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maks197457 [2]
3 years ago
13

A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equiva

lence point in a titration. What is the concentration of the naoh solution?
Chemistry
2 answers:
cestrela7 [59]3 years ago
7 0

Answer: The concentration of NaOH solution is 0.25 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of H_2SO_4 solution = 0.1053 M

V_1 = volume of H_2SO_4 solution = 17.88 ml

M_2 = molarity of NaOH solution = 0.46 M

V_2 = volume of NaOH solution = 15.0 ml

n_1 = valency of H_2SO_4 = 2

n_2 = valency of NaOH = 1

2\times 0.1053M\times 17.88=1\times M_2\times 15

M_2=0.25

Therefore, the concentration of  NaOH solution is 0.25 M

Sergeu [11.5K]3 years ago
3 0

Answer:

  • <u>0.1255 M</u>

Explanation:

<u>1) Data:</u>

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

<u>2) Chemical reaction:</u>

The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:

  • Acid + Base → Salt + Water

  • H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

<u>3) Balanced chemical equation:</u>

  • H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

<u>4) Stoichiometric mole ratio:</u>

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

  • 1 mole H₂SO₄ : 2 mole NaOH

<u>5) Calculations:</u>

a) Molarity formula: M = n / V (in liter)

   ⇒ n = M × V

b) Nunber of moles of acid:

  • nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

  • nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

  • nₐ = nb

  • 0.1053 × (17.88 / 1,000) =  Mb × (15.00 / 1,000)

  • Mb = 0.1053 × 17.88  / 15.00 = 0.1255 mole/liter = 0.1255 M
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Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al: 2 moles
  • H₂SO₄: 3 moles
  • Al₂(SO₄)₃. 1 mole
  • H₂: 3 moles

The molar mass of the compounds is:

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  • H₂SO₄: 98 g/mole
  • Al₂(SO₄)₃: 342 g/mole
  • H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al: 2 moles ×27 g/mole= 54 grams
  • H₂SO₄: 3 moles ×98 g/mole= 294 grams
  • Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
  • H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield} x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know:

  • actual yield= 75.26 grams
  • theorical yield= 88.67 grams

Replacing in the definition of percent yields:

percent yield=\frac{75.26 grams}{88.67 grams} x100

Solving:

<u><em>percent yield= 84.88%</em></u>

Finally, the percent yield for the reaction is 84.88%.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

#SPJ1

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