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Nataly [62]
2 years ago
10

Phosphoric acid (H3PO4) is a triprotic acid because each of its three hydrogen atoms may dissociate in water. What mathematical

relationship exists among the acid-dissociation constants (Ka1, Ka2, and Ka3) for phosphoric acid?
Chemistry
1 answer:
Juli2301 [7.4K]2 years ago
6 0

The formula that connects Ka1, Ka2, and Ka3 is the relation;

Ka = Ka1 * Ka2 *  Ka3

<h3>What is the Ka?</h3>

The Ka is defined as the acid dissociation constant of a solution. Now we have the triprotic acid called phosphoric acid. this implies that we have Ka1, Ka2, and Ka3.

Thus the formula that connects Ka1, Ka2, and Ka3 is the relation;

Ka = Ka1 * Ka2 *  Ka3

Learn more about Ka:brainly.com/question/16035742

#SPJ1

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4 0
3 years ago
Please help with this , thanks you very much
netineya [11]

Explanation:

reaction A is exothermic reaction as a substance is produced from two substances

and

reaction B is endothermic as two substances are formed from one complex structure

7 0
3 years ago
Name two fluid technologies that make use of air
nika2105 [10]

Answer: Refrigerator and fire extinguisher

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7 0
3 years ago
To measure the amount of iron in a certain type of iron ore, an analytical chemist dissolves a sample in strong acid and titrate
Fiesta28 [93]

Answer:

\% Fe=11.4\%

Explanation:

Hello!

In this case, for the described chemical reaction, we find it is:

8H^+(aq)+5Fe^{2+}(aq)+MnO_4^-(aq)\rightarrow 5Fe^{3+}(aq)+Mn^{2+}(aq)+4H_2O(l)

Because it says that the iron is dissolved in a strong acid which provides addition hydrogen ions to the reaction media. Thus, for the questions attached on the figure we find:

- This a REDOX reaction because we see iron is being oxidized from 2+ to 3+ and manganese reduced from +7 to +2.

- Since it is a redox reaction and the oxidized species is that undergoing an oxidation number increase, we evidence iron goes from +2 to +3, which means iron is the oxidized species.

- In this case, for the used 59.2 mL (0.0592 L) of the 0.2000 M solution of potassium permanganate, we can compute the consumed grams of iron via stoichiometry including the 5:1 mole ratio between them in the chemical reaction:

m_{Fe^{2+}}=0.0592L*0.2000\frac{molMnO_4^-}{L}*\frac{5molFe^{2+}}{1molMnO_4^-}  *\frac{55.85gFe^{2+}}{1molFe^{2+}} \\\\m_{Fe^{2+}}=3.31gFe^{2+}

It means that the percent of iron in that sample is:

\% Fe=\frac{3.31g}{29.00g} *100\%\\\\\% Fe=11.4\%

Best regards.

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