If a buffer solution contains 1.1 M acetic acid and 1.4 M sodium acetate, what is the pH of the buffer if the Ka of the acetic a

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If a buffer solution contains 1.1 M acetic acid and 1.4 M sodium acetate, what is the pH of the buffer if the Ka of the acetic a

cid is 1.8 x 10*. b.What is the pH of the solution if 0.15 mol of NaOH is added to 1.00 L of the solution? (Assume there is no volume change.) 5.Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1 °C. What is the molar mass of this substance? (Benzene's K= 5.12 ° C/m, Tr 5.5 °C).

Chemistry

1 answer:

WINSTONCH [101] · Answer 1 · 2022-03-20T11:16:37+03:00

Answer:

a) pH = 4.635

b) 0.15 mol NaOH is added:

⇒ pH = 4.698

c) molar mass (Mw) lauryl alcohol in benezen solution:

⇒ Mw = 46.545g C12H26O / mol C6H6

Explanation:

CH3COONa → CH3COO- + Na+

∴ C CH3COONa = 1.4 M

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ H3O+ ] * [ CH3COO- ]) / [ CH3COOH ]

∴ C CH3COOH = 1.1 M

2H2O ↔ H3O+ + OH-

∴ Kw = [ H3O+ ] * [ OH- ] = 1 E-14

a) pH sln:

mass balance:

⇒ C CH3COOH + C CH3COONa = [ CH3COO- ] + [ CH3COOH ] = 1.1 + 1.4 = 2.5 M

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ].......[ OH- ] is neglected, comes from water.

∴ [ Na+ ] ≅ C CH3COONa = 1.4 M

⇒ [ H3O+ ] + 1.4 = [ CH3COO- ]

replacing in Ka:

⇒ 1.8 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 1.4 ) ) / ( 2.5 - ( [ H3O+ ] + 1.4 ) )

⇒ [ H3O+ ]² + 1.4 [ H3O+ ] - 1.98 E-5 = 0

⇒ [ H3O+ ] = 2.314 E-5 M

⇒ pH = 4.635

b) 0.15 M NaOH is added:

CH3COOH + NaOH ↔ CH3COONa + H2O

∴ m NaOH = 0.15 mol

∴ C CH3COOH = (( 1 * 1.1)mol CH3COOH - 0.15mol NaOH ) / 1 L

⇒ C CH3COOH = 0.95 mol/L

∴ C CH3COONa = (( 1 * 1.4 ) + 0.15 ) / 1 L

⇒C CH3COONa = 1.55 mol/L

⇒ Ka = ([ H3O+ ] * ( 1.55 + [ H3O+ ] )) / ( 2.5 - ( [ H3O+ ] + 1.55)) = 1.8 E-5

⇒ [ H3O+ ]² + 1.55 [ H3O+ ] - 1,71 E-5 = 0

⇒ [ H3O+ ] = 2.003 E-5 M

⇒ pH = 4.698

c) C12H26O + C6H6 ↔ C18H32O

⇒ sln: 5g C12H26O / 0.100 Kg C6H6 = 50 gC12H26O / kgC6H6

∴ benzene:

Tr = K * m

∴ m (molality) ≡ moles solute / Kg solvent ≡ mol/Kg

⇒ m benzene = 5.5 Kg°C / 5.12 °C/m

⇒ m Benzene = 1.074 mol/Kg

⇒ Mw C12H26O = 50g C12H26O / Kg C6H6 * ( Kg C6H6 / 1.074 mol C6H6)

⇒ Mw C12H26O = 46.545 gC12H26O / mol C6H6