HSC 2009 Multiple Choice Q (1 Viewer)

[mtsmahia]

Can someone explain how to do Q 14 from multiple choice, with an equation.

Citric acid, the predominant acid in lemon juice, is a triprotic acid. A student titrated
25.0 mL samples of lemon juice with 0.550 mol L–1 NaOH. The mean titration volume was 29.50 mL. The molar mass of citric acid is 192.12 g mol–1.
What was the concentration of citric acid in the lemon juice?

(A)
1.04 g L–1
(B)
41.6 g L–1
(C)
125gL–1
(D)
374gL–1
–

Thanks.

 

[MetroMattums]

Is the answer B?

EDIT: Checked it's B.

So basically whenever the question mentions something about the acid being monoprotic etc, you assume that the acid fully ionises.

So you know: n[NaOH] = c.v = (0.550) x (0.02950) ≈ 0.0162 mol

So assuming full ionisation: n[citric acid] = (n[NaOH]) / 3 ≈ 0.00541 mol

So: m[citric acid] = n.M = (0.00541) x (192.12) ≈ 1.04 g

So the concentration is: m/V = (1.04) / (0.025) ≈ 41.6 g/L

 

[mtsmahia]

Is the answer B?

EDIT: Checked it's B.

So basically whenever the question mentions something about the acid being monoprotic etc, you assume that the acid fully ionises.

So you know: n[NaOH] = c.v = (0.550) x (0.02950) ≈ 0.0162 mol

So assuming full ionisation: n[citric acid] = (n[NaOH]) / 3 ≈ 0.00541 mol

So: m[citric acid] = n.M = (0.00541) x (192.12) ≈ 1.04 g

So the concentration is: m/V = (1.04) / (0.025) ≈ 41.6 g/L

Okay, so, does that mean that whenever it is triprotic, we divide by 3, and diprotic we divide by 2 ? cuz.. in the case of Sulphuric acid- since it is diprotic, would we divide by 2? in a similar situation

 

[adomad]

this is how you do it
For something monoprotic:




For something Diprotic as




Hence, with something triprotic, you divide the moles of the triprotic acid by 3 to get the moles of the H ions. so that is why you divide when the acid/ base is not monoprotic

 

[mtsmahia]

this is how you do it
For something monoprotic:




For something Diprotic as




Hence, with something triprotic, you divide the moles of the triprotic acid by 3 to get the moles of the H ions. so that is why you divide when the acid/ base is not monoprotic

so how does balancing work into that ?, or is it already balanced when you divide it by 2 (for Sulphuric acid).
Because, I would balance the equation normally, without considering the acids donating ability.

so, does that mean

we consider the donating ablity of the acid (mono, diprotic) into balancing only when we are dealing with Titrations.

And we balance it normally when we are writing a normal equation?