Question about acid and base 2

[匿名]

2010-07-04 8:36 pm

1)60mL 0.42M acetic acid solution and 50mL 1.35M sodium hydroxide solution are run from burette into a 200.00mL volumetric flask, and the solution made up to the mark. What is the pH of the solution? (Ka for acetic acid 1.75X10^-5)

2)What volume(in mL) of 0.40M NaOH solution must be added to 458.8mL of 0.20M acetic acid solution to give a buffer solution with pH 4.27? ( Ka for acetic acid 1.75X10^-5)

3)What is the percent ionization of the acid in a 0.001M solution of hydrofluoric acid (HF) (Ka 1.0X10^-3)

4)1.35g of ammonium chloride is dissolved in water and the solution is made up to 250.00mL in a volumetric flask. What is the pH of the solution? (Kb for ammonia 1.80 X 10^-5)

回答 (1)

andrew

2010-07-04 10:12 pm

✔ 最佳答案

1)
No. of moles of CH3COOH added = 0.42 x (60/1000) = 0.0252 mol
No. of moles of OH¯ added = 1.35 x (50/1000) = 0.0675 mol

CH3COOH + OH¯ → CH3COO¯ + H2O
After mixing the solution:
[OH¯] = (0.0675 - 0.0252)/(200/1000) = 0.2115 M

pH = 14 - [-log(0.2115)] = 13.33

2)
CH3COOH + OH¯ → CH3COO¯ + H2O

Let V cm³ be the volume of NaOH added.

CH3COOH ⇌ CH3COO¯ + H⁺ … (Ka)
pH = pKa- log([CH3COOH]/[CH3COO¯])
4.27 = -log(1.75 x 10¯⁵) - log[0.4V/(0.4V - 0.2x458.8)]
V = 340.3

Volume of NaOH added = 340.3 mL

3)
Let α (α ≤ 1) be the percentage ionization.

HF ⇌ H⁺ + F¯ … (Ka)
At eqm:
Ka = (0.001α)²/[0.001(1 - α)] = 1 x 10¯³
Percentage ionization = 61.8%

4)
Molar mass of NH4Cl = 14.01 + 1.008 x 4 + 35.45 = 53.49
[NH4⁺] = [NH4Cl] = (1.35/53.49)/(250/1000) = 0.101 M

NH4⁺ ⇌ NH3 + H⁺ … (Kh)
At eqm: Let [H⁺] = y M
Kh = y²/(0.101 - y) = (1 x 10¯¹⁴)/(1.8 x 10¯⁵)
y = 7.49 x 10¯⁶
pH = -log(y) = 5.13

參考: andrew

👍 0 👎 0