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jeka57 [31]
3 years ago
5

A 27.86 ml sample of 0.1744 m hno3 is tirades with 29.4ml of a job solution. What is the molarity of the koh

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer:

0.165 mol·L⁻¹

Explanation:

1. Write the <em>chemical equation</em> for the reaction.

HNO₃ + KOH ⟶ KNO₃ + H₂O

===============

2. Calculate the <em>moles of HNO₃</em>

c = n/V                                               Multiply each side by V and transpose

n = Vc

V = 0.027 86 L

c = 0.1744 mol·L⁻¹                             Calculate the moles of HNO₃

Moles of HNO₃ = 0.027 86 × 0.1744

Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃

===============

3. Calculate the <em>moles of KOH </em>

1 mol KOH ≡ 1 mol HNO₃                 Calculate the moles of KOH

Moles of KOH = 4.859 × 10⁻³× 1/1

Moles of KOH = 4.859 × 10⁻³ mol KOH

===============

4. Calculate the <em>molar concentration</em> of the KOH

V = 29.4 mL = 0.0294 L                   Calculate the concentration

c = 4.859 × 10⁻³/0.0294

c = 0.165 mol·L⁻¹

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Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

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c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

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ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

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