Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
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From the calculation, the molar mass of the solution is 141 g/mol.
<h3>What is the molar mass?</h3>
We know that;
ΔT = K m i
K = the freezing constant
m = molality of the solution
i = the Van't Hoft factor
The molality of the solution is obtained from;
m = ΔT/K i
m = 3.89/5.12 * 1
m = 0.76 m
Now;
0.76 = 26.7 /MM/0.250
0.76 = 26.7 /0.250MM
0.76 * 0.250MM = 26.7
MM= 26.7/0.76 * 0.250
MM = 141 g/mol
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Explanation:
<u>The first one is a base</u>
<u>The second one is an acid</u>
A base has a pH above 7
An acid has a pH below 7
Hope I helped!!! Have a great day!
D.
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