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3 years ago

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The Ka1 value for oxalic acid is 5.9 x10-2 , and the Ka2 value is 4.6 x 10-5 . What are the values of Kb1 and Kb2 of the oxalate

ion

1 answer:

andre [41]3 years ago

6 0

Answer:

2.17x10⁻¹⁰ = Kb1

1.69x10⁻¹³ = Kb2

Explanation:

Oxalic acid, C₂O₄H₂, has two intercambiable protons, its equilibriums are:

C₂O₄H₂ ⇄ C₂O₄H⁻ + H⁺ Ka1 = 5.9x10⁻²

C₂O₄H⁻ ⇄ C₂O₄²⁻ + H⁺ Ka2 = 4.6x10⁻⁵

Oxalate ion, C₂O₄²⁻, has as equilibriums:

C₂O₄²⁻ + H₂O ⇄ C₂O₄H⁻ + OH⁻ Kb1

C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2

Also, you can know: KaₓKb = Kw

<em>Where Kw is 1x10⁻¹⁴</em>

Thus:

Kw = Kb2ₓKa1

1x10⁻¹⁴ =Kb2ₓ4.6x10⁻⁵

<h3>2.17x10⁻¹⁰ = Kb1</h3>

And:

Kw = Kb1ₓKa2

1x10⁻¹⁴ =Kb1ₓ5.9x10⁻²

<h3>1.69x10⁻¹³ = Kb1</h3>

<em>That is because the inverse reaction of, for example, Ka1:</em>

<em>C₂O₄H⁻ + H⁺ ⇄ C₂O₄H₂ K = 1 / Ka1</em>

<em>+ H₂O ⇄ H⁺ + OH⁻ K = Kw = 1x10⁻¹⁴</em>

<em>= </em>

<em>C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2 = Kw × 1/Ka1</em>

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The balanced chemical reaction is:

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$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}$

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Molar mass of $CaCO_3$ = 100 g/mol

$\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol$

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$\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M$

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

$CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}$

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Concentration of calcium ion = Concentration of $CaCO_3$ = 0.131 M

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Answer:

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Explanation:

<h3>1.</h3>

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

$5.6792\;\rm m$ has four decimal places.
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<h3>2.</h3>

Similarly:

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$2.9133\; \rm g$ has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be $\rm g$ (same as the unit of the two inputs.)

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