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8090 [49]
3 years ago
5

A student makes a standard solution of sulfuric acid by taking 10.00 mL of a super-concentrated stock solution and diluting it t

o 5.00 L. He then standardizes the diluted solution by titrating 30.00 mL with 34.68 mL of a 0.2458 M solution of KOH. What is the concentration of the stock solution and the standard solution?
Chemistry
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

Standard solution: 0.1421M

Stock solution: 71.04M

Explanation:

The reaction of H₂SO₄ with KOH produces:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

In titration, moles of KOH are:

0.03468L ₓ (0.2458mol / L) = 0.008524 moles of KOH. As 1 mole of acid reacts with 2 moles of KOH, moles of titrated acid are: <em>0.008524/2 = 0.004262 moles of H₂SO₄</em>.

Thus, concentration of standard solution is:

0.004262 moles of H₂SO₄ / 0.03000L = <em>0.1421M standard solution</em>

Now, standard solution was prepared as a dilution of 10.00mL to 5.00L of stock solution. That means concentration of stock solution is:

0.1421M × (5000mL / 10.00mL) = <em>71.04M stock solution</em>

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A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

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How many atoms are there in each of the following ?a.)1.50 mol Na c.) 0.250 mol. Sib.)6.755 mol Pb
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