Titration of a triprotic acid with strong base NaOH

In summary, an unknown triprotic acid is titrated to its equivalence point using 35.2 ml of 0.106 M NaOH. The molar mass of the acid is calculated to be in g/mole.
  • #1
xpatelsxownage
13
0

Homework Statement


A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.


Homework Equations





The Attempt at a Solution


I just don't know where to start with this question.
35.2 ml * 1L/1000ml * .106M NaOH/1L * 39.9969gNaOH/1mole NaOH = 120.430 g NaOH
I don't know where to go after this part
 
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  • #2
A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.
Not "Its" equivalence point; WHICH equivalence point? The first, second, or third? The question is about a triprotic acid.
 
  • #3
xpatelsxownage said:

Homework Statement


A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.


Homework Equations





The Attempt at a Solution


I just don't know where to start with this question.
35.2 ml * 1L/1000ml * .106M NaOH/1L * 39.9969gNaOH/1mole NaOH = 120.430 g NaOH
I don't know where to go after this part

You don't need to calculate the number of grams of NaOH (which you did incorrectly). You set it up right but probably entered it in your calculator incorrectly. You should have gotten something like 0.149 g NaOH if this was what the question asked, which it wasn't.

Why don't you calculate the number of moles of NaOH used and find a relationship between that and the number of moles of triprotic acid. Once you have that, you will have the number of moles of acid and the mass. From that you can calculate formula weight.
 
  • #4
xpatelsxownage said:

Homework Statement


A 0.307g sample of an unknown triprotic acid is titrated to its equivilence point using 35.2ml of 0.106 M NaOH. Calculate the molar mass of the acid. Answer is in g/mole.


Homework Equations





The Attempt at a Solution


I just don't know where to start with this question.
35.2 ml * 1L/1000ml * .106M NaOH/1L * 39.9969gNaOH/1mole NaOH = 120.430 g NaOH
I don't know where to go after this part



It should equal to .149g NaOH
then I think you just subtract from the total.
 
  • #5
yaho8888 said:
...then I think you just subtract from the total.

no, nope, nein, nyet.
 
  • #6


A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 mL of 0.106 M NaOH. Calculate the molar mass of the acid.

What if the problem is to the Third equivalence point?
 

1. What is the purpose of titrating a triprotic acid with strong base NaOH?

The purpose of titrating a triprotic acid with strong base NaOH is to determine the concentration of the acid. This is done by gradually adding a known concentration of NaOH to the acid until the reaction reaches its equivalence point, where the moles of acid equal the moles of base. This allows for the determination of the acid's concentration through stoichiometric calculations.

2. How is the equivalence point determined in a titration of a triprotic acid with strong base NaOH?

The equivalence point is determined by using an indicator, such as phenolphthalein, which changes color at the point where all the acid has been neutralized by the base. Alternatively, a pH meter can be used to detect the sudden rise in pH at the equivalence point.

3. What is the difference between the first, second, and third equivalence points in a titration of a triprotic acid with strong base NaOH?

The first equivalence point corresponds to the neutralization of the first acidic hydrogen ion, the second equivalence point corresponds to the neutralization of the second acidic hydrogen ion, and the third equivalence point corresponds to the neutralization of the third acidic hydrogen ion. Each equivalence point will have a different pH and volume of base required to reach it.

4. How does the concentration of the triprotic acid affect the titration curve?

The concentration of the triprotic acid will affect the shape of the titration curve. A higher concentration of acid will require a larger volume of base to reach the equivalence point, resulting in a steeper curve. Additionally, a higher concentration of acid will also result in a lower pH at each equivalence point.

5. What other factors can affect the titration of a triprotic acid with strong base NaOH?

Other factors that can affect the titration of a triprotic acid with strong base NaOH include the accuracy and precision of the equipment used, the strength and concentration of the base, and any impurities present in the acid or base solutions. It is important to control these factors as much as possible to obtain accurate and reliable results.

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