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Molodets [167]
3 years ago
7

0.0400 M solution of HClO2, pH = 1.80 solve for Ka

Chemistry
1 answer:
umka2103 [35]3 years ago
6 0
Ka for weak acids is ---> Ka= (X)²/ (initial- X)

initial= 0.0400 M

to find X, you can find the H concentration using pH

[H+]= 10^-pH
[H+]= 10^-1.80= 0.016 which is X

ka = (0.016)²/ (0.0400 - 0.016) = 0.011
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Explanation:

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So, it means that an increase in the value of pH will show that there occurs a decrease in concentration of hydrogen ions.

Therefore, the solution becomes basic in nature.

On the other hand, a decrease in the value of pH will show that there occurs an increase in the concentration of hydrogen ions.

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Answer:

(molecular) 3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) +  6 NH₄Cl(aq)

(ionic) 3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)

(net ionic) 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)

Explanation:

The molecular equation includes al the species in the molecular form.

3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) +  6 NH₄Cl(aq)

The ionic equation includes all the ions (species that dissociate in water) and the species that do not dissociate in water.

3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)

The net ionic equation includes only the ions that participate in the reaction and the species that do not dissociate in water. In does not include <em>spectator ions</em>.

3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)

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