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jacquilynfoster7
02.10.2019 •
Chemistry
Calculate the ph for each of the following cases in the titration of 25.0 ml of 0.150 m acetic acid (ka = 1.75x10-5) with 0.150 m naoh(aq). (1 point each) (a) before addition of any naoh (b) after addition of 5.0 ml of naoh (c) after addition of 15.0 ml of naoh (d) after addition of 25.0 ml of naoh (e) after addition of 40.0 ml of naoh (f) after addition of 60 ml of naoh
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Ответ:
a) pH = 2.793
b) pH = 4.280
c) pH = 4.933
d) pH = 8.816
e) pH = 8.861
f) pH = 8.891
Explanation:
a) VNaOH = 0 mL
∴ CH3COOH ↔ CHECOO- + H3O+
⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]
mass balance:
⇒ C CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M
charge balance:
⇒ [ H3O+ ] = [ CH3COO- ]
⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )
⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0
⇒ [ H3O+ ] = 1.61146 E-3 M
⇒ pH = - Log [ H3O+ ] = 2.793
b) after 5.0 mL NaOH:
∴ CH3COOH + NaOH ↔ CH3COONa + H2O
⇒ C NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M
⇒ C CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M
mass balance:
⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M
charge balance:
⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]
⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]
⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5
⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0
⇒ [ H3O+ ] = 5.26 E-5 M
⇒ pH = 4.28
c) after 15 mL NaOH:
⇒ C CH3COOH = 0.0375 M
⇒ C NaOH = 0.05625 M
mass balance:
⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]
charge balance:
⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]
⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])
⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]
⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0
⇒ [ H3O+ ] = 1.1662 E-5 M
⇒ pH = 4.933
d) after 25 mL NaOH:
⇒ C NaOH = 0.075 M
⇒ C CH3COOH = 0 Mequiv. point
⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])
⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0
⇒ [ OH- ] = 6.5463 E-6 M
⇒ pOH = 5.184
⇒ pH = 8.816
e) after 40 mL NaOH:
⇒ C NaOH = 0.0923 M
⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0
⇒ [OH-] = 7.2624 E-6 M
⇒ pOH = 5.139
⇒ pH = 8.861
f) after 60 mL NaOH:
⇒ C NaOH = 0.106 M
⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0
⇒ [OH-] = 7.7782 E-6 M
⇒ pOH = 5.11
⇒ pH = 8.891
Ответ:
PbBr₄
C₂O₆
Al₂S₃
Explanation:
Lead (IV) bromide
This is an ionic compound. Ionic compounds do not tell you how many of each atom you have in the name. You have to figure out how much there is based on the charges of the atoms.
The (IV) means that lead has a charge of +4. The charge is put in parentheses since lead is a transition metal, and the charge of a transition metal can vary. Bromine, on the other hand, is always assumed to have a charge of -1. When making a molecular formula, you need to have enough of each atom so that the charges cancel out.
Dicarbon hexoxide
This is a covalent compound. Covalent compounds will tell you how much of each atom you have in the name.
Dicarbon means two carbons. Hexoxide means six oxygens.
Aluminum sulfide
Aluminum sulfide is an ionic compound.
Aluminum has a charge of +3. Sulfur has a charge of -2.