pH sampel darah tersebut?
pH + pOH = 14,00
pOH = -log [OH-] = -log (2,5 x 10-7) = 6,60
pH = 14,00 – pOH = 14,00 – 6,60 = 7,40
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Menghitung [H3O+], pH, [OH-], dan pOH
Soal: 1. Seorang kimiawan mengencerkan asam klorida pekat untuk membuat dua larutan: (a) 3,0 M dan (b) 0,0024 M. Hitunglah [H3O+], pH, [OH-], dan pOH dari kedua larutan tersebut pada suhu 25°C.
2. Berapakah [H3O+], [OH-], dan pOH dari suatu larutan yang memiliki pH = 3,67? dan pH = 8,05?
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elektrolit kuat – 100% terdisosiasi
NaCl (s) Na+ (aq) + Cl- (aq)H2O
elektrolit lemah – tidak terdisosiasi sempurna
CH3COOH CH3COO- (aq) + H+ (aq)
Asam kuat adalah elektrolit kuat
HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)
H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)
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HF (aq) + H2O (l) H3O+ (aq) + F- (aq)
Asam lemah adalah elektrolit lemah
HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)
HSO4- (aq) + H2O (l) H3O+ (aq) + SO4
2- (aq)
H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)
Basa kuat adalah elektrolit kuat
NaOH (s) Na+ (aq) + OH- (aq)H2O
KOH (s) K+ (aq) + OH- (aq)H2O
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O
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F- (aq) + H2O (l) OH- (aq) + HF (aq)
Basa lemah adalah elektrolit lemah (NH3)
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
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asam kuat asam lemah
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Berapakah pH dari larutan 2 x 10-3 M HNO3?
HNO3 adalah asam kuat – 100% terdisosiasi.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
pH = -log [H+] = -log [H3O+] = -log(0,002) = 2,7
Awal
Akhir
0,002 M
0,002 M 0,002 M0,0 M
0,0 M 0,0 M
Berapakah pH dari larutan 1,8 x 10-2 M Ba(OH)2?
Ba(OH)2 adalah basa kuat – 100% terdisosiasi.
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
Awal
Akhir
0,018 M
0,018 M 0,036 M0,0 M
0,0 M 0,0 M
pH = 14,00 – pOH = 14,00 + log(0,036) = 12,56
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HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Asam Lemah (HA) dan Konstanta Ionisasi Asam
HA (aq) H+ (aq) + A- (aq)
Ka =[H+][A-]
[HA]
Ka adalah konstanta ionisasi asam
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Berapakah pH dari larutan 0,5 M HF (pada 250C)?
HF (aq) H+ (aq) + F- (aq)Ka =[H+][F-]
[HF]= 7,1 x 10-4
HF (aq) H+ (aq) + F- (aq)
Awal (M)
Perubahan (M)
Akhir (M)
0,50 0,00
-x +x
0,50 - x
0,00
+x
x x
Ka =x2
0,50 - x= 7,1 x 10-4
Ka x2
0,50= 7,1 x 10-4
0,50 – x 0,50Ka << 1
x2 = 3,55 x 10-4 x = 0,019 M
[H+] = [F-] = 0,019 M pH = -log [H+] = 1,72
[HF] = 0,50 – x = 0,48 M
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Kapan kita boleh menggunakan aproksimasi?
0,50 – x 0,50
Jika x lebih kecil daripada 5% dari konsentrasi awal.
x = 0,0190,019 M
0,50 Mx 100% = 3,8%
Lebih kecil dari 5%
Aproksiomasi ok.
Berapakah pH dari larutan 0,05 M HF (pada 250C)?
Ka x2
0,05= 7,1 x 10-4 x = 0,006 M
0,006 M
0,05 Mx 100% = 12%
Lebih dari 5%
Aproksimasi tidak ok.
Harus menggunakan persamaan kuadrat atau metode pendekatan berjenjang untuk mencari nilai x.
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Berapakah pH dari 0,122 M asam monoprotik yang nilai Ka-nya 5,7 x 10-4?
HA (aq) H+ (aq) + A- (aq)
Awal (M)
Perubahan (M)
Akhir (M)
0,122 0,00
-x +x
0,122 - x
0,00
+x
x x
Ka =x2
0,122 - x= 5,7 x 10-4
Ka x2
0,122= 5,7 x 10-4
0,122 – x 0,122Ka << 1
x2 = 6,95 x 10-5 x = 0,0083 M
0,0083 M
0,122 Mx 100% = 6,8%
Lebih dari 5%
Aproksimasi tidak ok.
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Ka =x2
0,122 - x= 5,7 x 10-4 x2 + 0,00057x – 6,95 x 10-5 = 0
ax2 + bx + c =0-b ± b2 – 4ac
2ax =
x = 0,0081 x = - 0,0081
HA (aq) H+ (aq) + A- (aq)
Awal (M)
Perubahan (M)
Akhir (M)
0,122 0.00
-x +x
0,122 - x
0,00
+x
x x
[H+] = x = 0,0081 M pH = -log[H+] = 2,09
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persen ionisasi = konsentrasi asam Ionisasi pada kesetimbangan
konsentrasi awal asamx 100%
Untuk asam monoprotik HA
Persen ionisasi = [H+]
[HA]0
x 100% [HA]0 = konsentrasi awal
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NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Basa Lemah dan Konstanta Ionisasi Basa
Kb =[NH4
+][OH-][NH3]
Kb adalah konstanta ionisasi basa
Selesaikan soal-soal basa lemah seperti asam lemah namun di sini kita mencari [OH-] bukan [H+].
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Konstanta disosiasi-basa Kb mengacu pada kesetim-bangan yang terjadi ketika basa lemah ditambahkan ke dalam air.
