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jeyben [28]
2 years ago
5

Using chemical equations, show how the triprotic acid h3po4 ionizes in water. Phases are optional.

Chemistry
1 answer:
ludmilkaskok [199]2 years ago
5 0

The acid that contains three hydrogen ions are called triprotic acid.

<h3>The triprotic acid h3po4 ionizes in water:</h3>

White solid pure anhydrous phosphoric acid melts into a viscous liquid at 42.35 degrees Celsius. Phosphoric acid, which contains three ion hydrogen atoms, acts as a triprotic acid in an aqueous solution. The hydrogen ions disappear one at a time.

H₃PO₄ (aq)  ⇄  H⁺ (aq) + H₂PO₄⁻ (aq)          Kₐ₁ = 7.5 × 10⁻³

H₂PO₄⁻ (aq)  ⇄  H⁺ (aq) + H₂PO₄²⁻ (aq)       Kₐ₂ = 6.2 × 10⁻³

HPO₄²⁻ (aq)  ⇄  H⁺ (aq) + PO₄³⁻ (aq)            Kₐ₃ = 7.5 × 10⁻³

The first dissociation constant of phosphoric acid indicates that it is not an especially potent acid. It is a weaker acid than hydrochloric acid and sulfuric acid, but stronger than acetic acid. The ease with which each subsequent dissociation stage happens decreases. As a result, H₂PO₄⁻  is a relatively weak acid but HPO₄²⁻ is an extremely weak acid.

Leran more about triprotic acid here:

brainly.com/question/3286728

#SPJ4

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For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             Q_{1} = mC_{1} \Delta T_{1}

Putting the given values into the above equation as follows.

           Q_{1} = mC_{1} \Delta T_{1}

                      = 27.3 g \times 1.70 J/g K \times 41

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           Q_{2} = energy required = mL_{v}

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Therefor, calculate the value of energy required as follows.

             Q_{2} = mL_{v}

                         = 27.3 \times 444

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            Q_{3} = mC_{2} \Delta T_{2}

Value of C_{2} = 1.06 J/g,    \Delta T_{2} = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             Q_{3} = mC_{2} \Delta T_{2}

                       = 27.3 g \times 1.06 J/g \times 34 K

                       = 983.892 J

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            Q = Q_{1} + Q_{2} + Q_{3}

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

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