Ask question

Ask question

All categories

3 years ago

13

Phosphoric acid (h3po4) has three acid dissociation constants ( ka). the first dissociation constant has the greatest value, and

it applies to the reaction described by which equation?

2 answers:

Nuetrik [128]3 years ago

6 0

Missing question:
<span>A. H3PO4 (aq)<--> H+ (aq) + H2PO4- (aq).
B. H2PO4- (aq)<--> H+ (aq) + H2PO4-^2- (aq).
C. H3PO4 (aq)<--> 2H+ (aq) + HPO4^2-(aq).
D. HPO4^2- (aq)<--> H+ (aq) + PO4^3- (aq).
</span>Answer is: A. H3PO4 (aq)<--> H+ (aq) + H2PO4- (aq).
<span>The first dissociation constant :</span>Ka₁ = [H+]·[H₂PO₄⁻] / [H₃PO₄<span>].
In first dissociation phosphoric acid lost one proton (hydrogen ion).

</span>

ryzh [129]3 years ago

4 0

Ans: H₃PO₄ ↔ H⁺ + H₂PO₄⁻

Phosphoric acid is a weak triprotic acid. The three acidic protons are released in a stepwise manner and are represented in terms of their dissociation constants, ka values.

Step 1: H₃PO₄ ↔ H⁺ + H₂PO₄⁻ Ka₁ = 7.1 *10⁻³

Step 2: H₂PO₄⁻ ↔ H⁺ + HPO₄²⁻ Ka₂ = 6.3 *10⁻⁸

Step 3: HPO₄²⁻ ↔ H⁺ + PO₄³⁻ Ka₃ = 4.5 *10⁻¹³

Since Ka1 is the greatest, the reaction is described by step 1.

You might be interested in

Identify the chemical reactions as endothermic, exothermic, or neither. An egg cooking: A candle burning: Plaster and water comb

lianna [129]

Answer :

An egg cooking is an endothermic reaction.

A candle burning and Plaster & water combining and becoming warm are exothermic reaction.

Salt being added to water, with no change in temperature neither endothermic nor exothermic reaction.

Explanation :

Endothermic reaction : It is a type of reaction in which energy is absorbed in the form of heat from the surroundings.

Exothermic reaction : It is a type of reaction in which energy is released in the form of heat from the system.

An egg cooking : In this process, the energy is absorbed in the form of heat.

A candle burning and Plaster & water combining and becoming warm : In both the process, the energy is released in the form of heat.

Salt being added to water, with no change in temperature : In this process, neither energy released nor absorbed.

5 0

3 years ago

Read 2 more answers

What is the volume in liters of hydrogen gas that would be produced by the reaction of 40.0 g of Al with excess HCl at STP accor

DaniilM [7]

The volume in liters of Hydrogen gas = 49.28 L

<h3>Further explanation </h3>

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

Reaction

2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

mol Al :

$\tt mol=\dfrac{mass}{Ar}=\dfrac{40}{27}=1.481$

mol H₂ :

$\tt \dfrac{3}{2}\times 1.481=2.2$

At STP, 1 mol =22.4 L

so the volume of hydrogen :

$\tt 2.2 \times 22.4=49.28~L$

8 0

2 years ago

Read 2 more answers

The number of protons + number of neutrons is<br> equal to the

max2010maxim [7]

Answer: mass

Explanation: the protons and neutrons make up the nucleus and nearly the entire mass of the atom

5 0

2 years ago

HELP ME PLS <br> Someone give me the answers to the SELF-CHECK!<br> pls...

Ivenika [448]

Sorry its tooo blurry

6 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

2 years ago