A 25.0 mL sample of a 0.260 M solution of aqueous trimethylamine is titrated with a 0.325 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25°C.
10.0ml=9.81
20.0ml=?
30.0ml=?
I got the first one but the rest are giving me trouble. thank u
Need help with chem. thanx :)?
2017-11-06 1:17 pm
回答 (1)
2017-11-06 2:39 pm
Initial n((CH₃)₃N) = (0.260 mol/L) × (25.0/1000 L) = 0.0065 mol
After 20.0 mL 0.325 M HCl is added :
Each mole of HCl added reacts with 1 mole (CH₃)₃N to form 1 mole of (CH₃)₃NH⁺ ion.
m(HCl) added = (0.325 mol/L) × (20.0/1000 L) = 0.0065 mol
Hence, after 20.0 mL 0.325 M HCl is added :
n((CH₃)₃N) = (0.0065 - 0.0065) mol = 0 M
n((CH₃)₃NH⁺) = 0.0065 mol
[(CH₃)₃NH⁺] = (0.0065 mol) / [(25.0 + 20.0)/1000 L] = 0.144 M
Consider the dissociation of (CH₃)₃NH⁺ :
____________ (CH₃)₃NH⁺(aq) __ ⇌ __ (CH₃)₃N(aq) __ + __ H⁺(aq) ___ pKa = 14.00 - 4.19 = 9.81
Initial: _______ 0.144 M ____________ 0 M ___________ 0 M
Change: _______ -y M _____________ +y M __________ +y M
At eqm: _____ (0.144 - y) M _________ y M ___________ y M
As Ka of (CH₃)₃NH⁺ is very small, the dissociation of (CH₃)₃NH⁺ would be to a very small extent.
It is assumed that 0.144 ≫ y, i.e. [(CH₃)₃NH⁺] at equilibrium = (0.144 - y) M ≈ 0.144 M
Ka = [(CH₃)₃N] [H⁺] / [(CH₃)₃NH⁺]
10⁻⁹·⁸¹ = y² / 0.144
y = √(0.144 ×10⁻⁹·⁸¹) = 4.72 × 10⁻⁶
pH = -log(4.72 × 10⁻⁶) = 5.33
====
After 30.0 mL 0.325 M HCl is added :
Each mole of HCl added reacts with 1 mole (CH₃)₃N to form 1 mole of (CH₃)₃NH⁺ ion.
m(HCl) added = (0.325 mol/L) × (30.0/1000 L) = 0.00975 mol
Hence, after 30.0 mL 0.325 M HCl is added :
n((CH₃)₃N) = (0.0065 - 0.0065) mol = 0 M
n((CH₃)₃NH⁺) = 0.0065 mol
n((HCl)) = (0.00975 - 0.0065) mol = 0.00325 mol
As HCl is a strong acid, 0.00325 mol of HCl completely dissociates to give 0.00325 mol of H⁺.
[H⁺] = (0.00325 mol) / [(25.0 + 30.0)/1000 L] = 0.0591 M
As Ka of (CH₃)₃NH⁺ is very small, [H⁺] formed from the dissociation of (CH₃)₃NH⁺ is negligible.
pH = -log[H⁺] = -log(0.0591) = 1.23
After 20.0 mL 0.325 M HCl is added :
Each mole of HCl added reacts with 1 mole (CH₃)₃N to form 1 mole of (CH₃)₃NH⁺ ion.
m(HCl) added = (0.325 mol/L) × (20.0/1000 L) = 0.0065 mol
Hence, after 20.0 mL 0.325 M HCl is added :
n((CH₃)₃N) = (0.0065 - 0.0065) mol = 0 M
n((CH₃)₃NH⁺) = 0.0065 mol
[(CH₃)₃NH⁺] = (0.0065 mol) / [(25.0 + 20.0)/1000 L] = 0.144 M
Consider the dissociation of (CH₃)₃NH⁺ :
____________ (CH₃)₃NH⁺(aq) __ ⇌ __ (CH₃)₃N(aq) __ + __ H⁺(aq) ___ pKa = 14.00 - 4.19 = 9.81
Initial: _______ 0.144 M ____________ 0 M ___________ 0 M
Change: _______ -y M _____________ +y M __________ +y M
At eqm: _____ (0.144 - y) M _________ y M ___________ y M
As Ka of (CH₃)₃NH⁺ is very small, the dissociation of (CH₃)₃NH⁺ would be to a very small extent.
It is assumed that 0.144 ≫ y, i.e. [(CH₃)₃NH⁺] at equilibrium = (0.144 - y) M ≈ 0.144 M
Ka = [(CH₃)₃N] [H⁺] / [(CH₃)₃NH⁺]
10⁻⁹·⁸¹ = y² / 0.144
y = √(0.144 ×10⁻⁹·⁸¹) = 4.72 × 10⁻⁶
pH = -log(4.72 × 10⁻⁶) = 5.33
====
After 30.0 mL 0.325 M HCl is added :
Each mole of HCl added reacts with 1 mole (CH₃)₃N to form 1 mole of (CH₃)₃NH⁺ ion.
m(HCl) added = (0.325 mol/L) × (30.0/1000 L) = 0.00975 mol
Hence, after 30.0 mL 0.325 M HCl is added :
n((CH₃)₃N) = (0.0065 - 0.0065) mol = 0 M
n((CH₃)₃NH⁺) = 0.0065 mol
n((HCl)) = (0.00975 - 0.0065) mol = 0.00325 mol
As HCl is a strong acid, 0.00325 mol of HCl completely dissociates to give 0.00325 mol of H⁺.
[H⁺] = (0.00325 mol) / [(25.0 + 30.0)/1000 L] = 0.0591 M
As Ka of (CH₃)₃NH⁺ is very small, [H⁺] formed from the dissociation of (CH₃)₃NH⁺ is negligible.
pH = -log[H⁺] = -log(0.0591) = 1.23
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