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The Common Ion EffectAcid-Base SolutionsCommon IonThe Common Ion Effect- the shift in
equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
pH
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The Common Ion Effect
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
common ion
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Henderson-Hasselbach EquationRelationship between pKa and Ka.
pKa = -log Ka
Henderson-Hasselbalch equation
pH = pKa + log[conjugate base]
[acid]
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pH CalculationsWhat is the pH of a solution containing 0.30
M HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30
-x
0.30 - x
0.00 0.52
+x +x
x 0.52 + x
HCOOH pKa = 3.77
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pH CalculationspH = pKa + log [HCOO-]
[HCOOH]
pH = 3.77 + log[0.52][0.30]
Common ion effect
0.30 – x 0.30
0.52 + x 0.52
= 4.01
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Buffer SolutionsA buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
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= 9,20
Hitunglah pH dari larutan buffer 0,30 M NH3/0,36 M NH4Cl. Berapa pH-nya setelah penambahan 20,0 mL NaOH(0,050 M ) ke dalam 80,0 mL larutan buffer tersebut? pKa = 9,25
NH4+ (aq) H+ (aq) + NH3 (aq)
pH = pKa + log[NH3]
[NH4+]
pKa = 9,25 pH = 9,25 + log[0,30]
[0,36]= 9,17
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
awal (mol)
akhir (mol)
0,029 0,001 0,024
0,028 0,0 0,025
pH = 9,25 + log[0,25]
[0,28][NH4
+] = 0,028
0,10
volume akhir = 80,0 mL + 20,0 mL = 100 mL
[NH3] = 0,025
0,10
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Effectiveness of Buffer Solutions
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SoalPada suhu 25°C, 0,01 M larutan ammonia
terionisasi 4,1%. Hitunglah:a. Konsentrasi ion OH- dan NH4
+.
b. Konsentrasi ammonia.c. Konstanta ionisasi larutan ammonia (Kb).d. pH larutan setelah 0,009 mol NH4Cl
ditambahkan ke 1 L larutan diatas.e. pH larutan yang disiapkan dari 0,01 mol NH3
dan 0,005 mol HCl per L.
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Acid-Base TitrationsIn a titration a solution of accurately known
concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Types of Titrations
Those involving a strong acid and a strong base
Those involving a weak acid and a strong base
Those involving a strong acid and a weak base
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Acid-Base TitrationsIndicator – substance that changes color
at (or near) the equivalence point
Slowly add baseto unknown acid
UNTIL
The indicatorchanges color
(pink)
Equivalence point – the point at which the reaction is complete
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Acid-Base Titrations
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Strong Acid-Strong Base TitrationsStrong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)
OH- (aq) + H+ (aq) H2O (l)
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Strong Acid-Base Titrations
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Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7):
CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)
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Weak Acid-Strong Base Titrations
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Strong Acid-Weak Base TitrationsStrong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq)
H+ (aq) + NH3 (aq) NH4Cl (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)
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Strong Acid-Weak Base Titrations
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Asam monoprotik: Asam yang hanya Asam monoprotik: Asam yang hanya memberikan satu protonmemberikan satu proton
Asam diprotik: Asam yang memberikan dua Asam diprotik: Asam yang memberikan dua protonproton
Asam triprotik: Asam yang memberikan tiga Asam triprotik: Asam yang memberikan tiga protonproton
Asam poliprotik: Secara umum, asam yang Asam poliprotik: Secara umum, asam yang memberikan dua proton atau lebihmemberikan dua proton atau lebih
Asam Poliprotik
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Asam PoliprotikReaksi disosiasi dan tetapan kesetimbangan
asam konjugat H2B dan HB-
H2B + H2O H3O+ + HB- Ka1
HB- + H2O H3O+ + B2- Ka2
Reaksi disosiasi dan tetapan kesetimbangan basa konjugat B- dan HB-
B2- + H2O HB- + OH- Kb1
HB- + H2O H2B + OH- Kb2
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Hubungan antara tetapan asam dan basa:
wKKbKa 21
wKKbKa 12
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Hidrolisis
Hidrolisis adalah reaksi antara ion-ion garam dgn air. Menurut Bronsted reaksi ini merupakan reaksi asam-
basa.
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GARAM TERHIDROLISISCH3COOK :Dalam air :
CH3COOK K+ + CH3COO-
H2O H+ + OH-
CH3COO- + H2O CH3COOH + OH-
Kh
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[CH3COOH] [OH-] [H+]
K = ----------------------------- x ----------- [CH3COO-] [ H2O] [H+]
[CH3COOH] [OH-][H+]
K [H2O] = Kh = --------------------- x ------------
[CH3COO-] [H+]
Kw
1/Ka
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KwKh= -------- Ka
[CH3COOH] [OH-] [OH-]2 Kh = ------------------------------- = ------------------
[CH3COO-] [CH3COO-]
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[OH-]2 = Kh [CH3COO-]
[OH-] = { Kw /Ka [CH3COO-] }1/2
pOH = ½ pKw - ½ pKa - ½ log [CH3COO-]
pH = ½ pKw +½ pKa + ½ log [CH3COO-]
pH =½ pKw +½ pKa + ½ log [garam]
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Buktikan bahwa untuk NH4Cl
Kh = Kw/Kb
pH =½ pKw - ½ pKb - ½ log [garam]
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Calculate the percent hydrolysis in a 0.01 M solution of KCN? Ka = 6.2 x 10-10
The basic ionization for hydrazin, N2H4, is 9.6 x 10-7. What would be the precent hydrolysis of 0.1 M N2H5Cl, a salt containing the acid ion conjugate to hydrazin base?