Why is it more accurate to say that the concentration of a solution of acetic acid is 0.01 f rather than 0.01 m?

Chemistry

2 answers:

Mandarinka [93]3 years ago

5 0

Answer:

The formal concentration accounts for the moles of the original chemical formula into the solution rather than the concentration of the species that actually exist in solution which is accounted for by the molarity.

Explanation:

Hello,

In this case, the difference between formal and molar units of concentration lies on the fact that the formal concentration accounts for the moles of the original chemical formula into the solution rather than the concentration of the species that actually exist in solution which is accounted for by the molarity. In such a way, since acetic acid is partially dissociated, formality better explains there are 0.01 moles of acetic acid added to the solution including both the non-dissociated and dissociated constituents. However, molarity just accounts for the 0.01 moles of acetic acid that could or could not have been dissociated into the solution.

As you see, this is just a by concept difference, as they lead to suitable calculations.

Best regards.

olasank [31]3 years ago

4 0

A 1 F solution stands 1 formula unit per litre and 0.01 F describes the concentration of solution with no deliberation for the real form of existence of species.

Molarity is defined as the ratio of moles of solute to the volume of solution in litres and it is used to describe formality. For accuracy, it is essential to expressed molarity of each species. In case of acetic acid, the molarity of acetic acid molecules is less than 0.01 M due to dissociation.

Thus, it is more precise to say that the concentration of a solution of acetic acid is 0.01 F instead of 0.01 M.

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Read 2 more answers

Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.

lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i) 0.115M 0 0

ΔC -x +x +x

C(eq) 0.115M - x x x

≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

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2 years ago

A certain compound was found to have the molecular formula C5H12O2. To which of the following compound classes could the compoun

IgorC [24]

Answer : The given compound belongs to ether and alcohol.

Explanation :

The chemical formula of the given compound is, $C_5H_{12}O_2$