Need help, this question has me completely stumped. A student used a standardize

Question

Need help, this question has me completely stumped.

A student used a standardized 0.205 M NaOH solution to determine the mass percent of citric acid (H3C6H5O7) in a mixture of citric acid and inert potassium chloride. The pertinent reaction is H3C6H5O7(aq) + 3NaOH (aq) rightarrow Na3C6H5O7(q) + 3H2O(l) Sample masses and titration data are given in the table below. Do the following calculations for each titration and enter your answers in this table. Show clear calculations for Run #1 to receive full credit for your completed data table. This includes units and clear setup.

Explanation / Answer

A titration is between the reactants (the citric acid and NaOH)

The first thing you need to do is calculate moles of NaOH. moles NaOH=mole citric acid

So you need to you the volume and molarity to calculate moles of NaOH

0.205 mol/L x 0.01952 L= 0.00400 moles NaOH Remember to change mL to L

Next, change moles NaOH to moles of citric acid. You will use the mole ratio in the balanced equation. There 1 mole of citric acid in 3 moles of NaOH.

0.00400 mole NaOH x 1 mole citric acid/ 3 mole NaOH= 0.00133 mole citric acid

To find the mass of citric acid you simply use the molar mass of citric acid and the moles to convert to grams.

Molar mass: 192.06 g/mol

0.00133 moles x 192.06 g/mol= 0.255 g citric acid

The mass percent = mass of citric acid/mass of mixture x 100

(0.255g/0.356g) x100=71.6%

And the mean is the average of your 3 trials.

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