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mod 6 question (1 Viewer)

jimmysmith560

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Original Poster:

thank youu!!! do you know what textbook that is ??
No worries! This version of the question actually appeared in the James Ruse Agricultural High School 2013 Chemistry trial exam.
 

jazz519

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If I was marking that above answer I wouldn't give it full marks.

The H3A is coming from citric acid being triprotic. However, you can't just ignore the actual formula of the citric acid (C6H8O7) and make up your own formula of H3A. A mark would be lost in the writing of the equation since it's not an equation including the acid provided.

Here is the correct working:

C6H8O7(aq) + 3NaOH(aq) --> Na3C6H5O7(aq) + 3H2O(l)

v(C6H8O7) used in titration = 25.00 mL = 0.025 L
c(NaOH) = 0.1045 mol/L, v(NaOH) = 23.05 mL = 0.02305 L

Always start calculations like this by writing an equation and assigning the values BEFORE trying to find moles. Unless you have built up a good knowledge on how to do the questions through lots of practice then this is the best way to do it. It clarifies all the information and puts it in an easy to process way vs trying to read the values in the question and doing it that way.

Find moles of NaOH as we have c and v
n(NaOH) = cv = (0.1045)(0.02305) = 0.002408... moles

Do molar ratio with C6H8O7 (citric acid)
n(C6H8O7) = 1/3 n(NaOH)
as there is a 1:3 ratio as seen in the chemical equation

n(C6H8O7) = 1/3(0.002408...) = 0.0008029... moles

This is the moles in 25.0 mL of the diluted juice solution

so finding the concentration:
c(C6H8O7) diluted solution = n / v = 0.0008029... / 0.025 = 0.0321... mol/L

However, this is the diluted juice. The question asks for the undiluted.

The original solution was diluted from 25.00 mL to 500.0 mL so use c1v1 = c2v2 (the dilution formula)

c1v1 = c2v2
c1(0.025) = (0.0321...)(0.5)
c1 = 0.6423 mol/L (4 sig figs)

which is the final answer
 

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