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Tanya [424]
2 years ago
15

A student carried out a titration using HC2H3O2(aq) and NaOH(aq). The net ionic equation for the neutralization reaction that oc

curs during the titration is represented above. The NaOH(aq) was added from a buret to the HC2H3O2(aq) in a flask. The equivalence point was reached when a total of 20.0mL of NaOH(aq) had been added to the flask. How does the amount of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compare to the amount of HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq), and what is the reason for this result?
Chemistry
1 answer:
cluponka [151]2 years ago
7 0

Answer:

The  ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .

Explanation:

HC2H3O2 is  CH₃⁻ COOH, which is also known as Acetic acid.

IUPAC name of this compound is Ethanoic acid.

Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.

Given that, equivalence point was reached when 20.0mL of NaOH is added.

let the normality of acetic acid is N₁ and that of NaOH is N₂.

    volume of acetic acid is V₁ and that of NaOH is V₂.  

 Equivalence point occurs when, N₁ × V₁ =  N₂ × V₂.

⇒  N₁ × V₁ =  N₂ × 20.

after the addition of 5.0mL of NaOH(aq), remaining N₁ × V°  = N₂ × (20 - 5).

                                                                                                  = N₂ × 15.

after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).

                                                                                                  = N₂ × 19.

⇒ V° : Vˣ = 15 : 19 .

⇒

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Answer:

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Explanation:

An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:

HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)

And Ka, the acid dissociation constant is:

<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />

When base:

HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)

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<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>

6 0
1 year ago
A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me
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Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

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c = 4.186

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E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0
2 years ago
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Density of a substance is defined as mass per unit volume, thus volume can be calculated as:

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The half cell reactions for the above reaction follows:

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