You need to use the Ka for the acetic acid and the equilibrium equation.
Ka = 1.85 * 10^ -5
Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+)
Ka = [CH3COO-][H+] / [CH3COOH]
Molar concentrations at equilibrium
CH3COOH CH3COO- H+
0.50 - x x x
Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)
Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50
=> Ka ≈ x^2 / 0.50
=> x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6
=> x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030
pH = - log [H+] = - log (x) = - log (0.0030) = 2.5
Answer: 2.5
Answer:
Explanation:
her is the answer hope this helps
Answer:
47.5 mL
Solving:
M1 = 4.00 M
V1 = ?
M2 = 0.760 M
V2 = 0.250 L
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M1 * V1 = M2 * V2
V1 = ( M2 * V2 ) / M1
V1 = ( 0.760 * 0.250 ) / 4.00
V1 = ( 0.190 ) / 4.00
V1 = 0.0475 L
Answer:
1.86% NH₃
Explanation:
The reaction that takes place is:
- HCl(aq) + NH₃(aq) → NH₄Cl(aq)
We <u>calculate the moles of HCl that reacted</u>, using the volume used and the concentration:
- 32.27 mL ⇒ 32.27/1000 = 0.03227 L
- 0.1080 M * 0.03227 L = 3.4852x10⁻³ mol HCl
The moles of HCl are equal to the moles of NH₃, so now we <u>calculate the mass of NH₃ that was titrated</u>, using its molecular weight:
- 3.4852x10⁻³ mol NH₃ * 17 g/mol = 0.0592 g NH₃
The weight percent NH₃ in the aliquot (and thus in the diluted sample) is:
- 0.0592 / 12.949 * 100% = 0.4575%
Now we <u>calculate the total mass of NH₃ in the diluted sample</u>:
Diluted sample total mass = Aqueous waste Mass + Water mass = 23.495 + 72.311 = 95.806 g
- 0.4575% * 95.806 g = 0.4383 g NH₃
Finally we calculate the weight percent NH₃ in the original sample of aqueous waste:
- 0.4383 g NH₃ / 23.495 g * 100% = 1.86% NH₃
Answer:
70.6 %
Explanation:
First step, we define the reaction:
2P + 3Br₂ → 2PBr₃
We determine the moles of reactant:
35 g . 1mol / 159.8 g = 0.219 moles
We assume, the P is in excess, so the bromine is the limiting reagent.
3 moles of Br₂ can produce 2 moles of phophorous tribromide
Then, 0.219 moles may produce (0.219 . 2) /3 = 0.146 moles of PBr₃
We convert moles to mass:
0.146 mol . 270.67 g /mol = 39.5 g
That's the 100 % yield reaction, also called theoretical yield. The way to determine the % yield is:
(Yield produced / Thoeretical yield) . 100
(27.9 / 39.5) . 100 = 70.6 %
