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Nataliya [291]
3 years ago
6

The reaction between NaOH(aq) and H2SO4(aq) was studied in a constant-pressure calorimeter: 100.0 mL portions of 1.00 M aqueous

NaOH and H2SO4, each at 24.0C, were mixed. The volume of the mixture was 200.0 mL. The maximum temperature achieved was 30.6C. Neglect the heat capacity of the calorimeter and the thermometer, and assume that the solution of products has a density of 1.00 g/mL and a specific heat capacity of 4.18 J/(g·K). Calculate H, the heat (enthalpy) of reaction, in kJ per mole of acid. (Hint: You may not have the equivalent amounts of NaOH and H2SO4 in this experiment.)
Chemistry
1 answer:
Anettt [7]3 years ago
3 0

Answer:

110,4 kJ / mole of acid

Explanation:

The reaction of NaOH with H₂SO₄ is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Moles of NaOH and H₂SO₄ are:

0,1000L×1,00M = 0,100 moles of NaOH and 0,100 moles of H₂SO₄. As 2 moles of NaOH reacts with 1 mole of  H₂SO₄, moles of acid that react are 0,100/2 = <em>0,0500moles of acid</em>.

The produced heat is:

Q = C×m×ΔT

Where C is specific heat capacity (4,18J/gK)

m is mass: 200,0mL×(1,00g/mL) = 200g

ΔT is change in temperature: 30,6°C - 24,0°C = 6,6°C = 6,6K

Thus, Q is:

Q = 4,18J/gK×200g×6,6K

Q = 5518J = <em>5,52kJ</em>

ΔH in kJ per mole of acid:

5,52kJ / 0,0500moles of acid =<em> 110,4 kJ / mole of acid</em>

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I hope it helps!

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Hello,

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Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0
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