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inysia [295]
3 years ago
10

Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titra

tion of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH (iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH Which statement is most likely to be true?
Chemistry
2 answers:
Goshia [24]3 years ago
8 0

Answer:

Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH (iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH. Which statement is most likely to be true?

(a) All three titrations require the same volume of NaOH to reach their first equivalence point.

(b) All three titrations have the same initial pH.

(c) All three titrations have the same pH at their first equivalence point.

The correct anser is (a) All three titrations require the same volume of NaOH to reach their first equivalence point.

Explanation:

In titration, an equivalent point is the point at which there is enough titrant to react completely with the solution under analysis that is the number of moles and the number of moles of the acid are equal and the compounds in the solution are the produced salt and water

It is the stoichiometric point of the reaction where the number of moles of reactant acid required to neutralize the base is reached which may be in a ratio of acid to base other than 1:1 depending on the stoichiometry of the chemical reaction equation

A weak polyprotoic acid has more than one equivalence point depending on the number of ionized hydrogen ions that can react with the base in the titration

Thus the 25ml titration of sodium hydroxide solution consists of \frac{25}{1000} × 0.1 ,oles of NaOH = 0.0025mols of NaOH which gives

NaOH(aq)→ Na^{+} (aq) + OH^{-}(aq) which consists of one mole of Positive sodium ion and one mole of negatively charged hydroxide ion OH^{-} available for reaction with an hydrogen ion hence all three titrations require the same volume of NaOH to reach their first equivalence point.

Setler79 [48]3 years ago
6 0

Complete Question:

Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH (iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH. Which statement is most likely to be true?

(a) All three titrations require the same volume of NaOH to reach their first equivalence point.

(b) All three titrations have the same initial pH.

(c) All three titrations have the same pH at their first equivalence point.

Answer:

(a) All three titrations require the same volume of NaOH to reach their first equivalence point.

Explanation:

(i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH

number of moles of acid = \frac{25}{1000} dm^{3}  * 0.1 M = 0.0025 moles

ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH

number of moles of acid = \frac{25}{1000} dm^{3}  * 0.1 M = 0.0025 moles

                     

(iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH

number of moles of acid = \frac{25}{1000} dm^{3}  * 0.1 M = 0.0025 moles

Therefore, all the acids require the same number of moles of NaOH to reach their first equivalence points

Note that the concentration of the base NaOH are also the same, therefore the volume of NaOH required to reach equivalence point would also be the same for all the three titrations.

All three titrations don't have the same initial and equivalence point pH because they all have different acidic properties.

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(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S
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    Question 2:

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Explanation:

<u>Question 1:</u>

The<em> neutralization</em> reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) <u>Word equation:</u>

  • sulfuric acid + potassium hydroxide → potassium sulfate + water

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

<u>2) Skeleton equation (unbalanced)</u>

  • H₂SO₄ + KOH → K₂SO₄ + H₂O

<u>#) Balanced chemical equation (including phases)</u>

  • H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

<u>Question 2:</u>

<u>1) Mol ratio:</u>

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

  • 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

<u>2) Moles of H₂SO₄:</u>

  • V = 0.750 liter
  • M = 0.480 mol/liter
  • M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

<u>3) Moles of KOH:</u>

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  • M = 0.290 mol/liter
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<u>4) Determine the limiting reagent:</u>

a) Stoichiometric ratio:

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b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

<u>5) Amount of H₂SO₄ that reacts:</u>

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         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

<u>6) Concentration of H₂SO₄ remaining:</u>

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  • Total volume = 0.700 liter + 0.750 liter = 1.450 liter

  • Concetration = M

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

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